# Time Dependent Perturbations

Consider two particles in the following (identical) time dependent potentials: the infinite square well for $t<0$, a constant perturbed system for $0<t<T$, and a return to the infinite square well for $t>T$. Recall that the energies of the infinite square well are given by $n^2 E_1$ ($n=1,2,3,…$).

Particle 1 is initially in the state with $n=2$, particle 2 is initially in the state with n=4.

If the energy of both particles is measured at some time $t>T$, which particle is more likely to be found with the ground state energy? Explain.

Consider each particle initial wave function (particle 1 on $E2$ and particle 2 on $E4$)

$ψ_1 (x,t<0)=∑_(n=1)^∞ a_n ϕ_n (x)*e_n^{-(i E_n)/ℏt}$ with all $a_n=0$ but $a_2=1$

$ψ_2 (x,t<0)=∑_(n=1)^∞ b_n ϕ_n (x)*e_n^{-(i E_n)/ℏt}$ with all $b_n=0$ but $b_4=1$

$ϕ_n (x)=√(2/L)*\sin (nπ/L*x)$ the wave function of n-th energy level (L width of the well)

Applying the time-dependent perturbation theory one finds for energy level 1 that

$a_1 (t)=-i/ℏ ∫_0^t H_{21}’ exp(iω_{21} t)d t=-(-i/ℏ) H_{21}’ [1/(iω_{21})] *exp(iω_{21} t)$

with $ω_{21}=ω_2-ω_1$

and $b_1=-i/ℏ ∫_0^t H_{41}’ exp(-iω_{41} t)d t=⋯$..

The probability for particle 1 to undergo a transition $2->1$ (and be found on energy level E1) is

$P_{21} (t)=|a_1 (t) |^2=(|H_21′ |^2/ℏ^2) \frac{sin^2 [(ω_2-ω_1 )t/2]}{(ω_2-ω_1 )^2}$

$P_41 (t)=|b_1 (t) |^2=⋯$..

This is similar to equation 9.28 from the book (Griffiths) only that now the perturbation is constant in time (until t=T when is switched off).

To say exactly which particle is most probable to be found on energy level 1 one needs to compute also

$H_{m n}’=<ϕ_m |H’ | ϕ_n>$

But we can observe that all $H_{m n}’$ are about the same (like for example $π,π/2,π/4,…$) and what really counts in the probability of transition is just

$sin^2 [(ω_2-ω_1 )t/2]/(ω_2-ω_1 )^2 ~1/(ω_2-ω_1 )^2$

So that

$P_{21}~ℏ^2/(E_2-E_1 )^2 =ℏ^2/((2^2-1^2 )^2 E_1^2)=1/9*ℏ^2/(E_1^2)$

and $P_{41}~ℏ^2/((4^2-1^2 )^2 E_1^2 )=1/225*ℏ^2/(E_1^2 )$

Now the answer is clear: after $t>T$ (in fact at all times) it is more probable to find particle initially on $E2$, finally on $E1$.

Which particle is more likely to be found with the second excited-state energy (E_3)? Explain.

For the energy level E3 the probabilities of transition are

$P_{23}~ℏ^2/((4-9)^2 E_1^2 )=1/25*ℏ^2/(E_1^2 )$

and

$P_43~ℏ^2/((16-9)^2 E_1^2 )=1/49*ℏ^2/(E_1^2 )$

So again it is more probable to find particle initially on E2, finally on E3 (than particle initially on E4 finally on E3).

Suppose that you could choose the length of time, T, for which the perturbation is active. Is there a value of T that maximizes the probability that both particles are found with the ground state energy ($E_1$). If so determine at least one such value. Explain.

From what is discussed above the probabilities of transitions for the two particles (on levels E1, E2, E3 and E4) are

for particle 1 (initially on E2): $1/9,∞,1/25,1/144$

for particle 2 (initially on E4):$1/( 225),1/144,1/49,∞$

So the probability to find both particles on E1 is

$P_{tot_1}~1/9+1/225=0.1155$

And the probability to find both particles on E3 is

$P_{tot3}=1/25+1/49=0.06$

If one can chose the time T then the total probability is

$P_{21} (T)+P_{41} (T)\sim \sin^2[(4-1)*T/2*E_1/ℏ]/((9E_1)/ℏ)+$

$+\sin^2[(16-1)*T/2*E_1/ℏ]/((225E_1)/ℏ)$

Taking $E_1/ℏ=1$ and imposing the condition $(d P_{tot})/d T=0$ we have

$(2 *3/2 \sin (3T/2))/9+(2*15/2 \sin(15T/2))/225=0$

or

$1/3*\sin(3T/2)+1/15*\sin(15T/2)=0$

This is an equation from which one can find the time $T$.