today’s outline - august 31, 2015phys.iit.edu/~segre/phys405/15f/lecture_03.pdf · today’s...
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Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Today’s Outline - August 31, 2015
• Time-dependent solutions
• Problem 1.5
• Dirac bra-ket notation
• Solutions of the infinite square well
• Wavefunction orthogonality
• Wavefunction completeness
Office Hours: Monday, Wednesday, Friday 14:00-15:00
Reading Assignment: Chapter 2.3
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15
Constructing the time-dependent solution
Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.
Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.
Ψ1(x , t) = ψ1(x)e−iE1t/~
Ψ2(x , t) = ψ2(x)e−iE2t/~
...
Ψn(x , t) = ψn(x)e−iEnt/~
Ψ(x , t) =∞∑n=1
cnψne−iEnt/~
The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy
We generally leave the time-dependent portion as an implicit part of thesolution.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15
Constructing the time-dependent solution
Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.
Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.
Ψ1(x , t) = ψ1(x)e−iE1t/~
Ψ2(x , t) = ψ2(x)e−iE2t/~
...
Ψn(x , t) = ψn(x)e−iEnt/~
Ψ(x , t) =∞∑n=1
cnψne−iEnt/~
The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy
We generally leave the time-dependent portion as an implicit part of thesolution.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15
Constructing the time-dependent solution
Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.
Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.
Ψ1(x , t) = ψ1(x)e−iE1t/~
Ψ2(x , t) = ψ2(x)e−iE2t/~
...
Ψn(x , t) = ψn(x)e−iEnt/~
Ψ(x , t) =∞∑n=1
cnψne−iEnt/~
The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy
We generally leave the time-dependent portion as an implicit part of thesolution.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15
Constructing the time-dependent solution
Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.
Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.
Ψ1(x , t) = ψ1(x)e−iE1t/~
Ψ2(x , t) = ψ2(x)e−iE2t/~
...
Ψn(x , t) = ψn(x)e−iEnt/~
Ψ(x , t) =∞∑n=1
cnψne−iEnt/~
The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy
We generally leave the time-dependent portion as an implicit part of thesolution.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15
Constructing the time-dependent solution
Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.
Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.
Ψ1(x , t) = ψ1(x)e−iE1t/~
Ψ2(x , t) = ψ2(x)e−iE2t/~
...
Ψn(x , t) = ψn(x)e−iEnt/~
Ψ(x , t) =∞∑n=1
cnψne−iEnt/~
The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy
We generally leave the time-dependent portion as an implicit part of thesolution.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15
Constructing the time-dependent solution
Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.
Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.
Ψ1(x , t) = ψ1(x)e−iE1t/~
Ψ2(x , t) = ψ2(x)e−iE2t/~
...
Ψn(x , t) = ψn(x)e−iEnt/~
Ψ(x , t) =∞∑n=1
cnψne−iEnt/~
The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy
We generally leave the time-dependent portion as an implicit part of thesolution.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15
Constructing the time-dependent solution
Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.
Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.
Ψ1(x , t) = ψ1(x)e−iE1t/~
Ψ2(x , t) = ψ2(x)e−iE2t/~
...
Ψn(x , t) = ψn(x)e−iEnt/~
Ψ(x , t) =∞∑n=1
cnψne−iEnt/~
The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy
We generally leave the time-dependent portion as an implicit part of thesolution.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15
Constructing the time-dependent solution
Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.
Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.
Ψ1(x , t) = ψ1(x)e−iE1t/~
Ψ2(x , t) = ψ2(x)e−iE2t/~
...
Ψn(x , t) = ψn(x)e−iEnt/~
Ψ(x , t) =∞∑n=1
cnψne−iEnt/~
The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy
We generally leave the time-dependent portion as an implicit part of thesolution.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15
Problem 1.5
Consider the wave function
Ψ(x , t) = Ae−λ|x |e−iωt ,
where A, λ, and ω are positive real constants.
(a) Normalize Ψ.
