216
Solutions Physics to Problem Set 3
Spring 2010
where E2 > E1 . The quantities a and b are to be regarded as perturbations that are of the same order but small compared with E2 - E1 . We shall write write the Hamiltonian matrix as H = H (0) + H (1) , where E1 0 0 0 0 a H (0 ) = 0 E1 0 and H (1) = 0 0 b . 0 0 E2 a b 0
1. A system of three unperturbed states consisting of a degenerate pair of states of energy E1 and a non-degenerate state of energy E2 is subsequently perturbed, and is represented by the Hamiltonian matrix: E1 0 a H = 0 E1 b , (1) a b E2
for the unperturbed Hamiltonian and the perturbation, respectively. The unperturbed eigenvalues of H (0) are E1 (doubly degenerate) and E2 . The unperturbed eigenstates are eigenvectors of H (0) , which can be chosen as: 0 0 1 0 0 0 0 . 1 , 0 , 3 = 2 = 1 = 1 0 0
Note that there is no first order correction to the eigenvalues due to the perturbation, since 30 |H (1) |30 = 0 and the 2 2 matrix whose elements are i0 |H (1) |j 0 = 0 , i, j = 1, 2 .
Thus, we must employ second-order perturbation theory to find shifts in the unperturbed eigenvalues.
(a) Use second order nondegenerate perturbation theory to calculate the perturbed eigenvalues. Is this procedure correct? Suppose we apply nondegenerate perturbation theory (which is not the correct procedure given that the unperturbed Hamiltonian matrix possesses degenerate eigenvalues). Using, 0 (1) |m0 |2 | n |H (2) , (2) En = 0 0 En - Em m=n 1
we compute the following matrix elements: 10 |H (1) |20 = 20 |H (1) |10 = 0 , 10 |H (1) |30 = 30 |H (1) |10 20 |H (1) |30 = 30 |H (1) |20 Thus, eq. (2) yields: E1 =
(2)
= a, = b.
|a|2 , E1 - E2
E2 =
(2)
|b|2 , E1 - E2
E3 =
(2)
(2)
|a|2 + |b|2 . E1 - E2
Since |30 is a non-degenerate eigenstate, it follows that E3 is the correct secondorder energy shift corresponding to this eigenstate. However, as already noted, the procedure employed above is not correct when applied to the degenerate subspace.
(b) Use second order degenerate perturbation theory to calculate the perturbed eigenvalues. The correct procedure is to use second-order degenerate perturbation theory. First, we must compute the 2 2 matrix, Wji =
k=j
j 0 |H (1) |k 0 k 0 |H (1) |i0 , 0 0 Ej - Ek
j, i = 1, 2 ,
where the indices j and i run over the degenerate states. We have already computed W11 and W22 in part (a). Thus, we only need to compute
W12 = W21 =
ab . E1 - E2
The correct second-order energy shifts corresponding to the degenerate states |10 and |20 are obtained by computing the eigenvalues of W = 1 E1 - E2 |a|2 a b ab |b|2 .
The corresponding characteristic equation is (|a|2 - )(|b|2 - ) - |a|2 |b|2 = 0 . The solutions to this equation are the eigenvalues, = 0 , |a|2 + |b|2 . 2
Hence, one of the two eigenvalues corresponding to the unperturbed degenerate states is unshifted, whereas the other one is shifted by E
(2)
|a|2 + |b|2 =- E1 - E2
.
Finally, the energy shift corresponding the unperturbed non-degenerate state was correctly obtained in part (a), E3 =
(2)
|a|2 + |b|2 E1 - E2
.
(c) Calculate the eigenvalues exactly and compare with the results of parts (a) and (b). To compute the eigenvalues exactly, we simply diagonalize the matrix H. The corresponding characteristic equation is: (E1 - ) (E1 - )(E2 - ) - |b|2 - |a|2 (E1 - ) = 0 . We immediately see that one root is = E1 , which corresponds to the unshifted eigenvalue obtained in part (b). The other two eigenvalues are roots of the quadratic equation, (E1 - )(E2 - ) - |a|2 - |b|2 = 0 . Denoting the solution to this quadratic equation by , we find: =
1 2
E1 + E2
(E1 - E2 )2 + 4(|a|2 + |b|2 ) .
To compare with the results of part (b), we expand the square root above under the assumption that |a|2 + |b|2 (E1 - E2 )2 . Since E2 > E1 by assumption of the problem, 2(|a|2 + |b|2 ) 1 2 E1 + E2 (E2 - E1 ) 1 + . (E2 - E1 )2 That is, + E2 + |a|2 + |b|2 , E2 - E1 - E1 - |a|2 + |b|2 . E2 - E1
which reproduces the second-order energy shifts obtained in part (b).
3
2. A diatomic molecule behaves like a rigid rotator with a moment of inertia I = 2 Mr0 , where M is the reduced mass and r0 is the distance between the atoms. The Hamiltonian can be approximated by: L H= . 2I Assume that the molecule consists of two atoms of charge e, separated by a distance r0 . Suppose that a uniform electric field is present which points in a fixed direction. Compute the energy eigenvalues of the system, treating the electric field as a perturbation. Use second order perturbation theory. In this problem, the unperturbed Hamiltonian is H
(0) 2
L = , 2I
2
with corresponding unperturbed energy eigenvalues, Em =
(0) 2
( + 1) , 2I
= 0, 1, 2, 3, . . . m = -, - + 1, . . . , - 1, .
(3)
This, each energy level has a (2 + 1)-fold degeneracy. In the presence of a uniform electric field, the Hamiltonian contains an additional term due to the interaction of the charges e with the electric field. The diatomic molecule can be viewed as an electric dipole with dipole moment d = er0 . Hence, the perturbed term of the Hamiltonian is given by H (1) = -d E = -er0 E cos , where is the angle between the electric dipole moment and the electric field. We denote the magnitude of the electric field by E to avoid confusion with the symbol for energy. To employ perturbation theory, we first observe that the following matrix elements, m |H (1) |m = 0 unless m = m and | - | = 1 . (4)
To prove this result, we note that cos = z/r. Thus, we examine m |z|m . Since z is the 0th component of a spherical tensor of rank 1, it follows from the WignerEckart theorem that m |z|m = 0 unless m = m and | - | = 0 or 1. Parity considerations allow us to conclude that m |z|m = 0 unless - | is an odd integer. The latter follows from the fact that z is a parity-odd operator and |m is an eigenstate of parity with eigenvalue (-1) . Hence, eq. (4) follows. Since H(1) cannot connect states of different m, it follows that one can employ non-degenerate perturbation theory to solve for the perturbed energy eigenstates and eigenvalues, as long as one does not sum over degenerate intermediate states. 4
Since m |H (1) |m = 0, there is no first-order energy shift. Using second-order perturbation theory, Em = (er0 E)2
(2) | =
m| cos | m |2 Em - E m
(0) (0)
.
