# Sudden Approximation

You are given a particle that is in the ground state of the quantum mechanical infinite square well of width $a$. Suddenly you increase the size of the square well to $2a$ at $t_1$. Thus the wave function of the particle just after $t)1$ remains the same as it was before $t_1$. The new energy levels are $E_n’=n^2E_1’=n^2E_1/4$

a) Before time $t_1$ is the particle in an energy eigenstate? Which state? Please explain.

b) After time $t_1$ is the particle in an energy eigenstate? Which state? Please explain.

c) For $t_2>t_1$ with no changes, is the particle in an eigenstate? Which state? Please explain.

Discuss the possibility of measuring the energy of this particle again.Explain.

The texts says: the particle is initially in the ground state of the square well having width $a$. Therefore just before t1 the state is an eigenstate (bound state). The eigenvalue (the energy) of this state is E1. The wave function of the particle be written as (1):

$ψ(x,t)=A*cos⁡(k_1*x)*exp⁡(iω_1*t)$

where

$k_1=nπ/L=π/a$    and $ω_1=(2πE_1)/h=E_1/hbar$

The well width is increased to $2a$. At the same moment of time $t1$, just after $t1$, the all the wave function remains the same so also the part independent of time is the same. This part of the wave function that is independent of time is

$ψ(x)=A*cos⁡(k_1*x)=A*cos⁡(πx/a)=A*cos⁡(2πx/2a)=A*cos⁡(nπx/L)$

From above one can see the particle is on the energy level $n=2$, which is an energy eigenstate.

The time dependent Schrodinger equation is

$ihbar*dψ/dt=Hψ$

Since the wave function remains the same (1) we can write for a latter time $t_2>t_1$ (you make no changes to the system, so that the wave function of the particle is not modified):

$ihbar*iω_1*ψ(x,t)=Hψ(x,t)$

$E_1*ψ(x,t)=Hψ(x,t)$  that is   $4E1′ ψ=Hψ$  or equivalent $2^2*E1′ ψ(x,t)=Hψ(x,t)$

Therefore at $t2>t1$ the particle is in an energy eigenstate. It stays on the energy level 2, ground level of the second well.

Basically in quantum mechanics when one makes a measurement of a state, one implicitly modifies that state. As shown above the after time $t1$, the particle is found on the second energy level of the larger well, so that clearly the initial energy level is changed. But if the energy of state was not measured, the energy level could stay the same. I tend to say that for the new potential we have to use all energy eigenstates but to have all possible energies  in the well, this implies giving extra energy to the particle by measurement.

In my opinion there are possible after a measurement on the particle just energy levels 1 and 2. 2 is the initial state, 1 is the state after particle un-exciting. To determine the probability of energy states 1 and 2 you need to compute

$P=<ψ|H|ψ> =∫ ψHψ* dx=∫ ψE_n ψ^* dx$