Time-Independent, Non-Degenerate Perturbation Theory

Alexander S. Rinn

May 3, 2004

Contents

1 Theory
 1.1 What is Perturbation Theory?
 1.2 Degeneracy vs. Non-Degeneracy
 1.3 Derivation of 1st-order Eigenenergy Correction
 1.4 Derivation of 1st-order Eigenstate Correction
2 Hints
 2.1 For Eigenenergy Corrections
 2.2 For Eigenstate Corrections
3 Worked Examples
 3.1 Example of a First Order Energy Correction
 3.2 Example of a First Order Eigenstate Correction
 3.3 Energy Shift Due to Gravity in the Hydrogen Atom
4 Further Reading

1 Theory

1.1 What is Perturbation Theory?

Though some simple problems in quantum mechanics can be solved analytically, those problems that most accurately represent the physical world almost invariably rely on approximation methods. For example, one can analytically solve for the eigenvalues and the eigenstates corresponding to the Hamiltonian of the finite square well potential, but this is not a very physically relevant problem. Instead, consider a system, like a hydrogen atom, and then expose this system to some outside force, like an electric field. The electric field alters the Hamiltonian of the initial system, which in turn alters the corresponding eigenenergies and eigenstates. To illustrate how this works, consider the simplified example of a finite square well potential in which there is a slight deviation (or perturbation) to the potential somewhere within the well. This perturbation alters the Hamiltonian and therefore the corresponding eigenvalues and eigenstates from what they were in the simple case of the unperturbed square well potential. Perturbation theory allows one to find approximate solutions to the perturbed eigenvalue problem by beginning with the known exact solutions of the unperturbed problem and then making small corrections to it based on the new perturbing potential. The limit of the infinite summation of corrections to the unperturbed solution is the exact solution to the perturbed problem. Of course, this infinite sum can never be calculated; the summation must be truncated at some point--hence the approximate nature of the solutions produced by perturbation theory. Luckily, subsequent corrections to the Hamiltonian become smaller and smaller, so the series can usually be truncated after only a few corrections.

One must be careful when using perturbation theory that the perturbing potential does not change the number of bound states in the system. As will be shown, perturbation theory relies on the assumption that the unperturbed states form a complete set, so the corrected states may be expressed as linear combinations of the unperturbed states. For example, if the perturbing potential changes the Hamiltonian of the system such that the number of bound states is increased by one, this new state must have come from the unbounded region. This is a problem because the unbounded region contains a continuum of energies, and perturbation theory requires (as will be shown) division by the difference between subsequent energy states. With a continuum of energies, this is clearly not possible. So while perturbation theory is a very useful tool, it is not always the method of choice.

1.2 Degeneracy vs. Non-Degeneracy

Recall that degeneracy in quantum mechanics refers to the situation when more than one eigenstate corresponds to the same energy. Conversely, non-degeneracy occurs when each eigenstate corresponds to a unique energy. Take for example,the hydrogen atom: in the absence of any external field, and ignoring spin, an electron in the nth energy level can have orbital quantum numbers

l  (-  {0, 1,...,n - 1}

and magnetic quantum numbers

mz   (-  {- (n - 1),-(n - 2),...,(n - 2),(n - 1)}.

Each combination of these quantum numbers corresponds to a particular eigenstate, but all of these eigenstates correspond to the same energy. However, in the presence of an external field, each of these eigenstates would correspond to a unique energy--the degeneracy is removed.

In non-degenerate perturbation theory there is no degeneracy of eigenstates; each eigenstate corresponds to a unique eigenenergy. One must only be concerned with the slight effects of the perturbing potential on the eigenenergies and eigenstates. However, the situation is not so simple in degenerate perturbation theory: the perturbing potential removes the degeneracy and alters the individual eigenstates. For more on degenerate perturbation theory see LINK LINK LINK.

1.3 Derivation of 1st-order Eigenenergy Correction

In the following derivations, let it be assumed that all eigenenergies and eigenfunctions are normalized.