Using eqs. (3) and (4), only two terms survive in the sum above, Em =
(2)
2I(er0 E)2
2
| m| cos | + 1m |2 | m| cos | - 1m |2 + ( + 1) - ( + 1)( + 2) ( + 1) - ( - 1)
d Ym (, ) cos Y+1,m(, ) .
.
(5)
Our final task is to evaluate the two matrix elements above. First, we examine m| cos | + 1 m = To perform this integral, we first write: cos Y+1,m(, ) = =
LM
(6)
4 Y10 (, )Y+1,m(, ) 3
2 + 3 + 1 m; 1 0|L M; + 1 1 + 1 0; 1 0|L 0; + 1 1 YLM (, ) , 2L + 1
after making use of eq. (17-36) on p. 365 of Baym, where we have employed Shankar's notation for the Clebsch-Gordon coefficients. Inserting this result into eq. (6), the remaining integrals are easily evaluated using the orthonormality relations,
d Ym (, )Ym (, ) = mm .
The end result is then given by: m| cos | + 1m = 2 + 3 + 1 m; 1 0| m; + 1 1 + 1 0; 1 0| 0; + 1 1 . 2 + 1
This can be evaluated by using the following general result for the Clebsch-Gordon coefficient, 1/2 L2 - M 2 L M , 1 0|L- 1 M , L 1 = - . L(2L + 1) Hence, ( + 1)2 - m2 + 1 m; 1 0| m; + 1 1 + 1 0; 1 0| 0; + 1 1 = ( + 1)(2 + 3) and we end up with: m| cos | + 1 m = 5 ( + 1)2 - m2 . (2 + 1)(2 + 3) (7)
1/2 1/2
+1 2 + 3
,
Second, we examine m| cos | - 1 m =
d Ym (, ) cos Y-1,m (, ) .
But, there is no need to perform an independent computation. Simply note that m| cos | - 1 m = - 1 m| cos |m
= - 1 m| cos |m ,
since the matrix elements above are clearly real. Thus, we can use the results of eq. (7) with replaced by - 1. Hence, m| cos | - 1 m = 2 - m2 . (2 - 1)(2 + 1) (8)
Inserting eqs. (7) and (8) into eq. (5), and simplifying the resulting expression yields Em = Hence,
2 (2)
I(er0 E)2 [( + 1) - 3m2 ] . 2 ( + 1)(2 - 1)(2 + 3)
Em
( + 1) I(er0 E)2 [( + 1) - 3m2 ] + 2 2I ( + 1)(2 - 1)(2 + 3)
2
Another way to write the same result is: er0 E (0) Em Em 1 + (0) Em
We expect perturbation theory to be accurate if the uniform electric field is weak, i.e. (0) er0 E Em .
( + 1) - 3m . 2(2 - 1)(2 + 3)
2
3. Positronium is a bound state of two spin-1/2 particles: an electron (e- ) and a positron (e+ ). Consider the Hamiltonian for the system, where we focus only on the spin degrees of freedom. In the presence of a uniform external magnetic field, we may take: H = A(I - 1 2 ) + B (1 - 2 ) B , (9) where A is a constant and B is the Bohr magneton. The labels 1 and 2 refer to the electron and positron respectively. The total spin operator is S S1 + S2 , where Si 1 i (i = 1, 2). Solving for 2 S1 S2 , one obtains: S1 S2 = 1 S 2 - S12 - S22 . 2 6
Hence, eq. (9) can be rewritten as: H =A I- 2
2
S 2 - S12 - S22
+ B (1 - 2 ) B .
(a) In zero magnetic field, a transition is observed to occur from the 3 S1 state to the 1 S0 state (which is the ground state). The emitted photon is observed to have a frequency of 2 105 MHz. Evaluate the constant A. If B = 0, the Hamiltonian is given by: H =A I- 2
2
S 2 - S12 - S22
.
Thus, H can be chosen to be simultaneous eigenstates of S 2 , Sz , S12 and S22 . These eigenstates are denoted by |s , ms , where the possible values of s and ms are: s = 1 and ms = +1 , 0 , -1 corresponding to the triplet spin state, and s = 0 and ms = 0 corresponding to the singlet spin state. In the case of zero orbital angular momentum ( = 0), the triplet state is denoted by 3 S1 and the singlet state by 1 S0 . The latter is the ground state of the system. Using S 2 |s , ms = S12 |s , ms = S22 |s , ms =
2
s(s + 1) |s , ms , |s , ms , |s , ms ,
3 2 4 3 2 4
where the latter two results above are a consequence of the fact the the electron and positron are both spin- 1 particles, it follows that 2 H |s , ms = A 1 - 2 s(s + 1) -
3 2
|s , ms =
0 4A |s , ms
, ,
for s = 1 , for s = 0 .
That is, the energies of the 3 S1 and 1 S0 states are given by E= 0 4A , , for the for the S1 state , 1 S0 state .
