2018 Fall Physics 491: Problem Set 3

Problem 2

Contents

1 More examples in momentum space 1

1.1 Normalization constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Momentum space wave function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Position and momentum uncertainties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Superposition of two Gaussians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4.1 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4.2 Momentum space wave function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4.3 Position and Momentum Uncertainties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1 More examples in momentum space

We consider the wave function (x) = Ax exp⇣� x

2

4�

2

⌘.

It is useful to recall the form of a normalized Gaussian probability distribution and its properties. The distri-bution with mean x

0

and variance �2 is as follows.

PGauss

=1p2⇡�2

exp

✓� (x� x

0

)2

2�2

◆(1)

The moments of this distribution are as follows. The zeroth moment gives the normalization.

Z+1

�1dxx0P

Gauss

=

Z+1

�1dxP

Gauss

= 1 (2)

The first moment gives the mean.

Z+1

�1dxx1P

Gauss

=

Z+1

�1dxxP

Gauss

= x

0

(3)

The second moment gives the variance.

Z+1

�1dxx2P

Gauss

=

Z+1

�1dxx2P

Gauss

= �

2 + x

2

0

(4)

1.1 Normalization constant

We find the normalization by requiring that the integral of the square of the absolute value of the wave functionintegrated over all space is 1.

Z+1

�1dx| (x)|2 = 1 =) |A|2

Z+1

�1dx

����x exp✓� x

2

4�2

◆����2

= 1 (5)

For our wave function we have the following.

Z+1

�1dx| (x)|2 = |A|2

Z+1

�1dx���x exp

⇣� x

2

4�

2

⌘���2

= |A|2p2⇡�2

1p2⇡�2

Z+1

�1dxx2 exp

⇣� x

2

4�

2

⌘= |A|2

p2⇡�2

�

2 (6)

1

Therefore the condition for normalization give the following expression for |A|2.

|A|2p2⇡�2

�

2 = 1 =) |A|2 =1

(2⇡�6)1/2=) |A| = 1

(2⇡)1/4�3/2

(7)

We choosing the overall phase to be zero, that is, choose A to be real and positive.

A =1

(2⇡)1/4�3/2

(8)

Therefore the normalized wavefunction is

(x) =1

(2⇡)1/4�3/2

x exp⇣� x

2

4�

2

⌘(9)

It is useful to factor out a Gaussian wavefunction.

(x) =1

(2⇡)1/4�3/2

x exp⇣� x

2

4�

2

⌘=⇣x

�

⌘ 1

(2⇡�2)1/4exp

⇣� x

2

4�

2

⌘=⇣x

�

⌘

Gauss

(x) (10)

1.2 Momentum space wave function

We find the momentum space wave function �̃(p) by doing a Fourier transform from position space to momentumspace.

�̃(p) =

Z+1

�1

dxp2⇡~

(x) exp��i

px

~�=

Ap2⇡~

Z+1

�1dxx exp

⇣� x

2

4�

2

⌘exp

��i

px

~�

(11)

To do this integral, we use the following trick.

d

dx

⇣exp

⇣� x

2

4�

2

⌘⌘= � x

2

2�2

exp⇣� x

2

4�

2

⌘(12)

Therefore the momentum space wavefunction is as follows.

�̃(p) = �A

2�2

p2⇡~

Z+1

�1dx

d

dx

⇣exp

⇣� x

2

4�

2

⌘⌘exp

��i

px

~�

(13)

We use integration by parts and use the fact that the Gaussian function goes to zero as |x| ! 1.

