perturbation theory
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Explanatory note by Phil dated 1.19.05, written after repeatedly being tripped up by this point in Jim's paper. It follows Schiff (pp. 245-246): expands eigenvectors and eigenvalues in a small parameter, derives the first-order eigenvalue shift (i, T' i) and the coefficients a_ik, and gives the unnormalized form. It then offers a variational derivation for real symmetric T and a check via orthogonality.
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How to do First Order Perturbation Theory PhL 1.19.05
After being tripped up 4 times on the same thing in Jim's paper on 4 different occasions, the time has now come to write down how this works. I will use Schiff page 245 as a basis.
Let's start with this problem, where the i are orthonormal (m, n) = m,n
T i = i i (1)
where i are eigenvalues of T and i are the eigenvectors. Now make this perturbation to T
T T + T' (2)
where is a smallness parameter. Assume the solutions now look like this
(T + T') i = i i (3)
Now do the following expansions
i = i + jaijj (4) // note that the sum includes some i
where we are now assuming that i i At the same time, let us assume that
i = i + Li (5) // how the eigenvalue changes
where again we are assuming that i i. Our equation (3) becomes
(T + T') [ i + j aijj ] = ( i + Li ) [ i + j aijj ] (6)
We can now group terms by powers of :
Ti = i i 0 (7)
T' i + j aij Tj = Li i + i j aijj 1 (8)
But Tj = j j in the second term, so we then have
T' i + j aij j j = Li i + i j aij j 1 i = 1,2,3... (9)
which we can rewrite as
T' i = Li i + i j (i - j) aij j (10)
Suppose we now close with (k , then we have [ not specifying Hilbert space here! ]
(k, T' i) = Li k,i + (i - k) aik (11)
Now separate this into the two cases:
(i, T' i) = Li k = i (11a)
(k, T' i) = (i - k) aik k i (11b)
The first result here is "the classic result" [ see Schiff page 246 eq 31.8 ] for the shift in an energy level, say. The second equation looks like Schiff (31.10).
Now one extra fact. Suppose we go from i to some other fi = kii which are not normalized. Then clearly we would replace our "classic result" with
Li = (i, T' i) = (1/ki)2 (fi, T' fi) = (fi, T' fi) / (fi, fi) (12)
We can summarize the "classic result" in this way. If we have Tfi = ifi then
i = (fi, T fi) / (fi, fi) // eigenvalue shift due to T perturbation (13)
Alternative Approach. Suppose we start with this eigenvalue equation where i need not be normalized, and where T is real symmetric,
T i = i i
Apply variation to get
(T) i +T(i) = (i)i + i (i) (13)
or
(i)i = (T) i + [ T - i 1] (i) (14)
Now close this in a scalar product with (i , to get
(i)(i , i) = (i , (T) i ) + (i , [ T - i 1] (i))
The last scalar product can be written as ( [ T - i 1]i , (i)) as long as T is Hermitian. But then we know that [ T - i 1]i = 0, so we find that (i , [ T - i 1] (i)) = 0 and we get our desired result
(i)(i , i) = (i , (T) i ) => (i) = (i , (T) i ) / (i , i) QED.
So we do NOT claim that (i) ~ 2 or anything like that! We need both the (i) terms and we only get these terms to "go away" when we do the scalar product.
We are done, but just for fun here is another way to get this same result:
From above we have
(i) = jaijj (15)
so that
[ T - i 1] (i) = [ T - i 1] jaijj = jaij [ T - i 1]j = j aij (j - i) j = ji aij (j - i) j (16)
Now watch what happens when we close with ( i,
( i, [ T - i 1] (i) ) = ( i, ji aij (j - i) j ) = ji aij (j - i) ( i,j ) = 0
In other words, the quantity [ T - i 1] (i) is orthogonal to i to this order of . Then when we do this same closure onto (14) we get
(i) = ( i, (T) i )
which agrees with the "classic result".