3
(10)
A photon with the a frequency of 2 105 MHz is observed in a transition from the 3 S1 state to the 1 S0 ground state. Since, E E(3 S1 ) - E(1 S0 ) = h = -4A , it follows that
1 A = - 4 h = - 1 (4.136 10-15 eVsec)(2 1011 sec-1 ) = -2.068 10-4 eV . 4
7
(b) Now turn on the magnetic field. Assume it points in the z-direction. Treating the magnetic field as a perturbation, compute the energy eigenvalues to second order in B and the energy eigenstates to first order in B. We take B = B z . The unperturbed Hamiltonian is given by ^ H (0) = A I - and the perturbation is given by H (1) = eB (S1z - S2z ) . m 2
2
S 2 - S12 - S22
,
As shown in part (a), the unperturbed eigenstates are given by |s , ms with s = 1 for the 3 S1 triplet states and s = 0 for the 1 S0 singlet state. First, consider the first-order energy shifts. For the 1 S0 singlet state, E|0,0 =
(1)
eB 0 , 0| S1z - S2z |0 , 0 . m
To evaluate this, we convert from the total spin basis to the product basis. Recall that: 1 |0 , 0 = | - | , 2 1 |1 , 0 = | + | . 2 Then, eB H (1) |0 , 0 = S1z | - | 2m e B = | + | 2m = - S2z | - | e B |1 , 0 . m
(1)
(11) (12)
(13)
Since |0 , 0 and |1 , 0 are orthogonal states, it follows that E|0,0 = 0. Next, consider the 3 S1 triplet states. These are three degenerate states with respect to H (0) , so we must use degenerate perturbation theory. Thus, we shall evaluate the matrix Wms ,ms 1 , ms | H (1) |1 , ms , which is a 3 3 matrix. To evaluate these matrix elements, we again rewrite the unperturbed eigenstates in terms of the product basis, using eq. (12) and |1 , 1 = | , 8 |1 , -1 = | .
We compute: H (1) |1 , 1 = H (1) |1 , -1 = and eB H (1) |1 , 0 = S1z | + | 2m e B = | - | 2m = - S2z | + | e B |0 , 0 . m eB (S1z - S2z ) | = 0 , m eB (S1z - S2z ) | = 0 , m
Since the |s , ms states are orthonormal, it follows that Wms ,ms = 0 , for all ms , ms = -1 , 0 , +1 .
Thus, we have proven that all first-order energy shifts vanish. Next, we consider the perturbed energy eigenstates at first order in the perturbation expansion. In class, we derived: n
(1)
= n
(0)
+
m=n
m(0)
m(0) H (1) n(0) En - Em
(0) (0)
,
(14)
where we exclude the degenerate states from the sum. Since H (1) |1 , 1 = 0 and H (1) |1 , -1 = 0, it follows that there is no first order shift in the wave functions for |1 , 1 and |1 , -1 . On the other hand, there are first-order shifts in the wave functions for |1 , 0 and |0 , 0 . Using eq. (14), it follows that (1) (1)
|1,0
= |1 , 0 + = |0 , 0 +
|0 , 0 0 , 0| H (1) |1 , 0 , 0 - 4A |1 , 0 1 , 0| H (1) |0 , 0 , 4A - 0
|0,0
where we have used the unperturbed energies obtained in eq. (10). Evaluating these expressions using results obtained above, one finds: (1) (1)
|1,0
= |1 , 0 - = |0 , 0 +
e B |0 , 0 , 4Am e B |1 , 0 . 4Am
(15) (16)
|0,0
The second order energy shifts can now be determined using E (2) = n(0) H (1) n(1) . 9 (17)
Due to H (1) |1 , 1 = 0 and H (1) |1 , -1 = 0, it follows that there are no second-order energy shifts for |1 , 1 and |1 , -1 . On the other hand, there are second-order energy shifts for |1 , 0 and |0 , 0 . Using eq. (17), one finds:
(2) E|1,0
= 1, 0 H
(1)
(1) |1,0
e B 1 =- 1 , 0| H (1) |0 , 0 = - 4Am 4A = e B 1 0 , 0| H (1) |1 , 0 = 4Am 4A
e B m
2
2
,
E|0,0 = 0 , 0 H (1) |0,0
(2)
(1)
e B m
.
The results obtained above are summarized in the following table: state |1 , 1 e B |1 , 0 - |0 , 0 4Am |1 , -1 |0 , 0 + e B |1 , 0 4Am 4A + energy 0 1 - 4A e B m 0 1 4A e B m
2 2
where A < 0 as shown in part (a). REMARK: Since the degeneracy was not resolved at first order, one might worry that we have not yet identified the correct unperturbed degenerate eigenstates. In class, I indicated that to resolve the degeneracy at second order, one should solve the eigenvalue problem for the matrix W (2) , which is defined by:
(2) Wji
=
k=ni
nj
(0)
H (1) nk
(0)
nk
(0)
H (1) ni
(0)
En(0) - Ek(0)
,
where ni are the degenerate states. Applying this result to the present problem, the sum over k consists of a sum over one unperturbed state, |0 , 0 , which is not degenerate with the set of degenerate states |1, , ms , ms = -1 , 0 , +1. Thus, the 3 3 matrix W (2) is given by:
(2) Wji
(0)
=
nj
(0)
H (1) 0 , 0
0 , 0 H (1) ni
(0)
0 - 4A
=-
1 4A
e B m
2
nj
(0)
1, 0
1 , 0 ni
(0)
,
after using H (1) |0 , 0 = (e B/m) |1 , 0 obtained in eq. (13). Since the states |1 , ms , ms = -1 , 0 , +1 are orthonormal, it follows that W (2) is a diagonal matrix in the total spin basis. Thus, the triplet states of the total spin basis are the correct unperturbed degenerate eigenstates. Moreover, the eigenvalues of W (2) are the correct second-order 10
energy shifts--two zero energy eigenvalues corresponding to |1 , 1 and |1 , -1 , and a third energy eigenvalue given by -[1/(4A)](e B/m)2 . The corresponding energy eigenstate is also correctly computed in eq. (15), as we have employed the correct unperturbed degenerate eigenstates.
