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Unformatted text preview: HIDTEZM SoLqTiou ‘4 ET We start with an ndimensional Hilbert space spanned by vectors M + for
2' m 0,    ,n and I take these to be orthonormal as (A+i)\+j) :61} (if I hadn’t started with such a basis I always could have constructed one
using the GramSchmidt procedure). We know that these are eigenstates of the A operator as
AlA—l—i) = (A+i)lA+i) (1) Now we deﬁne some operator B so that
lRM=aB m where 0 5 a S n is integer.
Let’s ﬁgure out how B acts on the basis states. We’ll call the state that
you get when you act on a basis state with B by the name Mn) as BlA ‘l' ’3 W1) and then try to ﬁgure out what lwi) is in terms of basis states. To do this
take [B, A] M + = aBlA + All/1i) (A + i _ a) la) Comparing to (1) it’s clear that BI/\+z') = W2.) is proportional‘ to /\+ i ~a)
so we write BIAH) =c,/\+z'—a) (3)
Z
for some constants ci. (We don’t have enough information ot compute Ci.) Equation (3) tells us that B acts as a lowering operator.
What about the Hamiltonian? We’re told that Hza—m 1A lot of people concluded that BIA +1) 2: [A +i  a), which is overly restrictive. There
is no reason, in general, for B to return a properly normalized state. In particular, the
action of some well known lowering operators like a and L does not return a normalized
state. l'Notz vomi K mmnakbcmrvibwu, aWMCé will YW Lon/AW 1% in WC» git/ow HUB“? grace. M6 “0%
on; ma Nob Vimafdfz: 4 an1l m acédw. ‘A+a‘——d> with Q = (3". Recalling that H = Hl we can write B+B"
2 then we can compute H in this basis since we know how B acts on the states.
The matrix elements are H: Hi3» = (A+iH)\+j) Clearly
EH 2 (A +iHA+j>
= cj(A+iA+j—a)
= Cj5i,j—a
and (Ble = B}; so 1 air
Hij = ‘ lCi+a5iyj~a + Ciéi‘aijl
2 All this is for a 3i 0. When a = 0 then A, B commute so the basis states
are simultaneous eigenvectors of both operators. Then the Hamiltonian will
lie diagonal in this basis. :: X
— N’ (YH’ .x a #3!» M haw,
.. A” < N
" — . {a A (o rad __
.. . I. vac .‘ r ,. g“rv —:u Io ' _
'nL on .. V0444“ ma’ ‘
. a ' .. u‘ who , g A . ,
_ '«H um M
. 3 ~ ((PA £93 A=A+
A' a x r v
A 1 co A w s A «N
«v A, r. (a ‘V «WV
a .. . J 4 '
/\—— ., sa 5
“ma /\ = owe é‘Dro ‘vhw ’a
N v.. n .z ; A '. ‘ u r. A ‘ aw. _
____ﬂ ‘ ‘ v ' Wu.
‘ d¢maa+ H3 w ' “A d. _.
W. LWLLJ v “6% w“ :30 ; 004432 1’ ﬁlm =A. ..
_______ $35.! Aﬁng ngso _, <
[MM—m; éﬁﬁléﬁl‘VgZ "‘ (KIIQALQ :_ @’412 4‘4‘6 H’s) :0 (.1515 : ob<~b\ A www*._w_.___w . _ IL: A I.
Co a
* w
:n z: _, l I an; ~ + +J. ' w ,ﬂmwQ.
__,.,_, ft... : ,_
t+.L +1.4» — u; .._I,. ..
' o 77 m. Tr 31L
:— e o
i¢~rﬁ . ‘
:3 A 9!” r
g WQJP
I
w 46‘ \—~ och N .  3: ,, mﬂﬂ MW
M 2‘. m.
m 3 3 r 5‘ .
‘ : ___, 1" r . v
 “£15,” ,“SILW4‘ x 4 ,957 A“ __r._rL.s1ﬂ§ ,m“ ##M We can write the hamﬂtonian and perturbation in matrix representation. E0 0 0 H92 0 Eu 0 0 0 Eu 0 —a D W: a 0 ——a 0 —a 0 Thetotalhamﬂtonimiathen a) a O
H=Ho+Wn 6 En —a
0 —a 8‘9 Note that the eigenvalues of Ho are all Bo and the eigenvecton are: (1)
MP 0
O
'0
Wm
0
0
Mo)
1 Initially, the system is in the state IMO» :2 Ml). Now, we can compute the
eigenvalues and eigenvectora of H. 3133‘) 1 1
Wk) " 7; 01 Engo+af§ >(éi) Eaaﬁ'oax/i M» We can write the initial state in terms of the new eigenvectora.
W0» = —‘— w  1w + 11%)
ﬂ 2 2
Then, we can apply the evolution operator. Mt» = Um W0» = w — Na) + 3%?“ w» ‘ z  If .. 33' ~ ._ “i
. . . ‘ “W
U i K. 4 ' . 5!
Mb U I h
 * v :— AI<W v
* v > t . w  an
‘._. ‘ ‘ u
“ ~ wt ,_ c ' = M '. .... n
e o ‘ 4" I ’ 41. ’4 1 1 1 "‘l___L___—__._
. 3 A ____.____.—_..——————
’ A u; &  A ‘1  H’Al1.44 I . ‘ '..
‘
‘ o ,
,________.__._——
I  . .  m ‘5.
H u
1
H
m (Mun) = 04.] 4— z 6" (ml (MM) + 1 63A: m;>é_ n
=' A mu + 1 IGII‘A.‘ OHMz , I!) Mr”. «me MM ‘1'
a: .. (“as ,
M1~
w 4::  4x =§(A+a+x[~33)4x
Ho) ‘ “ ______
z
A. $9.32 Ho): (4+a)a+ 5‘ f, , .
I
. a w_
‘ F60 ‘ .; X z A : ~—— —‘~
9(4) . ’
(“53x + a; ram] _' M
9 F = 0
NM :‘1 a“‘ JV xal we haw F ze‘eg * " ‘ﬂT—M
Wham—L131 Note that for any observables, we have: am) i dt 3 ([H. Al) + at a (A) ~— If A is intrinsically time independent mean: 9&9 = 0. Now, remember that if [q.oz]ica.wohm H )I
ACIAOJZ—T According to our ﬁrst equation, we have: Thus, we get: _mi‘ﬁ = d, (H. Al , “,,,_w_ww..~y.m—mmm ...
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 Winter '10
 TamiPeregBarnea
 mechanics

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