To find the 1st-order energy correction due to some perturbing potential, begin with the unperturbed eigenvalue problem

 ^   (0)    (0) (0)
H0f  n =  En  fn .
(1)

If some perturbing Hamiltonian is added to the unperturbed Hamiltonian, the total Hamiltonian becomes

 ^    ^    ^ '
H  = H0 +  H ,
(2)

where it is required that this perturbing Hamiltonian be small compared to the unperturbed Hamiltonian. The perturbed eigenvalue problem then becomes

(H ^0 + ^H')fn =  Enfn;
(3)

where

fn = f(n0)+ Dfn,
(4)

i.e. the unperturbed eigenstate plus some small correction, and

        (0)
En  = E n +  DEn,
(5)

i.e. the unperturbed eigenenergy plus some small correction.

Now, if one were to make further corrections to either the eigenstate or the eigenenergy, one would simply continue to add smaller and smaller corrections to the unperturbed value. One could use some parameter c to keep track of the higher and higher order corrections to the unperturbed value, such that

       (0)      (1)    2 (2)
fn =  fn  + cf n + c  fn  + ...
(6)

        (0)      (1)    2  (2)
En  = E n  + cE n  + c E n  + ...
(7)

Thus, as c --> 0

        (0)
fn -->  f n
(8)

        (0)
En -->  E n .
(9)

Substituting the new expressions for fn and En into the perturbed eigenvalue problem yields

  ^      ^'   (0)     (1)   2  (2)
(H0 +  cH  )(f n +  cfn  + c f n +  ...) =
(10)

(E(0n)+  cE(1n)+  c2E(2n)+  ...)(f(0n)+  cf(n1)+ c2f(2n)+  ...).

Expanding and collecting like terms:

 ^   (0)    (0) (0)      ^  (1)   ^' (0)    (1) (0)    (0) (1)
[H0f  n -  En f n ] + c[H0f n + H  fn  - E n fn  - E n f n ]+
(11)

c2[ ^H0f(2n)+ ^H'f(n1)- E(0n)f(2n)-  E(n1)f(n1) - E(2n)f(n0)] + ...=  0.

Formally, this equation says

F (0) + cF (1) + c2F (2) + ...= 0,
(12)

which means that

F (0) = F (1) = F (2) = ...= 0.
(13)

Now, since

  (0)       (0)     (0) (0)
F   =  ^H0f n  - E n fn  = 0,
(14)

clearly we get the original unperturbed portion of the problem back:

 ^  (0)    (0) (0)
H0f n  =  En  fn .
(15)

But, more importantly for us, we get the corrections to the unperturbed problem due to the perturbing potential as well:

 ^  (0)    (0)  (0)   ^   (1)   ^ ' (0)    (1) (0)     (0) (1)
H0f n  - E n f n =  H0f n +  H f n -  En f n -  En  fn .
(16)

Collecting like terms yields

 ^      (0)  (1)     (1)    ^'  (0)
(H0  - E n )fn  = (E n  - H )f n .
(17)

Equation (17) tells us that H^0 operates on fn(1) suggesting that f n(1) may be expressed as the superposition of the eigenstates corresponding to  ^
H0; namely

  (1)    sum       (0)
f n =     Cnif i .
        i
(18)

Then, switching to Dirac notation

 (0)    (0)
fn  = |fn  >,
(19)

and multiplying from the left by <fj(0)|, equation (17) becomes

   (0)         (0)  sum        (0)      (0)   (1)     '  (0)
<f j |(H^0  - En  )    Cni|fi > = <fj |(En  -  ^H )|fn >.
                  i
(20)

Now the LHS of equation (20) can be separated into two terms

                 sum 
<f(0)|(H ^ - E(0))    C  |f(0)> =
  j    0     n       ni  i
                  i
(21)

  (0)     sum        (0)      (0)      sum        (0)
<fj |^H0    Cni |f i >-  <f j |E(n0)    Cni| f i >.
         i                       i

The first of these two terms can be re-written

   (0) ^  sum        (0)    sum         (0) ^   (0)
<f j |H0     Cni|fi  > =    Cni<f j |H0 |fi >,
          i              i
(22)

and this bra-ket yields the unperturbed energy in the j-basis; therefore

 sum        (0)     (0)      (0) sum        (0)  (0)
    Cni<fj  |H^0 |fi > = E j     Cni<fj  |f i >.
  i                          i
(23)