(c) Repeat the calculation of part (b), but now solve the problem exactly. Expand out your solutions in a power series in B, and verify that the results of part (b) are indeed correct. This problem can be solved exactly by diagonalizing the full 4 4 matrix Hamiltonian, s , ms | H (1) |s , ms , where |s , ms = |1 , 1 , |1 , -1 , |1 , 0 , |0 , 0 (18) indicates a convenient ordering of the entries in the rows and columns of the matrix representation of H. In fact, we have already computed all the matrix elements of H in part (a) of this problem. For example, H |1 , 1 = (H (0) + H (1) ) |1 , 1 = 0 , H |1 , -1 = (H (0) + H (1) ) |1 , -1 = 0 . Using results already obtained in problem 2 and in part (b) of this problem, the 4 4 matrix H with respect to the basis of eq. (18) is given by: 0 0 0 0 0 0 0 0 e B m 0
s , ms | H (1) |s , ms
0 = 0 0
0 e B . m 4A
This matrix is easy to diagonalize. All we have to do is focus on the lower 2 2 block, i.e. the subspace spanned by |1 , 0 and |0 , 0 , e B 0 m . e B 4A m The eigenvalues of this matrix are obtained by solving: -E(4A - E) - 11 e B m
2
= 0.
The solutions to this quadratic equation are denoted by E , where E = 2A 4A2 + e B m
2
.
(19)
To check that this reproduces the perturbative computation of part (a), we expand eq. (19) in a power series about B = 0. First rewrite: 1 e B E = 2A 2A 1 + 2 4A m Noting that A < 0, it follows that A2 = -A and so 1 E 2A 2A 1 + 8A2 2 1 e B - , 4A m = 4A + 1 e B 4A m e B m
2 2 1/2
.
for the state
2
|1,0 |1,0
(1)
(1)
, (20) ,
,
for the state
in agreement with the results obtained in part (b). To obtain the exact energy eigenfunction corresponding to the first eigenvalue E+ , we solve: e B cos cos 2 0 m , = 2A + 4A2 + e B (21) e B m sin sin 4A m since sin is the most general eigenvector of a real symmetric matrix that is normalized to unity. (Without loss of generality, one can take the energy eigenfunctions to be real.) Eq. (21) yields 2 e B e B sin = 2A + 4A2 + cos . m m
cos
It is convenient to solve for tan = sin / cos . One then obtains: 2 e B m 2A + 4A2 + . tan = e B m Using the identities, sin2 = tan2 , 1 + tan2 12 cos2 =
1 , 1 + tan2
one can compute: 2A + sin = 2 4A2 +
2
4A2
+
e B m e B m
2
2 2
,
cos =
-2A +
4A2
+
e B m
2
2
,
2 4A2 +
e B m
where we have used cos2 = 1 - sin2 . The above formulae look simpler when expressed in terms of the energy eigenvalues determined above. Using the results of eq. (19) and noting that: E+ - E- = 2 it then follows that: sin2 = E+ , E+ - E- cos2 = -E- . E+ - E- (22) 4A2 + e B m
2
,
Since A < 0, one sees that E+ > 0 and E- < 0, in which case the above results are consistent with a real angle that can be chosen to lie in the range 0 /2. Thus, the eigenstate corresponding to the eigenvalue E+ is given by: -E- E+ - E-
1/2
|1 , 0 +
E+ E+ - E-
1/2
|0 , 0 .
The eigenstate corresponding to the eigenvalue E- must be orthogonal to the eigenstate obtained above, and is thus given by: - E+ E+ - E-
1/2
|1 , 0 +
-E- E+ - E-
1/2
|0 , 0 .
To complete the diagonalization of H, we note that the other two energy eigenvalues and eigenstates are not shifted by the perturbation due to the zeros in the matrix. That is, the other two eigenstates are |1 , 1 and |1 , -1 with corresponding zero eigenvalues, to all orders in perturbation theory. To summarize, the exact energy eigenstates and eigenvalues are given in the following table: state |1 , 1
-E- E+ -E- 1/2
energy 0
E+ E+ -E- 1/2
|1 , 0 + |1 , -1
|0 , 0
E+ 0
-
E+ E+ -E-
1/2
|1 , 0 +
-E- E+ -E-
1/2
|0 , 0
E-
13
where E 2A 4A2 + e B m
2
.
As a final check, one can expand these results in a power series about B = 0 and show that one obtains the energy eigenfunctions at first order that match the results obtained in part(a). Using eqs. (20) and (22), one finds: e B E+ m sin2 = E+ - E- 1 e B -4A - 2A m 1 - 4A
2 2
2
1 = 16A2
e B m
+ O(B 4 ) ,
Since 0 /2 by convention (and A < 0), it follows that sin - 1 4A e B m + O(B 3 ) .
Finally, since cos2 = 1 - sin2 = 1 - O(B 2 ) , it follows that: cos = 1 - O(B 2 ) . That is, we identify: |1,0 |0,0
(1) (1)
= cos |1 , 0 + sin |0 , 0 |1 , 0 - = - sin |1 , 0 + cos |0 , 0 |0 , 0 +
e B 4Am e B 4Am
|0 , 0 , |1 , 0 ,
which reproduces the results obtained in part (b) by the first-order perturbation theory computation.