�̃(p) = �A

2�2

p2⇡~

hexp

⇣� x

2

4�

2

⌘exp

��i

px

~�i+1

�1+A

2�2

p2⇡~

Z+1

�1dx exp

⇣� x

2

4�

2

⌘ d

dx

�exp

��i

px

~��

= 0 +A

2�2

p2⇡~

Z+1

�1dx exp

⇣� x

2

4�

2

⌘✓� ip

~

◆exp

��i

px

~�

= 2A�2

✓� ip

~

◆1p2⇡~

Z+1

�1dx exp

⇣� x

2

4�

2

⌘exp

��i

px

~�

| {z }Momentum space wave function of a Gaussian

= 2A�2

✓� ip

~

◆✓2�2

⇡~2

◆1/4

exp

✓�p

2

�

2

~2

◆(14)

Putting the factors together, we have the following.

�̃(p) = �i

✓2p�

~

◆✓2�2

⇡~2

◆1/4

exp

✓�p

2

�

2

~2

◆(15)

Again, it is useful to factor out a Gaussian wavefunction.

�̃(p) = �i

✓2p�

~

◆�̃

Gauss

(p) (16)

Up to overall constant factors, the momentum space wavefunction has the same form as the position space wavefunction.

2

1.3 Position and momentum uncertainties

The uncertainties in position and momentum can be found using the following relationships.

�x =

qhx2i � hxi2 �p =

qhp2i � hpi2 (17)

For evaluating these expressions, the following property of Gaussian distribution is useful to note.

hx� x

0

inGauss

=

(0 n is odd

(n� 1)!!�n

n is even(18)

Here (n� 1)!! is the double factorial, (n� 1)!! = (n� 1)⇥ (n� 3)⇥ 3⇥ 1.

In particular the second moment hxi2 for a Gaussian distribution with mean x

0

and variance �2 is �2 + x

2

0

We note that hxi = 0 as | (x)|2 is even about x = 0. Moreover hpi = 0 as |�̃(p)|2 is even about p = 0

hxi =Z

+1

�1dxx| (x)|2 = 0 hpi =

Z+1

�1dp p|�̃(p)|2 = 0 (19)

The second moment for x is the following.

hxi2 =

Z+1

�1dxx2| (x)|2 =

Z+1

�1dxx2

���⇣x

�

⌘

Gauss

(x)���2

=1

�

2

Z+1

�1dxx4|

Gauss

|2 (20)

Using the property of Gaussian distribution, we find the following.

hxi2 =1

�

2

(4� 1)!!�4 = 3�2 (21)

Therefore, the uncertainty in x, �x is as follows.

�x =

qhx2i � hxi2 =

p3� (22)

The second moment for p is the following.

hpi2 =

Z+1

�1dp p2

����̃(p)���2

=

Z+1

�1dp p2

����

✓�i

2p�

~

◆�̃

Gauss

(p)

����2

=4�2

~2

Z+1

�1dx p4

����̃Gauss

���2

(23)

Again, we use the property of the Gaussian distribution. Recall that the variance of the momentum spaceGaussian is ~

2�

.

hpi2 =4�2

~2 (4� 1)!!

✓~2�

◆4

=3

4

~�

(24)

Therefore, the uncertainty in p, �p is as follows.

�p =

qhp2i � hpi2 =

p3

2

~�

(25)

The uncertainty product is as follows.

�x�p =p3�

p3

2

~�

=3

2~ (26)

This is more than the minimum uncertainty ~2

. Therefore this is not a wavefunction with a minimum uncertaintyproduct.

3

1.4 Superposition of two Gaussians

Now we consider the wave function

(x) = A

exp

��x+ a

2

�2

4�2

!+ exp

��x+ a

2

�2

4�2

!!(27)

This can be written an equal superposition of two normalized Gaussian wavefunctions.

(x) = B( +

(x) + �(x)) (28)

Here ± refer to normalized Gaussian wavefunctions centered at ±a

2

.

±(x) =1

(2⇡�2)1/4

exp

✓�(x⌥ a

2

)2

4�2

◆(29)

1.4.1 Normalization

Normalization leads to the following condition.

