Independent, Time Multi-Level Perturbation Theory Given solutions
H( 0 ) (n0 ) E (n0 ) (n0 )
(1) will be set to 1, the
Find the perturbed energy and wavefunctions. Noting that perturbed Hamiltonian and wavefunction are
H H( 0 ) H(1) 2H( 2 ) ...
n
(0) n
(1) n
2
(2) n
...
(2) (3)
which has energies
E n E (n0 ) E (n1) 2E (n2 ) ...
(4)
the Schroedinger equation is H(0 ) H(1) 2H( 2 ) ... (n0 ) (n1) 2 (n2 ) ... E
(0) n
E
(1) n
E ...
2 ( 2) n
(0) n
(1) n
2
( 2) n
...
(5)
multiplying-out, collecting terms in
0 H(0 ) (n0 ) E (n0 ) (n0 )
n
s s s
(6)
1 H( 0 ) (n1) H(1) (n0 ) E n (n1) E (n1) (n0 )
n
2 H( 0 ) (n2 ) H(1) (n1) H( 2 ) (n0 ) E (n0 ) (n2 ) E (n1) (n1) E (n2 ) (n0 ) ... 0
n
since
is arbitrary, each power of
is zero separately, giving (7) (8) (9)
H( 0 ) (n0 ) E n (n0 ) H( 0 ) (n1) H(1) (n0 ) E (n0 ) (n1) E (n1) (n0 ) H( 0 ) (n2 ) H(1) (n1) H( 2 ) (n0 ) E (n0 ) (n2 ) E (n1) (n1) E (n2 ) (n0 )
Write eq. (8) in bra-ket notation H(0 ) n(1) H(1) n(0 ) E(n0 ) n(1) E(n1) n(0 ) and multiply by n(0)
n(0 ) H(0 ) n(1) n(0 ) H(1) n(0 ) n(0 ) E (n0 ) n(1) n(0 ) E (n1) n(0 ) n(0 ) E (n0 ) n(1) n(0 ) H(1) n(0 ) n(0 ) E (n0 ) n(1) E (n1) n(0 ) | n(0 ) n(0 ) H(1) n(0 ) E(n1)
(10)
(11) (12) (13)
i.e. first-order correction to energy
to find first-order wavefunction multiply eq. (8) by m(0)
m(0 ) H(0 ) n(1)
for m n m(0 ) H(1) n(0 ) m(0 ) E (n0 ) n(1) m(0 ) E (n1) n(0 )
(14) (15) (16) (17)
(0 Em ) m(0 ) | n(1) m(0 ) H(1) n(0 ) E (n0 ) m(0 ) | n(1) E (n0 ) n,m
rearranging (0 m(0 ) H(1) n(0 ) E (n0 ) Em ) m(0 ) | n(1)
d
i
m( 0 ) H(1) n( 0 ) E d
(0) n
E
(0) m
i
(1 m(0 ) | n(1) c m)
so the first-order correction to the wavefunction is (1 n(1) c m) m( 0 )
m n
(18)
and the first-order wavefunction is n(0 ) n(1)
(19)
2nd Order Energy Correction put eq. (8) in Bra-Ket form.
( ( ( H ( 0) n ( 2) H (1) n (1) H ( 2) n ( 0) En0) n ( 2) En1) n (1) En2) n ( 0)
(19a)
multiply by
n (0)
n ( 0) H ( 0) n ( 2) n ( 0) H (1) n (1) n ( 0) H ( 2) n ( 0) n
( 0)
(19b)
E
( n
n
( 0) 2)
n
( 0)
E
(1) n
n
(1)
n
(0)
E
( 2) n
n
(0)
( E n0) n ( 0) n ( 2) n ( 0) H (1) n (1) n ( 0) H ( 2) n ( 0) ( ( ( E n0) n ( 0) n ( 2) E n1) n ( 0) n (1) E n2) n ( 0) n ( 0)
(19c)
cancelling first term in each, gives:
n( 0) n(1) n( 0)
m n
c
(1) m
(1 m( 0) cm ) n( 0) m( 0) m n
, and
n ( 0) n ( 0) 1
( ( n ( 0) H (1) n (1) n ( 0) H ( 2) n ( 0) En1) n ( 0) n (1) En2)
( ( E n2 ) n ( 0 ) H ( 2 ) n ( 0 ) n ( 0 ) H (1) n (1) n ( 0 ) H ( 2 ) n ( 0 ) n ( 0 ) H (1) c m1) m ( 0 ) mn
(19d)
solving for E(2)
n
(0)
H
( 2)
n
(0)
c
mn
(1) m
n
(0)
H
(1)
m
(0)
n
(0)
H
( 2)
n
(0)
m ( 0 ) H (1) n ( 0 ) n ( 0 ) H (1) m ( 0 )
( ( E n0) E m0 )
mn
n
(0)
H
( 2)
n
(0)
n ( 0) H (1) m ( 0)
( ( E n0) E m0)
2
(19e)
m n
Perturbation Theory and the Morse Oscillator
(2i 2)( ax)i V ( x) D e 1 e De i! i1 7 2 4 V ( x) Dea2 x 2 ax3 a x 12
ax 2
(20) (21)
L M N
O P Q
considered the truncated expansion of the Morse potential as a perturbed harmonic oscillator system
p2 7 2 4 H D ea 2 x 2 ax3 a x 2m 12
p2 H D ea 2 x 2 2m H(1) D ea3 x 3 7 H( 2 ) D ea 4 x 4 12
(0)
L M N
O P Q
(23) (24) (25)
(22)
where
2D ea 2 m
(26)
Mass not quantum no.
For Harmonic Oscillato...