4. Consider the hydrogen atom in the n = 2 state. It is placed in a uniform magnetic field B. The Hamiltonian for the hydrogen atom in the non-relativistic limit is a sum of the Coulomb interaction with minimal coupling, the fine structure and the hyperfine structure contributions. That is, H = HC + HFS + HHFS For this problem the hyperfine structure will be ignored. The fine structure Hamiltonian is given by: HFS me = c2 -
2 1 dV p 4 + 2 V (r) . L S + 8m3 c2 2m2 c2 r dr 8m2 c2 e e e
Here L and S are the orbital and spin angular momentum of the electron. 14
(a) Using first order degenerate perturbation theory, compute the energy levels as a function of B, assuming that the contributions due to the fine structure and due to the term in the Hamiltonian which is linear in B are roughly of equal strength. You may neglect the term in the Hamiltonian which is quadratic in B. Also, neglect the proton's spin and the associated hyperfine structure. For the hydrogen atom, V (r) = -e2 /r is the Coulomb potential. Thus, the Hamiltonian for this problem (neglecting the hyperfine interactions) is: H = H (0) + HFS + HB , where H is the Coulomb Hamiltonian, HFS = - p 4 e2 e2 2 3 + L S + (r) 8m3 c2 2m2 c2 r 3 2m2 c2 e e e
(0)
e2 = me c - r
2
is the fine-structure Hamiltonian, and HB = e2 e (B r)2 B (L + 2S) + 2me c 8me c2 (23)
is the spin-dependent Hamiltonian that describes the interaction of an electron (of charge -e) in an external magnetic field [cf. problem 1 of problem set #2]. In the case of a uniform magnetic field, B = B z which is not too strong, we may ^ 2 ignore the term in HB that is of O(B ). Hence, we may take HB = B B (Lz + 2Sz ) ,
where B e /(2me c) is the Bohr magneton. Note that we have set the g-factor of the electron equal to 2. We now compute the first-order perturbation due to HFS +HB . The eigenfunctions of H (0) can be chosen to be simultaneous eigenfunctions of J 2 , Jz , L 2 and S 2 . For the n = 2 level of hydrogen there are eight states as indicated in the following table: j 2s1/2 2p1/2 2p3/2
1 2 1 2 3 2
m 1 2 1 2 1 2 , 3 2
0 1 1
s
1 2 1 2 1 2
15
These are all degenerate eigenstates of H (0) . So in principle, we should compute the matrix elements of the 8 8 matrix W , Wji = j 0 |HFS + HB |i0 . First, we work out the diagonal terms of W . We recognize that n0 |HFS |n0 is just the first-order splitting due to the fine structure, which we worked out in class. Thus, we may employ the formula [cf. eq. (17.3.22) on p. 469 of Shankar], EFS = -
(1)
me c2 4 2n3
1 3 - j + 1/2 4n
,
where j is the total angular momentum quantum number. In this problem, n = 2 is 1 fixed and j = 2 or 3 . 2 Next, we consider n0 |HB |n0 . To evaluate this, we need to express the total angular momentum basis in terms of the product basis. This procedure will involve the following Clebsch-Gordon coefficients may be useful: m , m ,
1 2
ms | + ms | -
1 2
mj , mj ,
1 2
+ 2mj ms + = 2 + 1 = (-1)
1 ms + 2
1 2
1/2
,
1 2 1/2
1 2
1 2
1 2
- 2mj ms + 2 + 1
,
where mj = m + ms . Thus, the s1/2 , p1/2 and p3/2 states can be expressed in terms of the angular momentum product basis states as follows: s1/2 p1/2
1 2 1 2 1 1 mj ; 0 2 = 0 0; 2 mj , 1 mj ; 1 2 = m ,ms 1 2 3 2 1 1 m ; 2 ms 1 1 m ; 1 ms | 2 mj ; 1 1 2 2
=
ms 3 2
(-1)
ms +
- 2mms 3
1 1 ; 2 2
1/2 1 1 mj - ms ; 2 ms 3 2
= p3/2
3 2 1 mj ; 1 2 =
+ mj 3
1/2
1 mj +
-
1 2
-
- mj 3
1/2 1 1 mj - 1 ; 2 2 1 2
,
1 1 m ; 2 ms m ,ms 1 2 3 2
3 1 m ; 1 ms | 2 mj ; 1 1 2 2
=
ms 3 2
(-1)
ms +
+ 2mj ms 3
1 1 ; 2 2
1/2 1 1 mj - ms ; 2 ms 3 2
=
- mj 3
1/2
1 mj +
-
1 2
+
+ mj 3
1/2 1 1 1 mj - 2 ; 2 1 2
,
1 3 where mj = 2 for s1/2 and p1/2 , and mj = 2 , 1 for p3/2 . 2
16
Hence, s1/2 |Lz + 2SZ |s1/2 = 2mj , p1/2 |Lz + 2SZ |p1/2 = p3/2 |Lz + 2SZ |p3/2 =
1 3 1 3 3 2 3 2 1 for mj = 2 , 1 3 3 2 3 2 1 - mj (mj + 2 ) = 2 mj , 3 1 for mj = 2 , 3 for mj = 1 , 2 . 2
+ mj (mj - 1 ) + 2
1 - mj (m - 2 ) +
1 3
4 1 + mj (mj + 2 ) = 3 mj ,
One can now complete the calculation of the diagonal matrix elements of H (1) . We now turn to the off-diagonal elements. First, we note that k 0 |HFS |n0 = 0 , for k = n .
This can be easily proven by noting that HFS commutes with J 2 , Jz , L 2 and S 2 , and therefore must be diagonal in the total angular momentum basis.1 Thus, all we need to do is compute the matrix elements k 0 |HB |n0 . The computations are similar to the diagonal elements evaluated above. In particular, many of the off-diagonal elements are zero: s1/2 mj |Lz + 2Sz |s1/2 mj = 2mj mj mj , p1/2 mj |Lz + 2Sz |p1/2 mj = 2 mj mj mj , 3 p3/2 mj |Lz + 2Sz |p3/2 mj = 4 mj mj mj , 3 s1/2 mj |Lz + 2Sz |p1/2 mj = s1/2 mj |Lz + 2Sz |p3/2 mj = 0 . (24) (25) (26) (27)
Eq. (27) is a consequence of the fact that neither Lz nor Sz can change the orbital angular momentum quantum number . Thus eq. (27) follows from the orthogonality of states of different orbital angular momentum. The only possible non-zero off1 diagonal matrix elements are between states of equal and mj . Thus, for mj = 2 , p1/2 mj |Lz + 2Sz |p3/2 mj = p3/2 mj |Lz + 2Sz |p1/2 mj =
1 3 3 2 1 2 1 2 1/2 1/2
+ mj
9 4
3 2
= -1 3
- mj
1/2
- mj mj - = -1 2 . 3
- mj +
All other non-diagonal matrix elements are zero. Collecting our results, we see that k 0 |H (1) |n0 is diagonal in the subspace spanned 3 by the states 2s1/2 (mj = 1 ) and 2p3/2 (mj = 2 ). Thus, we may immediately write 2 down the first-order energy shifts:
5 E (1) (2s1/2 ) = - 128 me c2 4 + 2mj B B , 4 1 E (1) (2p3/2 ) = - 128 me c2 4 + 3 mj B B ,
1
1 for mj = 2 ,
for mj = 3 . 2
In particular, if we write L S = 1 (J 2 - L 2 - S 2 ), it is easy to check that L S commutes with 2 J 2 , Jz , L 2 and S 2 .