Z+1

�1dx| (x)|2 = 1 =) |B|2

Z+1

�1dx|

+

(x) + �(x)|2 = 1

=) |B|2Z

+1

�1dx|

+

(x)|2 + |B|2Z

+1

�1dx| �(x)|2 + 2|B|2

Z+1

�1dxRe (

+

(x) �(x)) = 1 (30)

The integral with each normalized Gaussian wave functions is 1. Therefore the normalization condition is asfollows.

2|B|2✓1 +

Z+1

�1dx

+

(x) �(x)

◆= 1 (31)

The cross term is as follows.

+

(x) �(x) =1

(2⇡�2)1/4

exp

✓�(x� a

2

)2

4�2

◆1

(2⇡�2)1/4

exp

✓�(x+ a

2

)2

4�2

◆=

1

(2⇡�2)1/2

exp

�(x2 � a

2

4

)

2�2

!(32)

The integral of the cross term can be evaluated using the integral of a normalized Gaussian.

Z+1

�1dx

+

(x) �(x)) =

Z+1

�1dx

1

(2⇡�2)1/2

exp

�(x2 � a

2

4

)

2�2

!

= exp

✓� a

2

8�2

◆1

(2⇡�2)1/2

Z+1

�1dx exp

✓� x

2

2�2

◆= exp

✓� a

2

8�2

◆(33)

Therefore, we can find |B|

2|B|2✓1 + exp

✓� a

2

8�2

◆◆= 1 =) |B| = 1r

2⇣1 + exp

⇣� a

2

8�

2

⌘⌘ (34)

We choose the overall phase to make the wavefunction real. Therefore the normalized wavefunction is thefollowing.

(x) =1r

2⇣1 + exp

⇣� a

2

8�

2

⌘⌘

1

(2⇡�2)1/4

exp

✓�(x� a

2

)2

4�2

◆+

1

(2⇡�2)1/4

exp

✓�(x+ a

2

)2

4�2

◆!(35)

Therefore the normalization constant A is the following.

A =1r

2⇣1 + exp

⇣� a

2

8�

2

⌘⌘1

(2⇡�2)1/4

(36)

The wave functions are plotted below. When a is large compared �, the two Gaussian are distinguishable.

4

1.4.2 Momentum space wave function

The position space wavefunction is an equal superposition of +

(x) and �(x). We can use the linearity of theFourier transform to write the momentum space wave function as equal superposition of the Fourier transforms�̃

+

(p) and �̃�(p)

�̃(p) = B

⇣�̃

+

(p) + �̃�(p)⌘

(37)

The functions +

(x) and �(x) are displaced versions of the Gaussian wavefunction centered at zero, Gauss

(x).

+

(x) =

Gauss

�x� a

2

� �(x) =

Gauss

�x+ a

2

�(38)

Therefore, we use the phase shift property of the Fourier transform to find the momentum space wave function.

5

�̃(p) = B

⇣�̃

+

(p) + �̃�(p)⌘= B

⇣�̃

Gauss

(p) exp⇣+i

pa

2~

⌘+ �̃

Gauss

(p) exp⇣�i

pa

2~

⌘⌘= 2B�̃

Gauss

(p) cos⇣pa

2~

⌘(39)

Recall that for a Gaussian wavefunction with variance in position � and mean position x = 0, the momentumspace wavefunction is as follows.

�̃

Gauss

(p) =

✓2�2

⇡~2

◆1/4

exp

✓�p

2

�

2

~2

◆(40)

Using this and the expression for B we found earlier, we have the following.

�̃(p) = 21r

2⇣1 + exp

⇣� a

2

8�

2

⌘⌘

✓2�2

⇡~2

◆1/4

exp

✓�p

2

�

2

~2

◆cos⇣pa

2~

⌘

=) �̃(p) =1r

1 + exp⇣� a

2

8�

2

⌘

✓8�2

⇡~2

◆1/4

exp

✓�p

2

�

2

~2

◆cos⇣pa

2~

⌘(41)

These are plotted below. When a is large compared �, the phase di↵erence between the two Gaussians leads tointerference fringes.