(b) Determine the expectation values of x and x2.
(c) Find the standard deviation of x and the probabilityof finding the particle outside of this range.
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 3 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx
+
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx
= A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx
+
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |e iωte−λ|x |e−iωt dx
= A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx
+
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx
= A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx
+
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx
+
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx
+
∫ ∞0
e−2λx dx
)
break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx
+
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx +
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx +
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞
− 1
2λe−2λx
∣∣∣∞0
)
= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx +
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞− 1
2λe−2λx
∣∣∣∞0
)
= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx +
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)
=A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx +
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx +
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (a)
The normalization integral is
1 =
∫ψ∗ψ dx = A2
∫ ∞−∞
e−λ|x |���e iωte−λ|x |���e−iωt dx = A2
∫ ∞−∞
e−2λ|x | dx
= A2
(∫ 0
−∞e+2λx dx +
∫ ∞0
e−2λx dx
)break this into two integrals
= A2
(1
2λe2λx
∣∣∣0−∞− 1
2λe−2λx
∣∣∣∞0
)= A2
(1
2λ+
1
2λ
)=
A2
λ
A =√λ
So the normalized wave function is
Ψ(x , t) =√λe−λ|x |e−iωt
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx
= 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx
= 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx
= 0
this vanishes due to an odd integrand.
The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand.
The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is
⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)
= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2,
du = 2x dx ,
dv = e−2λxdx ,
v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx ,
v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(− x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)
= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−�
����xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)
= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(− xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)
= +1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)
= +1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx
=1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (b)
The expectation value of x is
〈x〉 = λ
∫ ∞−∞
xe−2λ|x | dx = 0
this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩
= λ
(∫ 0
−∞x2e+2λx dx +
∫ ∞0
x2e−2λx dx
)= 2λ
∫ ∞0
x2e−2λx dx
this can be solved applying integration by parts twice, first using
u = x2, du = 2x dx , dv = e−2λxdx , v = − 1
2λe−2λx
⟨x2⟩
= 2λ
(−��
����x2e−2λx
2λ
∣∣∣∣∞0
+1
λ
∫ ∞0
xe−2λx dx
)= +2
∫ ∞0
x e−2λx dx
and with u = x , du = dx for the second
⟨x2⟩
= 2
(−��
���xe−2λx
2λ
∣∣∣∣∞0
+1
2λ
∫ ∞0
e−2λx dx
)= +
1
λ
∫ ∞0
e−2λx dx =1
2λ2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2
=1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx
= 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ
= e−√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Problem 1.5 - part (c)
The variance is now given by
σ2 =⟨x2⟩− 〈x〉2 =
1
2λ2− 0 =
1
2λ2
σ =1√2λ
λ
-2λ -σ 0 +σ +2λ
|Ψ|2
x
and the probability that the particle is outside the interval −σ < x < σ is
Pout = λ
∫ −σ−∞
e2λx dx + λ
∫ ∞σ
e−2λx dx = 2λ
∫ ∞σ
e−2λx dx
= 2λ
(−e−2λx
2λ
)∣∣∣∣∞σ
= e−2λσ
= e−2λ/√2λ = e−
√2 = 0.2431
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15
Dirac bra-ket notation
Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.
integral bra-ket
bra ψ(x)∗ 〈ψ|complex conjugate isimplicit
ket ψ(x) |ψ〉
normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1
expectationvalue
∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied
to the right
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15
Dirac bra-ket notation
Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.
integral bra-ket
bra ψ(x)∗ 〈ψ|complex conjugate isimplicit
ket ψ(x) |ψ〉
normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1
expectationvalue
∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied
to the right
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15
Dirac bra-ket notation
Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.
integral bra-ket
bra ψ(x)∗ 〈ψ|complex conjugate isimplicit
ket ψ(x) |ψ〉
normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1
expectationvalue
∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied
to the right
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15
Dirac bra-ket notation
Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.