17
The other four states, 2p1/2 (mj = 1 ) and 2p3/2 (mj = 1 ) break up into two pairs 2 2 of states, (i) (ii) 2p1/2 , mj =
1 2
, 2p3/2 , mj =
1 2
,
1 2p1/2 , mj = - 2 , 2p3/2 , mj = - 1 . 2
To obtain the corresponding first-order energy shifts, we must find the eigenvalues of the two 2 2 matrices, 2 5 - 1 2 B B - 128 me c2 4 + 3 mj B B 3 , for mj = 1 . (28) 2 4 1 1 2 4 - 128 me c + 3 mj B B - 3 2 B B In an appendix to these solutions, I compute the eigenvalues of a real symmetric ac 2 2 matrix. In particular, given the real symmetric matrix c b , the corresponding eigenvalues are 1 a + b (a - b)2 + 4c2 . 2 Hence, the eigenvalues of the matrix given in eq. (28) are:
3 - 128 me c2 4 + mj B B 1 m c2 4 64 e 1 + 3 mj B B 2
+ 2 2 B 2 , 9 B
for mj = 1 . 2 (29) (30)
2
We now summarize the above results. The first-order energy shifts are given by:
5 E (1) (2s1/2 ) = - 128 me c2 4 + 2mj B B , 1 E (1) (2p3/2 ) = - 128 me c2 4 + 4 mj B B , 3 3 E (1) (2p1/2 , 2p3/2 ) = - 128 me c2 4 + mj B B 1 for mj = 2 , 3 for mj = 2 , 1 m c2 4 64 e
+ 1 mj B B 3
+ 2 2 B 2 , 9 B (31)
for mj = 1 . 2
(b) Compute the limits of large and small B. Weak field limit In the weak field limit, we shall neglect O(B 2 ) terms in the first-order energy 1 shifts. The first-order energy shifts for 2s1/2 mj = 2 and 2p3/2 mj = 3 have no 2 O(B 2 ) corrections and are given by eqs. (29) and (30). The first-order energy shifts of the remaining states are given by eq. (31). In these cases, we expand out the square root,
1 m c2 4 64 e 1 + 3 mj B B 2
+ 2 2 B 2 9 B
1 m c2 4 64 e
+ 1 mj B B + O(B 2 ) . 3
That is, the off-diagonal terms in eq. (28) can be neglected, as they contribute only terms of O(B 2 ) to the energy shifts. Thus, in the weak field limit, 2p1/2 mj = 1 and 2 18
2p3/2 mj = 1 are approximate perturbed energy eigenstates with corresponding 2 first-order energy shifts given by:
5 2 1 E (1) (2p1/2 , mj = 2 ) = - 128 me c2 4 + 3 mj B B , 1 1 4 E (1) (2p3/2 , mj = 2 ) = - 128 me c2 4 + 3 mj B B .
(32) (33)
REMARK: Since off-diagonal matrix elements do not contribute to the O(B) corrections to the first-order energy shifts, it is easy to compute the shifts for arbitrary 1 n, j and . Denoting the eigenstates of H (0) by njmj ; 1 , where j = 2 , one can 2 easily derive the relation between the total angular momentum basis and the product basis,
1 jmj ; 2
m+ = 2 + 1
1 2
1/2
mj -
1 1 1 ; 2 2 2
+
m+ 2 + 1
1 2
1/2 1 1 mj + 2 ; 2 - 1 2
,
1 for j = 2 ,
(34)
following the same steps used on p. 16 of these solutions. It follows that: njmj ; 1 |Lz 2 +
1 2Sz |njmj ; 2
=
mj + 2 + 1
1 2
(mj - + 1) + ,
1 2
mj + 2 + 1
1 2
(mj + 1 - 1) 2
= mj 1
1 2 + 1
for j = 1 . 2
1 Thus, the first-order energy shift of the state njmj ; 2 in the weak field limit is given by:
E (1) = -
me c2 4 2n3
3 1 - j + 1/2 4n
+ B Bmj 1
1 (2 + 1)
,
for j =
1 2
(35) and the possible values of mj are mj = -j , -j + 1 , . . . , j - 1 , j. As usual, only j = 1 is permitted in the case of = 0, since the possible values of j and are 2 3 5 j = 1 , 2 , 2 , . . . and = 0, 1, 2, 3, . . ., respectively. One can easily check that for the 2 1 3 cases of n = 2, = 0, 1 and j = 2 , 2 , eq. (35) reproduces the results of eqs. (29), (30), (32) and (33). Strong field limit In the strong field limit, we take B B me c2 4 [although keep in mind that we are still assuming that the field is not so strong in order that one may still justifying dropping the O(B 2 ) terms in HB [cf. eq. (23)]. Then, eqs. (29)(31) yield: E (1) (2s1/2 ) 2mj B B = B B ,
4 E (1) (2p3/2 ) 3 mj B B = 2B B ,
(36) (37) (38)
E (1) (2p1/2 , 2p3/2 ) B B mj
1 3
2 + m2 = 1 B B(1 1) , j 2
19
where in the last expression, all four combinations of signs are possible. In this limit, the states are approximate eigenstates of L 2 , Lz , S 2 and Sz . Thus, the first-order energy shifts can be computed via E (1) B B
1 1 m ; 2 ms |Lz + 2Sz |m ; 2 ms = B B(m + 2ms )
(39)
Possible values of m + 2ms are 1 for 2s1/2 and 2, 1, 0, 0 for 2p1/2 , 2p3/2 . Hence, eq. (39) correctly reproduces the results of eqs. (36)(38)
(c) Sketch a picture of the energy levels found in part (a) as a function of B. Below is a very rough sketch of the p-state energies as a function of B. Unfortunately, the scanning resolution was rather poor. I will provide a more readable sketch at a later date.