Now the bra-ket is simply a delta function, meaning that the only value in the series that is non-zero occurs when i = j:

     sum                        sum 
E(0j)    Cni<f(j0)|f(i0)> = E(j0)    Cnidji = E(0j)Cnj.
      i                      i
(24)

The second term in the LHS can be re-written in a similar way:

  (0)  (0) sum        (0)      (0) sum             (0)
<f j |En      Cni| f i > = En      Cnidji = E n Cnj.
          i                   i
(25)

Therefore, the LHS of equation (20) becomes

          (0)       (0)            (0)    (0)
LHS   =  Ej Cnj -  En  Cnj = Cnj(E j  - E n ).
(26)

The RHS of equation (20) can be simplified in a similar way:

  (0)   (1)   ^ '  (0)      (0)  (1)  (0)      (0) ^ ' (0)     (1)     ^ '
<fj |(En  - H  )|fn > = <fj |E n |fn >-  <fj |H |fn > = E n djn - H jn,
(27)

where we use the shorthand notation

 ^'      (0) ^ ' (0)
H jn = <fj |H |fn >.
(28)

Now we re-write equation (20) in terms of the new LHS and RHS:

      (0)    (0)     (1)      ^'
Cnj(E j -  En  ) = En djn - H jn,
(29)

or

      (0)    (0)    ^'      (1)
Cnj(E j -  En  ) + H jn = E n djn.
(30)

When n = j, equation (30) becomes the final expression for the 1st-order energy correction due to a perturbing potential H^':

  (1)   ^'       (0) ^'  (0)
E n  = H nn = <f n |H  |f n >.
(31)

This equation states that the 1st-order correction to the energy corresponding to some non-degenerate energy level is simply the perturbing potential ^H' “averaged” over the unperturbed eigenstate of the system.

Thus the new eigenenergy is

        (0)     (0)   ' (0)
En  = E n +  <fn |^H |fn >.
(32)

1.4 Derivation of 1st-order Eigenstate Correction

To find the 1st-order eigenstate correction, remember equation (18), which says that fn(1) may be expressed as the superposition of the eigenstates corresponding to H^0; namely

  (1)     sum        (0)
|f n > =    Cni |f i >.
         i
(33)

Then for n /= j equation (30) becomes

       (0)    (0)     '
Cnj(E j  -  En ) + H^jn = 0,
(34)

or

          H'
Cnj =  -----jn----.
       E(n0)- E(0j)
(35)

But j is simply one particular state out of all i. So in general

            '
C   = ----H-in----.
  ni  E(0n) - E(0)
               i
(36)

Substituting this equation back into equation (33) yields

  (1)     sum   ---H'in----  (0)
|f n > =      (0)    (0)|f i >.
        i/=n En  - E i
(37)

Thus, the new eigenstate is

            sum         '
f  = f(0)+     ----H-in----f(0).
 n    n        E(0n) - E(0)  i
            i/=n          i
(38)

2 Hints

2.1 For Eigenenergy Corrections

When calculating the first order energy correction due to some perturbing potential, it must be kept in mind that the only part of the eigenstate that is used in the calculation is that part which is effected by the potential. For example, Figure 1 shows a finite well with a perturbing potential. In such a case, the eigenstate is seperated into three regions I, II, III, but clearly the only region effected by the perturbation is region II. Therefore, the calculation for the first order energy correction is really a modified form of equation (31):

E(1n) = <f(0n)II| V0|f(n0)II>.


psfig
Figure 1: Finite well with a perturbation.


2.2 For Eigenstate Corrections

Recall equation (37) which gives the first order eigenstate correction due to a perturbation

  (1)     sum   ---H'in----  (0)
|f n > =     E(0)- E(0) |f i >.
        i/=n  n      i

This is an infinite sum over all states i /= n, and this infinite sum must be retained. Just because it is usually sufficient to truncate the corrections to the eigenstate at first or second order, does not mean that it is sufficient to use only a few terms in the infinite sum for any single correction to the eigenstate. Section 4.2 shows an example of a first order correction to an eigenstate, and the infinite sum is left in the expression for the eigenstate correction.

