Quantum Harmonic oscillator problem

[humanist rho]

Homework Statement

A particle of mass m is placed in the ground state of a one-dimensional harmonic
oscillator potential of the form

V(x)=1/2 kx²

where the stiffness constant k can be varied externally. The ground state wavefunction
has the form ψ(x)$\propto$ exp(−ax²$\sqrt{k}$) where a is a constant. If, suddenly, the parameter k is changed to 4k, the probability that the particle will remain in the ground state of the new potential is;

(a) 0.47 (b) 0.06 (c) 0.53 (d) 0.67 (e) 0.33 (f) 0.94

2. The attempt at a solution

The system is in the ground state before changing k

ie, $\int$$\Psi$*$\Psi$dx = ($\pi/2a\sqrt{k}$)^1/2 =1
When the parameter is changed;let the wave function be $\Psi'$
the probability to be in ground state is;

$\int$$\Psi'*$$\Psi'$dx = ($\pi/4a\sqrt{k}$)^1/2 = $\frac{1}{\sqrt{2}}$$\times$($\pi/2a\sqrt{k}$)^1/2 =$\frac{1}{\sqrt{2}}$$\times$1=0.707

But this is not there in the option.
Could anybody pls check the steps and tell me where's the mistake or correct it?

 

[vela]

The wave function must evolve continuously, so the state of the system right after the change is the state of the system right before the change.

Also, your expression for the probability of finding the system in the new ground state is incorrect. The integral you have, which is the integral of the modulus squared of a wave function, should equal 1. You need to come up with the correct integral.

 

[humanist rho]

Thanks a lot for the reply.

If I take $\Psi(x)$=$\Psi'(x)$ at some x=x₀ just after the change;

Exp(-ax₀²$\sqrt{k}$)=Exp(-2ax₀²$\sqrt{k}$)

ie, -ax₀²$\sqrt{k}$=0 or x₀ = 0

What do this mean?
And how do this help me in solving the problem?
Is that mean that I should integrate the second integral from 0 to infinity ?

 

[vela]

I have no clue what you are doing.

 

[humanist rho]

:(

I'll try to explain what i have done.

In the question, the wavefunction is given as ;

$\Psi(x)$ $\propto$ Exp (-ax²$\sqrt{k}$)

On normalizing it I get ;

A²$\int$$\Psi$*$\Psi$dx = 1

A²$\int$ Exp (-2ax²$\sqrt{k}$)dx = 1

A² ($\pi/2a\sqrt{k}$)^1/2 = 1

A = (2a$\sqrt{k}$/$\pi$)^1/4

ie , the harmonic oscillator wavefunction before changing the value of k is ;

$\Psi(x)$ = (2a$\sqrt{k}$/$\pi$)^1/4 Exp (-ax²$\sqrt{k}$)

Now the parameter k is changed to 4k ;

Then the new wavefunction be $\Psi'(x)$.

I think,$\Psi'(x)$ can be obtained by replacing k by 4k in $\Psi(x)$ .

If so;

$\Psi'(x)$ = (4a$\sqrt{k}$/$\pi$)^1/4 Exp (-2ax²$\sqrt{k}$)

As you said wavefunction should be continuous. ie,$\Psi'(x)$ should be same as $\Psi(x)$ just after the change.

Then can i take $\Psi(x)$ = $\Psi'(x)$ ?

On taking so, i get x = 0 .

I just donno what this mean?
I have to find the probability of the system to be in the new ground state. How can i obtain this?What is the problem with my procedure? Isn't it the correct method?

 

[vela]

If the state doesn't change, the wave function doesn't change.

 

[humanist rho]

If the state doesn't change, the wave function doesn't change.

Could you please clarify?
Is that mean that the state of the system and probability remain same even after changing the parameters like k ?

 

[vela]

Remember the wave function contains all the information about the state of the particle. If the wave function changes, that implies the state changed, and vice versa. Immediately after the potential changes, the wave function of the particle is still$$\psi(x) = \left( \frac{2a\sqrt{k}}{\pi}\right)^{1/4} e^{-ax^2\sqrt{k}}$$What has changed, however, is the ground state of the system.

In other words, with the original potential, the particle was in a state that was also the ground state of the system. With the new potential, the particle is still in what was the ground state of the system, but the ground state of the new system is different.

 

[humanist rho]

Then how to find new ground state?

 

[humanist rho]

I didn't get the answer yet.Please help.

 

[vela]

At t=0^-, you have
$$\hat H_- = -\frac{\hbar^2}{2m}\frac{d}{dx} + \frac{1}{2}kx^2$$which has eigenstates $\phi_0(x)$, $\phi_1(x)$, ... where
$$\phi_0(x) = \left( \frac{2a\sqrt{k}}{\pi}\right)^{1/4} e^{-ax^2\sqrt{k}}$$is the ground state. Note that we haven't said anything about the state of the system. We're just talking about the eigenstates of the system.

At t=0⁺, the potential changes so that
$$\hat H_+ = -\frac{\hbar^2}{2m}\frac{d}{dx} + \frac{1}{2}(4k)x^2$$which has eigenstates $\chi_0(x)$, $\chi_1(x)$, ... You should be able to tell us what the ground state $\chi_0(x)$ is. Again, note that we have said absolutely nothing about what the actual state of the system is.

You're told that prior to the change, the system is in the ground state. That means $\psi(x) = \phi_0(x)$ at t=0^-. At t=0⁺, you must also have $\psi(x) = \phi_0(x)$ because the wave function evolves continuously.

Now what expression gives the probability amplitude of finding the system in the state $\chi_0(x)$ when it's in the state $\psi(x)$?