20
5. Consider the hydrogen atom in an excited n = 2 state, which is subjected to an external uniform electric field E. Do not neglect the spin of the electron. Assume that the field E is sufficiently weak so that eEa0 is small compared to the fine structure, but such that the Lamb shift (where = 1057 MHz) cannot be neglected. That is, treat the problem as a two-level system consisting of the the 2s1/2 and 2p1/2 states of hydrogen. In particular, you may ignore the 2p3/2 state of hydrogen and the hyperfine interactions. (a) Compute the Stark effect for the 2s1/2 and 2p1/2 states of hydrogen by solving the two-level system exactly. The Hamiltonian governing the system is: p2 e2 H= - + HFS - eEz + HLS , 2m r where HFS is responsible for the fine-structure and HLS is responsible for the Lamb shift. Since only the 2s1/2 and 2p1/2 states matter in this calculation, all we need to do is compute the eigenvalues of the 2 2 matrix, H 2p1/2 |H|2s1/2 2s1/2 |H|2s1/2 2p1/2 |H|2p1/2 2s1/2 |H|2p1/2 . (40)
We shall treat -eEz as the perturbation. That is, we denote: H
(0)
p2 e2 = - + HFS + HLS , 2m r
H (1) = -eEz .
By definition, 2s1/2 and 2p1/2 are eigenstates of H (0) . If the Lamb-shift is neglected, then the 2s1/2 and 2p1/2 states are degenerate with energy,
1 E (0) = - 2 me c2 2 - 5 m c2 4 , 128 e
as computed in class. Including the Lamb shift, 2s1/2 and 2p1/2 are no longer degenerate. Instead, we have E (0) (2s1/2 ) - E (0) (2p1/2 ) > 0 , which defines . We now compute the matrix H [cf. eq. (40)]. Note that parity considerations imply that 2s1/2 |H (1) |2s1/2 = 2p1/2 |H (1) |2p1/2 = 0 . Moreover, H (0) is diagonal with respect to the 2s1/2 and 2p1/2 states. That is, 2s1/2 |H (0) |2p1/2 = 2p1/2 |H (0) |2s1/2 = 0 .
21
Thus, all we need to evaluate is 2s1/2 |H (1) |2p1/2 . In order to compute this matrix element, we shall make use of eq. (34) to express the 2s1/2 and 2p1/2 states in terms of eigenstates of L 2 , Lz , S 2 and Sz . In the coordinate representation,
j 1 r | j = 2 , mj = Rn (r)Ymj (, ) ,
where we have used [cf. eq. (34)]:
1 j = 2 , mj =
where Rn (r) is the radial wave function of the hydrogen atom, and the spin spherical harmonics are defined by:2 1 mj + 2 Y,m - 1 (, ) 1 j= 1 j 2 Ymj 2 (, ) , |j = 1 , mj = , 2 1 2 + 1 mj + 2 Y,m + 1 (, )
j
2
1 1 1 1 1 (m+ 1 )1/2 , 2 ; m - 1 , 2 + (m+ 1 )1/2 , 2 ; m + 2 , - 1 2 2 2 2 2 + 1
.
Due to the Wigner-Eckart theorem, it is sufficient to evaluate the matrix element 1 2s1/2 |H (1) |2p1/2 for a fixed value of mj . Thus, for mj = 2 , r | 2s1/2 = R20 (r) where R20 (r) = 1 2a0
3/2
Y00 (, ) 0
,
-Y10 (, ) 1 r | 2p1/2 = R21 (r) 3 2 Y11 (, )
,
r 2- a0
e
-r/(2a0 )
,
1 R21 (r) = 3
1 2a0
3/2
r -r/(2a0 ) e , a0
and a0 is the Bohr radius. Hence, using H (1) = -eEr cos , eE 2s1/2 |H (1) |2p1/2 = 3
d3 r Y00 (, )R20 (r)Y10 (, )R21 (r) r cos 3 0
eE 1 1 2 = 3 3 2a0 3a0 = - 3 eEa0 , where we have used,
1
r 4 (2a0 - r)e-r/a0 dr
3 cos2 d 4
4 , cos d = 2 cos d cos = 3 -1
2 2
and
0
xn e-x dx = n! ,
with x r/a0 .
2
Note that the usual spherical harmonics can be written as: Ym (, ) , | , m .
22
Thus, the matrix H is given by: H= where Es 2s1/2 |H|2s1/2 , Es - 3 eEa0
- 3 eEa0 Ep
,
Ep 2p1/2 |H|2p1/2 ,
and Es - Ep = . Since H is a real symmetric matrix, it can be diagonalized by a real orthogonal matrix. In an appendix to these solutions, I provide details of the computation of the eigenvalues and eigenvectors. Denoting the two eigenvalues by E , we have: E =
1 2
Es + Ep
2 + 12e2 E 2 a2 . 0
The corresponding eigenvectors are: cos 2s1/2 + sin 2p1/2 , - sin 2s1/2 + cos 2p1/2 , where 0 < (by convention) with: -2 3eEa0 sin 2 = , 2 + 12e2 E 2 a2 0
cos 2 =
2 + 12e2 E 2 a2 0
,
which imply that 1 < < . Then, it follows that: 2 sin = 1 - cos 2 , 2 cos = - 1 + cos 2 . 2
The Lamb shift corresponds to /h = 1057 Mhz. (b) Show that for eEa0 h, the energy shifts due to the external electric field are quadratic in E, whereas for eEa0 h, they are linear in E. Determine the (perturbed) energy eigenstates in both limiting cases. If eEa0 , then we can expand the square root, 2 + 12e2 E 2 a2 + 0 It follows that E
1 2
6e2 E 2 a2 0 . .
Es + Ep + 23
6e2 E 2 a2 0
3e2 E 2a2 3e2 E 2 a2 0 0 , E- Ep - , As advertised, the energy shifts due to the external electric field are quadratic in E. The corresponding eigenstates are determined by: -2 3eEa0 6e2 E 2 a2 0 sin 2 , cos 2 1 - , 2 Hence, it follows that: 3eEa0 e2 E 2 a2 0 sin , cos -1 + O . 2 E+ Es + We conclude that the corresponding eigenstates are:3 3eEa0 3eEa0 2p1/2 , 2s1/2 + 2p1/2 . 2s1/2 - We see that there is an admixture of the 2p1/2 state in the 2s1/2 state when an external uniform electric field is present. If eEa0 , then we can simply set = 0 to first approximation. Then, 1 E 2 (Es + Ep ) 3eEa0 , which is linear in the external electric field as advertised. In this limit, cos 2 = 0, in which case = 3/4. Hence, sin = - cos = 1/ 2, and we find that the corresponding eigenvectors are: 1 2 2s1/2 2p1/2 .
