Infinite square well transitions

In summary, the conversation discusses a particle with mass m propagating freely in a box of length L and the corresponding energy states. The perturbation V=kx is applied at time t=0 and turned off at t=T. The summary then outlines the steps to calculate the transition amplitude to the state ϕ_2 and the transition probability, p_21, which has a maximum value when T = integer 2kmL^2/(3∏).
  • #1
jbay9009
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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
 
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  • #2
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.

jbay9009 said:
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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Related to Infinite square well transitions

What is an infinite square well?

An infinite square well is a theoretical model in quantum mechanics that represents a particle confined to a potential well with infinite potential walls on either side. This allows for the study of quantum phenomena such as energy levels and transitions.

What are "transitions" in the context of an infinite square well?

Transitions refer to the movement of a particle from one energy level to another within the infinite square well. This occurs when the particle absorbs or emits energy in the form of a photon.

How are energy levels determined in an infinite square well?

The energy levels in an infinite square well are determined by the size and shape of the well. The larger the well, the more energy levels there are and the closer they are spaced together. The energy levels are also quantized, meaning they can only take on certain discrete values.

What is the significance of the "infinite" potential walls in an infinite square well?

The infinite potential walls represent a barrier that the particle cannot penetrate. This confinement results in the quantization of energy levels and the ability to study transitions between them.

Can an infinite square well be applied to real-world systems?

While an infinite square well is a useful model in quantum mechanics, it is not directly applicable to real-world systems. However, it can provide insight and understanding into quantum phenomena that can be observed in real-world systems.

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