6

1.4.3 Position and Momentum Uncertainties

The position space wave function (x) is even about x = 0. Therefore hxi = 0. Futhermore, the momentum spacewave function �̃(p) is even about p = 0. Therefore hpi = 0.

The second moment of position,⌦x

2

↵is as follows. We write | (x)|2 using three terms as we did earlier.

⌦x

2

↵=

Z+1

�1dxx2| (x)|2 = B

2

Z+1

�1dxx2|

+

(x)|2 +B

2

Z+1

�1dxx2| �(x)|2 + 2B2

Z+1

�1dxx2

+

(x) �(x)

(42)

Using the properties of the Gaussian distribution we noted earlier, the first two integrals become �2 + a

2

4

.

Z+1

�1dxx2|

+

(x)|2 =

Z+1

�1dxx2 exp

��x� a

2

�2

2�2

!= �

2 +a

2

4(43)

Z+1

�1dxx2| �(x)|2 =

Z+1

�1dxx2 exp

��x+ a

2

�2

2�2

!= �

2 +a

2

4(44)

The integral of the cross term simplifies as follows.

Z+1

�1dxx2

+

(x) �(x) = exp

✓� a

2

8�2

◆1p2⇡�2

Z+1

�1dxx2 exp

✓� x

2

2�2

◆= exp

✓� a

2

8�2

◆�

2 (45)

Putting them together, we have the following for⌦x

2

↵.

⌦x

2

↵= B

2

✓✓�

2 +a

2

4

◆+

✓�

2 +a

2

4

◆+ 2 exp

✓� a

2

8�2

◆�

2

◆= B

2

✓✓2 + 2 exp

✓� a

2

8�2

◆◆+

a

2

2

◆(46)

Using the expression for B we found earlier, we get the following.

⌦x

2

↵=

1

2⇣1 + exp

⇣� a

2

8�

2

⌘⌘✓✓

2 + 2 exp

✓� a

2

8�2

◆◆�

2 +a

2

2

◆= �

2 +a

2

4⇣1 + exp

⇣� a

2

8�

2

⌘⌘ (47)

When a � �, hxi2 ! a

2

4

, as expected.Therefore the uncertainty in position is the following.

�x =

vuut�

2 +a

2

4⇣1 + exp

⇣� a

2

8�

2

⌘⌘ = �

vuut1 +a

2

4�2

⇣1 + exp

⇣� a

2

8�

2

⌘⌘ (48)

7

The calculation for second moment of momentum⌦p

2

↵is involved.

⌦p

2

↵=

Z+1

�1dp p2

����̃(p)���2

(49)

We could write �̃(p) as a superposition of �̃+

(p) and �̃�(p). However, we do the integral directly for practice.We will see that these are equivalent.

⌦p

2

↵=

Z+1

�1dp p2

2

6641r

1 + exp⇣� a

2

8�

2

⌘

✓8�2

⇡~2

◆1/4

exp

✓�p

2

�

2

~2

◆cos⇣pa

2~

⌘3

775

2

=1

1 + exp⇣� a

2

8�

2

⌘✓8�2

⇡~2

◆1/2

Z+1

�1dp p2 exp

✓�2p2�2

~2

◆cos2

⇣pa

2~

⌘

=1

1 + exp⇣� a

2

8�

2

⌘✓8�2

⇡~2

◆1/2

Z+1

�1dp p2 exp

✓�2p2�2

~2

◆1

2

⇣1 + cos

⇣pa

~

⌘⌘

=1

1 + exp⇣� a

2

8�

2

⌘✓2�2

⇡~2

◆1/2

Z+1

�1dp p2 exp

✓�2p2�2

~2

◆+

Z+1

�1dp p2 exp

✓�2p2�2

~2

◆cos⇣pa

~

⌘�(50)

The first integral is the second moment of momentum for a Gaussian wavefunction in momentum space with

zero mean momentum and momentum variance ~2

4�

2 . This is the term we would get from the integral of �̃+

(p) and

�̃�(p). Therefore⌦p

2

↵is the following.

