# Infinite square well transitions

jbay9009

## Homework Statement

A particle, mass m propagates freely in a box, length L. The energy states are:

ϕ_n(x) = (2/L)^(1/2)sin(n∏x/L)

and energies E_n = n^2∏^2/(2mL^2)

at time t=0 the system is in state ϕ_1 and the perturbation V=kx is applied (k constant) and turned off at t=T.

1) Calculate the transition amplitude to the state ϕ_2 to 1st order in perturbation theory
2)Calculate the transition probability, p_21.
3)Calculate the maximum value of p_21.

## Homework Equations

Transition amplitude, T_fi = -i ∫ ϕ*_f V ϕ_i d^4x
= -i ∫(-T/2 to T/2)∫(-T/2 to T/2)∫ [ϕ_f exp(-iE_f t)]* V [ϕ_i exp(-iE_i t)] dx dt

where V = V_fi = ∫

## The Attempt at a Solution

1) T_12 = -i ∫ ϕ*_f V ϕ_i d^4x
= -i ∫(-T/2 to T/2)∫ [ϕ_2 exp(-iE_2 t)]* V(x) [ϕ_1 exp(-iE_1 t)] dx dt
= -i ∫ (2/L) sin(2∏x/L) kx sin(∏x/L) dx ∫(-T/2 to T/2) [ϕ_i expi(E_f-E_i)t] dt
using E_f-E_i = E_2-E_1 = 3∏^2/(2mL^2)
= 16ikL/(9∏^2) [exp(i 3∏^2/(2mL^2))-exp(-i 3∏^2/(2mL^2))] (2mL^2)/(3∏^2)) |(-T/2 to T/2)
=64ikmL^3/(27x^4) sin (T 3∏^2/(2mL^2) )

and so the transition probability, p= |T_21|^2 = |64ikmL^3/(27x^4)|^2 sin^2 (T 3∏^2/(2mL^2) )

and so p= |T_21|^2 has maximum values when T = integer 2kmL^2/(3∏)

Does this seem like the correct way of doing this problem / a reasonable result? It feels like I have done it the wrong way but I can't think of how else to do it.

Thanks for any help

## Answers and Replies

Homework Helper
Gold Member
It appears to me that your method is correct. You decided to integrate over time between -T/2 and T/2 rather than 0 to T. I think that's OK since your potential function V is time independent, but it seems more natural to me to integrate from 0 to T since the perturbation is switched on at t = 0 and switched off at t = T.

and so the transition probability, p= |T_21|^2 = |64ikmL^3/(27x^4)|^2 sin^2 (T 3∏^2/(2mL^2) )

Note that the probability cannot depend on the variable x, so this expression is incorrect (maybe just a typographical error). Also, if I'm not mistaken, I believe your argument of the sin function is off by a factor of 2.

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