3 Worked Examples

3.1 Example of a First Order Energy Correction

Problem 13.4 of Liboff’s Introduction to Quantum Mechanics states, “Calculate the corrected second eigenenergy and eigenfunction to first order in the perturbation.” First, we will calculate the fisrt order energy correction, and in the next section we will calculate the first order correction to the eigenfunction. Figure (1) depicts the potential well and the perturbation.

We know that in an unperturbed well, the nth eigenenergy has the value

E(0) = n2E(0).
  n        1
(39)

The second eigenenergy corresponds to n = 2, so in our case

E(0)=  4E(0).
 2       1
(40)

We also know that the second eigenenergy in an unperturbed well corresponds to the sine term in the general equation to the eigenstate

  (0)        npx          npx
f n =  A sin ---- + B cos ----.
             L            L
(41)

So in our case

        V~ --
f(0) =   -2 sin 2px-.
  2      L      L
(42)

From equation (31) we know what the first order correction to nth eigenenergy is. In our case

  (1)     (0)    (0)
E 2  = <f2 |V0|f2 >.
(43)

Switching from Dirac notation to integral notation:

        a integral /2 V~ --           V~ --
  (1)         -2    2px-     2-   2px-
E 2 =        L  sin  L  V0   L sin  L  dx.
      - a/2
(44)

        a/2
     2   integral      2px
= V0 --    sin2 ----dx.
     L          L
      - a/2

Making a change of variables:

h = 2px-
     L
(45)

     2p
dh = ---dx
      L

               pa integral /L
E(1) = V0 2L--      sin2 hdh.
  2       L2p
              -pa/L
(46)

The indefinite integral of sin 2hdh is

 integral             h   sin2h
   sin2 hdh =  --  ------,
              2     4
(47)

therefore

E(12) = V0-[h--  sin2h]pa/L
        p 2      4   -pa/L
(48)

  V   ap    sin 2ap    V  - ap    sin -2ap
= --0[----  ----L-]-  -0[------  ----L--]
   p  2L      4       p   2L        4
(49)

               2ap             2ap-
= V0-{ap--  sin-L--+ -ap - sin--L-}
   p  2L       4     2L       4
(50)

   V0  ap    sin 2aLp
=  --{ ----  ------}
   p   L       2
(51)

Therefore, to first order

       (0)    (1)
E2 = E 2  + E 2
(52)

                           2ap
E  =  4E(0)+  V0{ap--- sin--L-}.
  2      1    p   L       2

3.2 Example of a First Order Eigenstate Correction

In this section we will calculate the first order correction to the eigenstate based on the potential given in problem 13.4a of Liboff’s Introduction to Quantum Mechanics.

Based on equation (38), we know that the corrected eigenstate will have the form

            sum       H'
f2 = f(20)+     ------i2----f(i0).
            i/=2 E(02) - E(0i)
(53)

Equation (42) gives f2(0), so we only need calculate f 2(1). To calculate f 2(1) we must sum over all states i /= 2, so there will be an even solution and an odd solution:

             V~  --
ieven : f(0)=   2-sin ipx
       i       L     L
(54)

             V~ --
iodd : f(0i) = -2 cos ipx
              L      L
(55)

Let us first concentrate on the even solution. We must calculate

H'  = <f(0)|^H'|f(0) >
 i2     i      2
(56)

        a/2
     2  integral      ipx    2px
= V0--     sin ----sin----dx.
    L          L      L
      -a/2

The general form of this integral is

 integral                      sin (a-  b)x    sin (a + b)x
  sin(ax) sin (bx)dx =  ------------  -----------.
                        2(a - b)       2(a + b)
(57)

So, equation (56) becomes

            ip-  2p           ip-   2p-
= V -2[sin[(-L----L )x]-- sin[(L-+-L-)x]-]a/2
   0L    2(ip - 2p)       2( ip-+ 2p)   -a/2
            L   L            L    L
(58)

     2 sin[(i - 2)ap]   sin[(i + 2)ap]
= V0 -[---2p------2L- - ---2p------2L-]
     L    L-(i - 2)        L-(i + 2)
(59)

    2    sin[(i- 2) ap-]   sin[(i + 2)-ap ]
- V0--[- --2p------2L--+  --2p-----2L-]
    L      L (i- 2)        L (i + 2)

  '      2  sin[(i- 2) ap2L-]   sin[(i + 2) ap2L-]
H i2 = V0--[--p-----------  --p----------].
         L    L (i - 2)        L (i + 2)
(60)

