stationary)perturbation)theory)nsmn1.uh.edu/cratti/phy6316-spring_2018_files/qm_lecture_11_cla… ·...

ì Stationary perturbation theory Lecture notes 11 (based on CT, Sec4on 11)

Introduction Ø  In the case of harmonic oscillator and hydrogen atom, their

Hamiltonians are simple enough for their eigenvalue equa4ons to be solved exactly

Ø  This happens only in a small number of problems

Ø  There exist approxima4on methods which enable us to find analy4cally approximate solu4ons of the eigenvalue equa4on in certain cases

Ø  Sta4onary perturba4on theory is widely used in quantum physics

Ø  One isolates the main effects and, when they have been understood, one tries to explain the finer details

Description of the method Ø  We can use perturba4on theory when the Hamiltonian of the

system being studied can be wriHen as H=H0+W, where we know the eigenstates and eigenvalues of H0 and W is small compared to H0 (its matrix elements are much smaller than those of H0)

Ø  H0 is 4me-‐independent and it is called the “unperturbed Hamiltonian”

Ø  W is called “the perturba4on”

Ø  If W is not 4me-‐dependent, we say that it is a “sta4onary perturba4on”

Ø  We need to find the modifica4ons in the eigenvalues and eigenstates due to the perturba4on

Description of the method Ø  We assume that W is propor4onal to a real, dimensionless

parameter λ:

Ø  Here, is an operator whose matrix elements are comparable to those of H0 in magnitude

Ø  Perturba4on theory consists of expanding the eigenvalues and eigenstates of H in powers of λ, keeping only a finite number of terms

Description of the method Ø  We assume the eigenvalues and eigenvectors of H0 to be known

Ø  We assume that the unperturbed energies form a discrete spectrum:

the index i labels the various eigenvectors associated with the energy level in the case of degeneracy

Ø  The set of vectors forms an orthonormal basis of the state space:

Ø  We can rewrite : the eigenvalues E(λ) of H(λ) generally depend on λ

Description of the method Example of varia4on of the eigenvalues E(λ) of H(λ) with respect to λ

Ø  For λ=0 we obtain the spectrum of H0

Ø  The eigenvalues and are doubly degenerate

Ø  The applica4on of the perturba4on removes the degeneracy in but not the one in

Ø  An addi4onal two-‐fold degeneracy appears at λ=λ1

Description of the method Ø  We are looking for the eigenstates |ψ(λ)> and the eigenvalues

E(λ) of the Hermi4an operator H(λ):

Ø  We assume that they can be expanded in powers of λ:

Ø  We replace the expression for H(λ):

Ø  We want this equa4on to be sa4sfied for λ small but arbitrary: we must equate the coefficients of successive powers of λ

Description of the method Ø  Thus we have:

Ø  0th-‐order terms

Ø  1st-‐ order terms

Ø  2nd-‐order terms

Ø  qth-‐order terms

Description of the method Ø  We will neglect terms of order higher than 2 in λ

Ø  We require that |ψ(λ)> is normalized, and we fix the phase by requiring that <0|ψ(λ)> is real

Ø  To 0th-‐order, this implies that |0> must be normalized: <0|0>=1

Ø  To 1st-‐order, we can write

which is =1 if the λ term is zero. But, since we chose the phase such that <0|1> is real, we must have <0,1>=<1,0>=0

A similar argument, for 2nd-‐order in λ yields <0|2>=<2|0>=-‐1/2<1|1>

Description of the method Ø  In general we have:

Ø  Therefore, confining ourselves to 2nd-‐order in λ, we have found the perturba4on equa4ons and we have to supplement them with the above condi4ons

Description of the method Ø  This equa4on expresses the fact that |0> is an eigenvector of H0

with eigenvalue ε0

Ø  This was to be expected

Ø  We therefore choose an eigenvalue of H0

Ø  There can exist one or many energies E(λ) of H(λ) which approach when λà0

Ø  The number of eigenstates corresponding to these E(λ) cannot vary discon4nuously when λ varies around 0. Therefore, this equals the degeneracy of

Perturbation of a non-‐degenerate level

Ø  Consider a non-‐degenerate level of the unperturbed Hamiltonian H0

Ø  The eigenvector associated with it is unique within a constant factor

Ø  We want to determine the change in the energy and vector by adding a perturba4on W to the Hamiltonian

Ø  We choose , which means that |0> is propor4onal to

Ø  We choose


Ø  We call En(λ) the eigenvalue of H(λ) which approaches as λà0

Ø  We assume that λ is small enough that this eigenvalue remains non-‐degenerate

Ø  We now calculate the first terms of the expansion of En(λ) and |ψn(λ)> in powers of λ