integral bra-ket
bra ψ(x)∗ 〈ψ|complex conjugate isimplicit
ket ψ(x) |ψ〉
normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1
expectationvalue
∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied
to the right
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15
Dirac bra-ket notation
Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.
integral bra-ket
bra ψ(x)∗ 〈ψ|complex conjugate isimplicit
ket ψ(x) |ψ〉
normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1
expectationvalue
∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied
to the right
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15
Dirac bra-ket notation
Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.
integral bra-ket
bra ψ(x)∗ 〈ψ|complex conjugate isimplicit
ket ψ(x) |ψ〉
normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1
expectationvalue
∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied
to the right
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0.
Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging
and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging
and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ
= −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ
= −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well
Consider a potential
V (x) =
{0, if 0 ≤ x ≤ a.
∞, otherwise
Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is
− ~2
2m
d2ψ
dx2= Eψ
This is the classical standing waveboundary value problem with gen-eral solution
V(x)
xa0
rearranging and substituting
k =√
2mE/~
d2ψ
dx2= −2mE
~2ψ = −k2ψ
ψ(x) = A sin(kx) + B cos(kx)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) + B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
ψ(x) = A sin(kx) + B cos(kx)
ψ(0) = ψ(a) = 0
for ψ(0) = 0
0 = ����A sin(0) + B cos(0)
B = 0
ψn(x) = A sin(nπ
ax)
En =⟨n∣∣Hn
⟩
Starting with the general solution,apply boundary conditions
for an infinite potential, the deriva-tive must be discontinuous
for ψ(a) = 0
0 = A sin(ka) +�����B cos(ka)
ka = nπ, n = 1, 2, 3, · · ·
k =nπ
an = 1, 2, 3, · · ·
all that remains is to calculate En
and normalize ψn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩
=n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2
=~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Infinite square well - solution
Normalization
1 = 〈n | n〉
=
∫ a
0|A|2 sin2(knx)dx
= |A|2 a2
A =
√2
a
ψn(x) =
√2
asin(nπ
ax)
State Energy
∣∣Hn⟩
= − ~2
2m
d2
dx2A sin
(nπax)
= +n2π2~2
2ma2A sin
(nπax)
=n2π2~2
2ma2|n〉
En =⟨n∣∣Hn
⟩=
n2π2~2
2ma2〈n|n〉
=n2π2~2
2ma2=
~2k2n2m
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15
Wavefunction properties
0 a
x
n=1
n=2
n=3
The first three wavefunctions areshown offset in the y direction
The states are alternating even andodd about the center of the infinitewell x = a/2
The energy and momentum arenaturally quantized
The states are mutually orthogonal
〈ψm |ψn〉 = δmn
The states form a complete basisset
f (x) =∞∑n=1
cnψn(x)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15
Wavefunction properties
0 a
x
n=1
n=2
n=3
The first three wavefunctions areshown offset in the y direction
The states are alternating even andodd about the center of the infinitewell x = a/2
The energy and momentum arenaturally quantized
The states are mutually orthogonal
〈ψm |ψn〉 = δmn
The states form a complete basisset
f (x) =∞∑n=1
cnψn(x)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15
Wavefunction properties
0 a
x
n=1
n=2
n=3
The first three wavefunctions areshown offset in the y direction
The states are alternating even andodd about the center of the infinitewell x = a/2
The energy and momentum arenaturally quantized
The states are mutually orthogonal
〈ψm |ψn〉 = δmn
The states form a complete basisset
f (x) =∞∑n=1
cnψn(x)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15
Wavefunction properties
0 a
x
n=1
n=2
n=3
The first three wavefunctions areshown offset in the y direction