Now, by substituting equations (39), (40), (42), (54), and (60) into equation (53) we arrive at the even solution:

      V~ --                          ap          ap   V~ --
        2    2px     sum     2 [sin[(pi-(i 2-)22)L]- sin[(pi+(2i+)22)L]]  2     ipx
f2 =   -- sin ----+     V0 -----L---(0)-----L(0)-----  --sin ----
       L      L     i/=2   L     4E 1  - i2E 1        L      L
(61)

                                     ap          ap
   V~ -2     2px     sum       2  [sin[(i-2)2L]- sin[(i+2)2L]]   ipx
=    -{sin ----+     V0 ---------(i-2)--------(i+2)--- sin ----}
     L      L     i/=2    E(10)p         4 - i2             L
(62)

Let us now turn to the odd solution. We again need to calculate H'i2, but for the odd solutions

            integral a/2
  '     2         ipx    2px
Hi2 = V0--     cos----sin----dx.
        L -a/2     L       L
(63)

The general form of this integral is

 integral 
                        cos(a---b)x-  cos(a-+-b)x-
   cos(ax) sin (bx)dx =   2(a - b)   -   2(a + b)  .
(64)

So, equation (56) becomes

         2  cos[(i-  2)ap]   cos[(i + 2)ap]
H'i2 = V0--[--2p------2L--  --2p------2L-]
         L    L (i- 2)         L (i + 2)
(65)

     2-cos[--(i--2)a2pL]   cos[--(i +-2)a2pL]
- V0 L[    2p-(i - 2)    +     2p-(i + 2)  ].
           L                 L

But, since cos (x) = cos (-x),

           ap                ap
cos[(i--2)-2L-]   cos[--(i---2)2L]
   2p(i- 2)   =     2p(i-  2)
   L                 L
(66)

and

cos[(i + 2) a2pL]   cos[- (i + 2) ap2L-]
---2p-------- ] = ---2p----------];
   L (i + 2)         L (i + 2)
(67)

therefore

H'i2 = 0.
(68)

This means that there is no first order correction to the eigenstate for odd energies. Thus the odd solution is simply

      V~  --
        2    2px
f2 =    --sin ----.
        L     L
(69)

3.3 Energy Shift Due to Gravity in the Hydrogen Atom

We know that the interaction between the proton and the electron in an hydrogen atom is not only electrostatic, but also gravitational; however, we also know that the gravitational effect is very small compared to the electrostatic. With perturbation theory we can calculate just how small the gravitational effect within the hydrogen atom actually is. In this example we will show the relative energy shift between the unperturbed 1s state and the corrected 1s state due to the gravitational potential between the proton and the electron in atomic hydrogen.

In Bransden and Joachain’s Introduction to Quantum Mechanics the wavefunction for the 1s state in the hydrogen atom is

      -1-- Z--3/2 -Zr/am
y1s =  V~ p-(a )   e      ,
            m
(70)

and the unperturbed energy is

             e2
E(01)s = - ----------,
         (4pe0)2am
(71)

where

           2
am = 4pe0h--,
       me2
(72)

and m is the reduced mass. In this example the perturbing potential will be the gravitational potential

  '       mMp---
H   = - G   r  ,
(73)

where m is the mass of the electron and Mp is the mass of the proton. Then the first order energy correction is

  (1)     (0)     (0)
E 1s  = <y1s |H'y| 1s >
(74)

                integral 
    GmMp   Z3    1  -2Zr/am
= - ---p---a3-   r-e
             m

=  - GmMp---
       am

Therefore, the relative energy shift between the unperturbed 1s state and the gravitationally perturbed 1s state is

  (1)
E1s- = 8pe0GmMp----
E(0)        e2
 1s
(75)

                  - 12            -11              31              -27
=  8p(8.85419--10---)(6.672--10---)(9.10953--10--)(1.67265---10---)-
                           (1.60219  10- 19)2

E(11s)           -40
  (0)  ~~  8.8 10   .
E1s
(76)

Our suspicions are confirmed. Clearly the gravitational interaction between the electron and the proton in the hydrogen atom contributes very little to the forces within the atom.

4 Further Reading