Ø  First order correc+ons

Ø  The equa4ons we need are:

and

Ø  We project the first equa4on onto

Ø  The first term is zero since is eigenvector of H0 with eigenvalue ε0


Ø  The equa4on then becomes

Ø  Therefore, to first order in the perturba4on we have:

Ø  The first-‐order correc4on to a non-‐degenerate energy is simply equal to the mean value of the perturba4on term W in the unperturbed state


Ø  We must now find the correc4on to the eigenvector

Ø  We project the equa4on onto all vectors of the basis other than

Ø 

Ø  The last term is zero due to the orthogonality of the eigenvectors

Ø  Besides, we can let H0 act on :


Ø  Therefore, the coefficients of the expansion of |1> on the unperturbed basis states is:

Ø  The last coefficient, , is zero because and <1|0>=0 due to the boundary condi4ons

Ø  Therefore:

Ø  To 1st-‐order in perturba4on theory we have:


Ø  Second order correc+ons

We need the following equa4ons:

And <0|2>=<2|0>=-‐1/2<1|1>

Ø  We project the first one onto

Ø  Again the first term is zero, as well as for the orthogonality of the states


Ø  We get:

namely, replacing the expression for vector |1>:

Ø  Therefore, the energy En(λ) can be wriHen, to second order, as:


Ø  If we limit the energy expansion to first order in λ, we can have an idea of the error involved by evalua4ng the second order term

Ø  Consider the expression for ε2. It contains a sum of terms whose numerators are posi4ve or zero.

Ø  We call ΔΕ the absolute value of the difference between and that of the closest level. We have:

Ø  This gives an upper limit for |ε2|


Ø  We can rewrite it as

Ø  Recalling the closure rela4on:

we get:


Ø  Mul4plying both sides by λ2 we obtain an upper limit for the second order term in the expansion of En(λ) as:

where ΔW is the root-‐mean-‐square devia4on of the perturba4on W in the unperturbed state

Ø  This indicates the order of magnitude of the error commiHed by taking only the first-‐order correc4on into account

Perturbation of a degenerate state Ø  We now take a level which is gn-‐fold degenerate

Ø  The corresponding eigensubspace of H0 is

Ø  In this case, the choice is not enough to determine the vector |0>

Ø  Under the ac4on of W, the level generally gives rise to several dis4nct sublevels

Ø  Their number fn is between 1 and gn

Ø  The total number of orthogonal eigenvectors of H associated with the fn sublevels is always equal to gn: if fn<gn, some of these sublevels are degenerate

Perturbation of a degenerate state Ø  We limit ourselves to 1st-‐order in λ for the energies and to 0th-‐

order for the eigenvectors

Ø  We need to determine ε1 and |0>

Ø  We project the equa4on

onto the gn basis vectors

Ø  Since the are eigenvectors of H0 with eigenvalue we obtain the gn rela4ons

Perturbation of a degenerate state Ø  We use the closure rela4on

Ø  The vector |0> is orthogonal to all the basis vectors for which p≠n

Ø  Therefore the sum over p reduces to a single term:

Ø  We have gn2 numbers : we arrange them in a gnxgn matrix of raw index i and column index i’

Perturbation of a degenerate state Ø  This square matrix is cut out of the matrix which

represents in the basis

Ø  The column vector of elements is an eigenvector of with the eigenvalue ε1

Ø  Effec4vely, this corresponds to solving the vector equa4on

Perturbation of a degenerate state

To calculate the eigenvalues (to first order) and the eigenvectors (to zeroeth order) of the Hamiltonian corresponding to a degenerate unperturbed state , diagonalize the matrix which represents the perturba4on W inside the eigensubspace associated with

/

Perturbation of a degenerate state Ø  We call ε1j the dis4nct roots of the characteris4c equa4on of

Ø  These eigenvalues are all real and the sum of their degrees of degeneracy is equal to gn

Ø  Each eigenvalue introduces a different energy correc4on

Ø  The degenerate level splits, to first order in λ, into fn(1) dis4nct sublevels whose energies can be wriHen as

Perturbation of a degenerate state Ø  If fn(1)=gn we say that, to first order, the perturba4on completely

removes the degeneracy of

Ø  If fn(1)<gn we say that the degeneracy, to first order, is only par4ally removed

Ø  We now choose an eigenvalue ε1j of

Ø  If it is non-‐degenerate, the eigenvector |0> is uniquely determined

Ø  Then there exists a single, non-‐degenerate eigenvalue E(λ) of H(λ) which is equal to