The states are alternating even andodd about the center of the infinitewell x = a/2
The energy and momentum arenaturally quantized
The states are mutually orthogonal
〈ψm |ψn〉 = δmn
The states form a complete basisset
f (x) =∞∑n=1
cnψn(x)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15
Wavefunction properties
0 a
x
n=1
n=2
n=3
The first three wavefunctions areshown offset in the y direction
The states are alternating even andodd about the center of the infinitewell x = a/2
The energy and momentum arenaturally quantized
The states are mutually orthogonal
〈ψm |ψn〉 = δmn
The states form a complete basisset
f (x) =∞∑n=1
cnψn(x)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15
Orthogonality
∫ψ∗mψndx
=2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
cos(α± β) = cosα cosβ ∓ sinα sinβ
cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)
− (������cosα cosβ − sinα sinβ)
= 2 sinα sinβ
sinα sinβ =1
2cos(α− β)− 1
2cos(α + β)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
cos(α± β) = cosα cosβ ∓ sinα sinβ
cos(α− β)− cos(α + β)
= (������cosα cosβ + sinα sinβ)
− (������cosα cosβ − sinα sinβ)
= 2 sinα sinβ
sinα sinβ =1
2cos(α− β)− 1
2cos(α + β)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
cos(α± β) = cosα cosβ ∓ sinα sinβ
cos(α− β)− cos(α + β) = (cosα cosβ + sinα sinβ)
− (������cosα cosβ − sinα sinβ)
= 2 sinα sinβ
sinα sinβ =1
2cos(α− β)− 1
2cos(α + β)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
cos(α± β) = cosα cosβ ∓ sinα sinβ
cos(α− β)− cos(α + β) = (cosα cosβ + sinα sinβ)
− (cosα cosβ − sinα sinβ)
= 2 sinα sinβ
sinα sinβ =1
2cos(α− β)− 1
2cos(α + β)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
cos(α± β) = cosα cosβ ∓ sinα sinβ
cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)
− (������cosα cosβ − sinα sinβ)
= 2 sinα sinβ
sinα sinβ =1
2cos(α− β)− 1
2cos(α + β)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
cos(α± β) = cosα cosβ ∓ sinα sinβ
cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)
− (������cosα cosβ − sinα sinβ)
= 2 sinα sinβ
sinα sinβ =1
2cos(α− β)− 1
2cos(α + β)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
cos(α± β) = cosα cosβ ∓ sinα sinβ
cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)
− (������cosα cosβ − sinα sinβ)
= 2 sinα sinβ
sinα sinβ =1
2cos(α− β)− 1
2cos(α + β)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
cos(α± β) = cosα cosβ ∓ sinα sinβ
cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)
− (������cosα cosβ − sinα sinβ)
= 2 sinα sinβ
sinα sinβ =1
2cos(α− β)− 1
2cos(α + β)
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral
∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]
= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .
. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0.
We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.
∫ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Orthogonality
∫ψ∗mψndx =
2
a
∫ a
0sin(mπ
ax)
sin(nπ
ax)dx
=1
a
∫ a
0
[cos
(m − n
aπx
)− cos
(m + n
aπx
)]dx
now proceed to evaluate the integral∫ψ∗mψndx =
[1
(m − n)πsin
(m − n
aπx
)− 1
(m + n)πsin
(m + n
aπx
)∣∣∣∣a0
=1
π
[sin[(m − n)π]
m − n− sin[(m + n)π]
m + n
]= 0 for m 6= n
If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫
ψ∗mψndx = δmn
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15
Completeness
Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.
The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.
f (x) =∞∑n=1
cnψn(x)
〈ψm|f (x)〉 =
∫ψ∗m(x)f (x)dx
=∞∑n=1
cn
∫ψ∗m(x)ψn(x)dx
=∞∑n=1
cnδmn = cm
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15
Completeness
Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.
The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.
f (x) =∞∑n=1
cnψn(x)
〈ψm|f (x)〉 =
∫ψ∗m(x)f (x)dx
=∞∑n=1
cn
∫ψ∗m(x)ψn(x)dx
=∞∑n=1
cnδmn = cm
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15
Completeness
Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.