Putting Es - Ep then yields:
(c) The critical field is defined as: Ec h , 3ea0
where the factor of 3 is conventional. The linear or quadratic behavior of the energy shifts obtained in part (b) depend on the magnitude of E as compared to Ec . Determine Ec in volts/cm. We finish off with some numerics. Putting hc = 12, 400 eV- a0 = 0.53 A, A, 6 -1 10 -1 = 1057 10 s and c = 3 10 cms , it follows that: (12, 400)(1057 106 ) Ec = volts/cm = 476 Vcm-1 . 3(3 1010 )(0.53)
For convenience, I have multiplied both eigenstates by an overall factor of -1, as the overall phase of the state is a matter of convention.
3
24
Appendix: The eigenvalues and eigenvectors of a real symmetric 2 2 matrix Consider a real symmetric matrix, A= a c c b ,
where a, b and c are arbitrary real numbers. The eigenvalues are the roots of the characteristic equation: a- c = (a - )(b - ) - c2 = 2 - (a + b) + (ab - c2 ) = 0 . c b- The two roots, 1 and 2 , can be determined from the quadratic formula. Noting that (a + b)2 - 4(ab - c2 ) = (a - b)2 + 4c2 , the two roots can be written as: 1 =
1 2
a+b+
(a - b)2 + 4c2
and 2 =
1 2
a+b-
(a - b)2 + 4c2
(41)
where by convention we take 1 2 . One can diagonalize A with a real orthogonal matrix S, S -1 AS = where S= 1 0 0 2 ,
cos sin
- sin cos
.
The columns of S correspond to the normalized eigenvectors of A, with the first and second column vectors correspond to the eigenvalues 1 and 2 , respectively. We now show how to determine in terms of a, b and c. The most straightforward approach is to compute S -1 AS explicitly. Since the off-diagonal terms must vanish, one obtains a constraint on the angle . S -1 AS = = cos - sin cos - sin sin cos sin cos a c c b cos sin - sin cos -a sin + c cos -c sin + b cos (b - a) cos sin + c(cos2 - sin2 ) a sin2 - 2c cos sin + b cos2 (42)
a cos + c sin c cos + b sin
=
a cos2 + 2c cos sin + b sin2 (b - a) cos sin + c(cos2 - sin2 ) 1 0 0 2 .
=
25
The vanishing of the off-diagonal elements of S -1 AS implies that: (b - a) cos sin + c(cos2 - sin2 ) = 0 . Using sin 2 = 2 sin cos and cos 2 = cos2 - sin2 , we can rewrite the above equation as 1 (b - a) sin 2 + c cos 2 = 0 . 2 It follows that: tan 2 = 2c a-b (43)
after writing tan 2 = sin 2/ cos 2. Let us now consider the range of the angle . You might think that 0 < 2. However, since cos( + ) = - cos , and sin( + ) = - sin ,
it follows that shifting + simply multiplies S by an overall factor of -1. Thus, S -1 AS is unchanged. Hence, without loss of generality, we may assume that 0 < . Unfortunately, eq. (43) does not determine whether 0 /2 or /2 < , since tan 2 = tan(2 + ) is unchanged if + /2. However, we have not yet used all the available information. In particular, the diagonal elements of eq. (42) also provide some information on the possible values of . Summing the diagonal terms of the matrices in eq. (42) yields: 1 + 2 = (a cos2 + 2c cos sin + b sin2 ) + (a sin2 - 2c cos sin + b cos2 ) = (a + b)(cos2 + sin2 ) = a + b , which is independent of . This is not surprising since we know that Tr A = 1 + 2 = a + b . However, 1 - 2 does depend on : 1 - 2 = (a cos2 + 2c cos sin + b sin2 ) - (a sin2 - 2c cos sin + b cos2 ) = (a - b)(cos2 - sin2 ) + 4c sin cos = (a - b) cos 2 + 2c sin 2 . From eqs. (41) and (44), we obtain 1 - 2 = Using eq. (43) to write: a-b= 2c cos 2 2c = , tan 2 sin 2 26 (a - b)2 + 4c2 = (a - b) cos 2 + 2c sin 2 . (45) (44)
and inserting this on the left hand side of eq. (45), the latter reduces to: (a - b) cos 2 + 2c sin 2 = 2c cos2 2 2c 2c cos2 2 + sin2 2 = + 2c sin 2 = . sin 2 sin 2 sin 2
Substituting this result back into eq. (45) and solving for sin 2, we find: sin 2 = 2c (a - b)2 + 4c2 (46)
We can also obtain cos 2 using eqs. (43) and (46): cos 2 = a-b (a - b)2 + 4c2 (47)
1 Eq. (46) tells us in which quadrant lives. If 0 < < 2 , then sin 2 > 0, which implies that c > 0. If 1 < < , then sin 2 < 0, which implies that c < 0. Thus, 2 the sign of c determines the quadrant of . Eq. (47) provides additional information. 1 For c > 0, the sign of a - b determines whether 0 < < 4 or 1 < < 1 . The 4 2 former corresponds to a - b > 0 while the latter corresponds to a - b < 0. Likewise, if 1 c < 0, the sign of a - b determines whether 2 < < 3 or 3 < < . The former 4 4 corresponds to a - b < 0 while the latter corresponds to a - b > 0. The borderline cases are likewise determined:
c = 0 and a > b c = 0 and a < b a = b and c > 0 a = b and c < 0
= = = =
= 0, = 1 , 2 = 1 , 4 = 3 . 4
If c = 0 and a = b, then A = I and it follows that S -1 AS = S -1 S = I, which is satisfied for any invertible matrix S. Consequently, in this limit is undefined. Finally, we can determine sin and cos from
1 sin2 = 2 (1 - cos 2) , 1 cos2 = 2 (1 + cos 2) .
By convention, sin 0. The sign of cos is determined by the sign of c as noted above. Taking the square root, we obtain: sin = 1 - cos 2 , 2 cos = sgn(c) 1 + cos 2 , 2
1 assuming that c = 0. If c = 0, then = 0 if a > b and = 2 if a < b as noted above.
27