# Infinite square well transitions

## 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

TSny
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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