Variational method Ø  Consider a physical system whose Hamiltonian H is 4me-‐

independent

Ø  We assume that the en4re spectrum of H is discrete and non-‐degenerate:

Ø  We know H but not its eigenvalues and eigenstates

Ø  The varia4onal method is useful when we do not know how to diagonalize H exactly

Variational method Ø  We choose an arbitrary ket |ψ> of the state space of the system

Ø  For sure we have:

where E0 is the ground state energy and the equality holds only when |ψ> is the eigenstate of H with eigenvalue E0

Ø  To prove the above inequality, we expand |ψ> on the basis of eigenstates of H:

Variational method Ø  We then have:

Ø  And of course:

which proves the inequality

Ø  This is a basis for a method of approximate determina4on of E0

Ø  We choose a family of kets |ψ(α)> which depend on a certain number of parameters α

Ø  We minimize <H>(α) with respect to the parameters α

Ø  This is a good approxima4on of the ground state E0

Variational method Ø  More generally, the mean value of the Hamiltonian H is sta4onary

in the neighborhood of its discrete eigenvalues

Ø  Consider as a func4onal of |ψ>

Ø  Calculate its increment δ<H> when |ψ> becomes |ψ>+|δψ>

Ø  To do so, we write: and differen4ate both sides:

Variational method Ø  Since <H> is a number, we can write:

Ø  The mean value <H> is sta4onary if

which means that

Ø  We set: and rewrite the above as

Variational method Ø  The last rela4on must be sa4sfied by any infinitesimal ket |δψ>

Ø  In par4cular, if we choose , where δλ is an infinitely small real number, we get:

Ø  The norm of the ket must be zero, therefore the ket itself is zero

Ø  This means that:

Ø  <H> is sta4onary if and only if the state vector |ψ> to which it corresponds is an eigenstate of H and the sta4onary values of <H> are the eigenvalues of the Hamiltonian

Variational method Ø  The varia4onal method can be generalized and applied to the

approximate determina4on of the eigenvalues of H

Ø  If the func4on <H>(α) obtained from the trial kets |ψ(α)> has several extrema, they give the approximate values of some of its energies En

Variational method Ø  We will now apply this method to the one-‐dimensional harmonic

oscillator

Ø  We will consider the Hamiltonian:

and solve its eigenvalue equa4on approximately by varia4onal calcula4ons

Ø  The Hamiltonian is even under parity transforma4ons, therefore its ground state is necessarily represented by an even wavefunc4on

Variational method Ø  We consider even trial func4ons

Ø  For example, we take the one-‐parameter family

Ø  The square of the norm of the ket is equal to:

Variational method Ø  We find:

Ø  so that:

Variational method Ø  The deriva4ve of the above func4on goes to zero when:

and we then have:

Ø  We find that the minimum value of <H>(α) is exactly equal to the energy of the ground state of the harmonic oscillator

Ø  This is due to the simplicity of the problem we are studying: the wavefunc4on of the ground state happens to be one of the func4ons of the trial family

Ø  The varia4onal method in this case gives the exact solu4on

Variational method Ø  To calculate the first excited state we have to choose

wavefunc4ons which are orthogonal to the ground state

Ø  We choose the trial family of odd func4ons:

Ø  In this case:

and

Variational method Ø  The above results yield:

Ø  This func4on, for the same value of α0 as above, presents a minimum equal to:

Ø  Again we find exactly the energy E1 and the associated eigenstate because the trial family includes the correct wavefunc4on

Variational method Ø  So far, the families we chose always included the exact

wavefunc4on

Ø  We now try a totally different type of wavefunc4on:

Ø  A simple calcula4on yields:

and:

Variational method Ø  The minimum of this func4on is obtained at:

and is equal to:

Ø  This minimum is 4mes the exact ground state energy

Ø  The error we commit with this method is:

Variational method Ø  We found a minimum of H which is rela4vely close to the real one

Ø  However, the corresponding approximate state is quite far from the true eigenstate

Ø  The wavefunc4on decreases too rapidly for small values of x and way too slowly at large x

Ø  We have to be very careful when physical proper4es other than the energy are calculated using the varia4onal method

Ø  For example we find that

which is not very far from the actual one,

Variational method Ø  However, the expecta4on value of X4 is infinite in the trial

func4on and finite for the real wavefunc4on

Ø  It is impossible to evaluate the error in a varia4onal calcula4on, if we do not know the exact solu4on of the problem

Ø  This method can be very flexible, and usually gives a good approxima4on for the energy

Ø  It is par4cularly valuable when physical arguments give us an idea of the qualita4ve or semi-‐quan4ta4ve form of the solu4ons

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