The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.
f (x) =∞∑n=1
cnψn(x)
〈ψm|f (x)〉 =
∫ψ∗m(x)f (x)dx
=∞∑n=1
cn
∫ψ∗m(x)ψn(x)dx
=∞∑n=1
cnδmn = cm
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15
Completeness
Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.
The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.
f (x) =∞∑n=1
cnψn(x)
〈ψm|f (x)〉 =
∫ψ∗m(x)f (x)dx
=∞∑n=1
cn
∫ψ∗m(x)ψn(x)dx
=∞∑n=1
cnδmn = cm
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15
Completeness
Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.
The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.
f (x) =∞∑n=1
cnψn(x)
〈ψm|f (x)〉 =
∫ψ∗m(x)f (x)dx
=∞∑n=1
cn
∫ψ∗m(x)ψn(x)dx
=∞∑n=1
cnδmn = cm
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15
Completeness
Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.
The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.
f (x) =∞∑n=1
cnψn(x)
〈ψm|f (x)〉 =
∫ψ∗m(x)f (x)dx
=∞∑n=1
cn
∫ψ∗m(x)ψn(x)dx
=∞∑n=1
cnδmn
= cm
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15
Completeness
Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.
The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.
f (x) =∞∑n=1
cnψn(x)
〈ψm|f (x)〉 =
∫ψ∗m(x)f (x)dx
=∞∑n=1
cn
∫ψ∗m(x)ψn(x)dx
=∞∑n=1
cnδmn = cm
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:
Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:
Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx
=
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx
=∞∑
m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn
=∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).
The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients
Ψ(x , t) =∞∑n=1
cn
√2
asin(nπ
ax)e−i(n
2π2~/2ma2)t
Ψ(x , 0) =∞∑n=1
cnψn(x)
cn =
√2
a
∫ a
0sin(nπ
ax)
Ψ(x , 0)dx
1 =
∫|Ψ(x , 0)|2 dx =
∫ ( ∞∑m=1
cmψm(x)
)∗( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcn
∫ψ∗m(x)ψn(x)dx =
∞∑m=1
∞∑n=1
c∗mcnδmn =∞∑n=1
|cn|2
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15
Arbitrary time-dependent wavefunction
Since the coefficients, cn, are independent of time, the expectation valueof the energy is also
⟨H⟩
=
∫Ψ∗HΨdx =
∫ ( ∞∑m=1
cmψm(x)
)∗H
( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcnEn
∫ψ∗mψndx =
∞∑n=1
|cn|2 En
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15
Arbitrary time-dependent wavefunction
Since the coefficients, cn, are independent of time, the expectation valueof the energy is also
⟨H⟩
=
∫Ψ∗HΨdx
=
∫ ( ∞∑m=1
cmψm(x)
)∗H
( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcnEn
∫ψ∗mψndx =
∞∑n=1
|cn|2 En
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15
Arbitrary time-dependent wavefunction
Since the coefficients, cn, are independent of time, the expectation valueof the energy is also
⟨H⟩
=
∫Ψ∗HΨdx =
∫ ( ∞∑m=1
cmψm(x)
)∗H
( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcnEn
∫ψ∗mψndx =
∞∑n=1
|cn|2 En
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15
Arbitrary time-dependent wavefunction
Since the coefficients, cn, are independent of time, the expectation valueof the energy is also
⟨H⟩
=
∫Ψ∗HΨdx =
∫ ( ∞∑m=1
cmψm(x)
)∗H
( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcnEn
∫ψ∗mψndx
=∞∑n=1
|cn|2 En
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15
Arbitrary time-dependent wavefunction
Since the coefficients, cn, are independent of time, the expectation valueof the energy is also
⟨H⟩
=
∫Ψ∗HΨdx =
∫ ( ∞∑m=1
cmψm(x)
)∗H
( ∞∑n=1
cnψn(x)
)dx
=∞∑
m=1
∞∑n=1
c∗mcnEn
∫ψ∗mψndx =
∞∑n=1
|cn|2 En
C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15