Advanced Quantum Mechanics

(3)

The energy eigen-states of such a system are stationary: the time enters only in the phases according to

-(4)

where a,'s are time-dependent constants and (r) is time-independent. v (1) = Σa, (0) (1) This equation repre-
sents solution of (3), therefore substituting value of y and II from (1) and (4) in (3). we get

[ 9 (0) 0 (r) e¹E/1 = (6+1) \E « (r)</ Zita (0) (0) (+) ZaEnn (1) E

or

where

an (t) = an

Using (2) i.c. Hon = En On we get

Zan En 9n (1) Ex +Σ an H 9n (1) e¯¡E Multiplying both sides by Q and integrating over configuration space,
we get

...(5)

(7)

(8)

But

or

Eibanon (r)e-¡E = Zalf (t) e-iE)/

Now using orthonormality condition of o's, i.c.

0 for n k

= 1 for n = k

(E) x Hdt.

we get

Because in L.H.S. all terms will be zero except kth term due to the properties of Keonecker delta 8

we have

....(6)

The integral 9" HQ, dt at right hand side is a matrix

<k\H\n>=H

(Ex-E) H

So

Ex-E

= Wkn is the Bohr's angular frequency.

Time dependent constants a,'s are given by

a = (i) = 4, Hi

= (in)¹ a<k\H\n>

If we replace in equation (1) H by H where 2. is the parameter, then coefficient a's can be expressed
in parameter 2 as follows: (0)

Substituting the value of a, in equation (9) we get

(10)

(0) 1 (0)

Comparing coefficients of different powers of 2 on both sides.

(a)

(b)

(0) Equating coefficient of 2. in a 0 Equating coefficient of in a = Ea Hane

In general

(e)

where

S = 0.1.2,...

(11)

So we can get desired order in the perturbation. Zeroth order calculation: From (11), we have

ina)=0 or a = 0 i.e.

Integrating, we obtain

constant in time.

For convenience without loss of generality, we may put

ak according as the initial state m is one of a discrete or continuous set.

Acoordingly

(0)
=1 for k=mie. a=1

(0) = 0 for kam

and

Thus in the sum we have only one term and equation (9) may be expressed as

a = (i)" ' <k \ H' \ m>{"

-(12)

1st order perturbation. Integration of (12), gives

(13)

The constant of integration is taken to be zero in onder that a" be zero at perburbation is applied).
(before the perturbation is applied).

(a) Perturbation constant in time 1 Let us consider a perturbation that is constant in time and that it

operates only during the time o to t, Le

for--< < 0

H₁ =<k Urm> <kilim> for 0 st si

Substituting (14) in (13), we get

-(14)

(1) « (0) = (n) [ <k \r \ m> dorm d

for > = (ih) <k\\IF \ m>

= (it) <k \H\\m> <k Hm>

0m h 1) Thus, to first order, the probability, of the system from with state kth state is given by

IN

(-1) 12

(15)

-(16)

Using the relation

lim
→0 (e" - 1) = 2ie sin = 2i sin (e" - 1)² = 4 sin

or

equation (16) takes the form

sin 20km

-(17)

as a

Physical Interpretation: In order to interpret eqution (17) physically, we plot sin (0/2)

20

function of 0 and find the curve as shown is fig. 9.1 The major maxima of probability curve occurs
at t=0 i.e. for E = E If we substitute =

sin 2

we note that

sin 201 2 2

if higher powers of are neglected due to their smaller values

-8 -6x-4xt

-2x/t O 2nt

sin (W2)

Fig. 9.1.

Thus the height of the main peak is proportional /4 and the intensity of probability curve touches the

00m axis at points where sin (xt/2)

This occurs at points where

= 0.

4t

where r is an integer (i.e. r = 0, 1,3,...)

Thus x=00km = ±

2πr

the area under the curve is proportional to t i.e. time of application of the perturbation. These results
derived under constant perturbation for a definite time are analogous to single slit diffraction experi-
ment and are applicable to problems of excitation and emission of radiation in elementary systems.

Now we seek the physical interpretation from the figure: For sufficiently large r, the area under the
central peak is much greater than that under the neighbouring peaks and hence most transitions take
place to states under the main peak.

If we have a system with the Hamiltonian Ho and if this system interacts with an external agency
with an interaction Hamiltonian H, then there are transitions in the system and it is observed that
there is a definite transition rate from a fixed initial state to a final state. Transition Probability:
Fermi - Golden Rule: Let us suppose that unperturbed Hamiltonian Ho has

continuous spectrum and the transitions are taking place to the state in the continuum. This is the
situation in most of physical problems of e.g. in scattering problem the scattered states belong to the
continuum eigen values of the Hamiltonian.

If the transition takes place to state & of energy between E and E+ dEs and the energy density of
states is given by p (4) at this stage, then the transition probability per unit time is defined as =\" (0)
1 p (k) dEx

where p (4) gives the number of final states in the energy interval from Es to Et des If is large
enough, the central peak in the probability curve is sharp and then the quantities <k Im> and p (k)
may be regarded as sufficiently independent of E, so that they may be taken out the integral, there-
fore

(0 =

41<k Him>

<k\H\m>p (k)

sin (0/2) p (k) dEx

4 sin (0/2) h do

[Since dE E-Edul

00= <k H° m > i² p (k) [! sin (0/2

we have the standard integral

Jinx dx dx = R

Substituting

x, i.e. Opm sin (Com 1/2 d OL= so dam sin x dx

sin x dx = 2n

(20)

2 p(k) | <k \H\ m>²

Substituting this in (19), we get

| <k !' \ m > |² p (k), 2 =

(21)

This in an important result of time dependent perturbation theory and is called the Fermi Golden
Rule. This relation has been successsfully employed to calculate the transition probabilities between
two states and their corresponding life times. In particular in B-decay it has given the results actu-
ally observed experimentally There may be several different groups of final states ki, ky, ky,... all of
which possess nearly the same

energy E = E + Mo for which the perturbation matrix elements <k, 11 m> and density of states

P(A) although nearly constant within each group, differ from one group to another. Then the transi-
tion

probability per unit time for th group (replacing k) by ky in (21) is given by

13 2x p(k) <k; \r\m>rij - 1.2....