⌦p

2

↵=

1

1 + exp⇣� a

2

8�

2

⌘"~24�2

+

✓2�2

⇡~2

◆1/2

Z+1

�1dp p2 exp

✓�2p2�2

~2

◆cos⇣pa

~

⌘#(51)

The integral with the cosine and the Gaussian and the quadratic term is complicated. This integral is of the

following form with A = 2�

2

~2 and B = a

~ .r

A

⇡

Z+1

�1dpp2e�Ap

2

cos (Bp) =1

2

rA

⇡

Z+1

�1dpp2e�Ap

2

e

�iBp +

rA

⇡

Z+1

�1dpp2e�Ap

2

e

+iBp

!(52)

Each of these integrals can be done as follows.

Z+1

�1dpp2e�Ap

2

e

±iBp = � @

@A

✓Z+1

�1dpe�Ap

2

e

±iBp

◆= � @

@A

✓⇣⇡

A

⌘1/2

exp

✓�B

2

4A

◆◆

=⇡

1/2

2A5/2

�A� 2B2

�exp

✓�B

2

4A

◆(53)

Therefore, our integral is as follows.

rA

⇡

Z+1

�1dpp2e�Ap

2

cos (Bp) =

✓A

⇡

◆1/2

⇡

1/2

2A5/2

�A� 2B2

�exp

✓�B

2

4A

◆=

1

A

2

�A� 2B2

�exp

✓�B

2

4A

◆

=

✓1

2A� B

2

A

2

◆exp

✓�B

2

4A

◆(54)

Plugging A and B back in, we have the following.

✓2�2

⇡~2

◆1/2

Z+1

�1dp p2 exp

✓�2p2�2

~2

◆cos⇣pa

~

⌘=

✓~24�2

� a

2

~2~44�4

◆exp

✓�a

2

~2~28�2

◆

= ~2✓

1

4�2

� a

2

4�4

◆exp

✓� a

2

8�2

◆(55)

8

Therefore the second moment of momentum,⌦p

2

↵is as follows.

⌦p

2

↵=

1

1 + exp⇣� a

2

8�

2

⌘~24�2

+ ~2✓

1

4�2

� a

2

4�4

◆exp

✓� a

2

8�2

◆�

=1

1 + exp⇣� a

2

8�

2

⌘~24�2

⇣1 + exp

⇣� a

2

8�

2

⌘⌘� ~2a2

4�4

exp⇣� a

2

8�

2

⌘�

=~24�2

� ~2a24�4

exp⇣� a

2

8�

2

⌘

1 + exp⇣� a

2

8�

2

⌘ =~24�2

� ~2a2

4�4

⇣exp

⇣+ a

2

8�

2

⌘+ 1⌘ (56)

Therefore the uncertainty in momentum is as follows.

�p =

vuut~24�2

� ~2a2

4�4

⇣exp

⇣+ a

2

8�

2

⌘+ 1⌘ =

~2�

vuut1� a

2

�

2

⇣exp

⇣+ a

2

8�

2

⌘+ 1⌘ (57)

When a � �, the first term is much larger and the uncertainty goes to ~2�

.The uncertainty product is

�x�p =~2

vuuut

0

@1 +a

2

4�2

⇣1 + exp

⇣� a

2

8�

2

⌘⌘

1

A

0

@1� a

2

�

2

⇣exp

⇣+ a

2

8�

2

⌘+ 1⌘

1

A (58)

For a ⌧ �, this approaches ~2

, the minimum uncertainty. For a � �, this approaches ~a4�

, which is much largerthan the minimum uncertainty.

Therefore, this wavefunction is not one of minimum uncertainty.

9