(22)

The spread of energy of final state to which transitions occur is connected with uncertainty relation

AE AI in the following manner: We can regard the perturbation if as a device that measures the fi-
nal energy of the system transferring it to one of the state &. The time available for the measure-
ment in r so that the uncertainty in energy perdicted by the relation is AE h/t and this is in agree-
ment with the width of the main peak. Thus there is no need to insert separate assumption for it. (b)
Harmonic Perturbation Let us now consider a different but physically important case when the per-
turbation is harmonic of frequency to, i.e.

10 for --< < 0

(23)

where <k Holm> is independent of time. (1) The first order amplitude a" (f) at timer, would then be
given as (1)

a" (r ≥ 0) = (m)' [ <k \H\m>df

= (in) 2 <k Holm> sin car!", 2 <k 1Holm>

<k Holm> h

<k Holm>

(00+ (0)

i (6+00)

(60-00)

<k\Hol/m>

The form of this equation suggests that

(24)

(i) The first order amplitude depends on perturbation duration t and not on instantaneous time r (ii)
The amplitude is appreciable only when the denominator of the one or the other terms is practically

zero.

The first term is important when to or E-Em-lo and the second term is important when

to or EE+ hoo. Thus the first order effect of a harmonic perturbation (e. perturbation that varies si-
nusoidally with time) with angular frequency to is to transfer or to receive from the system on
which it acts the Planck's energy quantum hoo.

In the special case in which the initial state m is a discrete bound state and the final state & is one of
the continuous set of dissociated states. Then E> Em and only the second term in (24) need be con-
sidered. In this case the first order probability of finding the system in k th state after the perturba-
tion is removed is given by

Fig.9.2

4 | <k Holm> 1 sin² (00-00)/ 2

(c) Second Order Perturbation: From equation (11), we have

= ΣHin an!
Puttings = 1, we have

(1) W! - 1) 4" (1) = Him (e

where

Setting k = n, we get

Hun (e 1100m 1) Σ

(2)

Therefore

Integrating this equation subject to initial condition a (0) = 0 (ie. at time = 0) gives for the

second order amplitude at time r

a² (1) = 1²² 2 HH

(26)

This equation shows that transition for which probability increases linearly with the time can be ob-
tained by putting either

W0 or = 0.

Suppose that the perturbation gives no transition in first order. This means that there is no state n in
the first order that conserves energy (i.e. m 0). For such state matrix element H0. Since 0 this means
also that

H O whenever 0 0.

so second term in equation (26) is never appreciable. Therefore second order may be calculated by
ag replaceing by second order matrix element.

HH Σ

E E Effect on first order transitions: The second order amplitude at time ris

H H E-E

It is still correct that the second term in the bracket of equation (15) is negligible for states n that
have energies appreciably different from E (or E), since w is large

E-E

...(27)
The energy states E. Em are all close together and neither Him nor H

is zero.

395

Ex-Em is small

Ex-En

is also small.

Since co is small, so second bracket term of (26) cannot be neglected without the summation. inte-
gration over n would have a singularity, when com is zero.

Ex-Em

0000km 00= It is easy to see that for any value of 0 (zero or otherwise). The entire bracket is pro-
portional to 6 when is small: this removes the cm in the denominator and makes the summand or in-
tegrand finite

Ex-En

E-E

where com = 0. We have to evaluate equation (27) if Σ can be replaced by an integral over E,, or co,
We divide the integral into parts according as 100 is large or not with respect to 1/1. In the first re-
gion the second

bracket term with equation (27) can be neglected since wo|=|- is also large as compared to

Thus we obtained this part of integral,

jon

-S

Hm Hm

p (n) h dom

Since p (n) dE, is the number of states of the particular group it under energy range dE,, the prime
over integration excludes the integration in the region,

(28)

where c is constant number that is large in comparison with unity.

In second region where am c/t, we assume that t is large enough so that H HP (n) can be treated as a
constant, taken outside and solved at co= 0. We must now take both terms in the bracket of equation
(27), in order that
the integrand be finite. This part of the integral is then

Hi HP()

i (0-0)

<-1 =0J e/t -1

don

...(29)

This equation (29) can be solved by considering the contour in the complex w plane shown in Fig.
93 which contains no pole of the integrand. Thus the integrand over closed contour is zero and the
integral in (28) is equal to integral around the semicircle of radius c/t taken in the counter clockwise
direction. The magnitude of to is great enough over this semicircle. Therefore contribution of sec-
ond term can be ignored in comparison with first. The integrand in

(29) is then easily solved and becomes

...(30)

For large time t, prime on the integral (28) is equivalent to taking its principal value. Therefore if
we substitute (30) in (29) and add the result to (28), we get an expression like (28) except that
primed

W PLANE +c/t REAL AXIS

IMAGINARY AXIS

Fig. 9.3. Closed Contour

En Em =

Equation (31) is to be used in place of (27) whenever summation can be represented by fp (n) dE