# &quot;Particle in a Box&quot; Measurement scenario

1. Apr 27, 2008

### IHateMayonnaise

[SOLVED] &quot;Particle in a Box&quot; Measurement scenario

Problem statement:

Measurement of the position of a particle in a 1D well with walls at x=0 and x=L finds the value at x=L/2. Show that in a subsequent measurement, the particle will be in any odd eigenstate with equal probability.

Relevant equations/attempt at solution:

Well, we know that if the energy is not sharp, then it is realized as an expectation value, so

$$<E>=\int_0^L\Psi^*(x)\hat{H}\Psi(x)dx$$

But, we know that any function, $$\Psi(x)$$ can be described as being a sum of "base states:"

$$\Psi(x)=\Sigma_{k=0}^\infty a_k\Psi_k(x)$$

where

$$<\Psi(x),\Psi_n(x)>=a_k \delta_{kn}=a_n$$

so, we can rearrange this to say that

$$a_n=<\Psi(x),\Psi_n(x)>$$

where even solutions take the form
$$\Psi(x)=\sqrt{\frac{2}{L}}Sin\left(\frac{(2n)\pi x}{L}\right)$$

and odd solutions

$$\Psi(x)=\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)$$

So, if we wanted to find the probability of detecting the particle in an odd eigenstate, we would calculate:

$$\left|a_{2n+1}\right|^2=<\Psi(x),\Psi_{2n+1}(x)>^2=\left<\sqrt{\frac{2}{L}}Sin\left(\frac{n\pi x}{L}\right),\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)\right >^2 =\left [\int_0^L \left(\sqrt{\frac{2}{L}}Sin\left(\frac{n\pi x}{L}\right)\right)^*\left(\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)\right)dx\right]^2$$

To my understanding, $$\left|a_n\right|^2$$ is the probability of finding a particle with a energy value of n, is this right? Or am I at least on the right track? Thanks all (as always...) :)

2. Apr 27, 2008

### G01

You are using the wrong initial wave function: $$\Psi (x)$$

Other than this, your work looks good.

Your particle is initially localized at x=L/2. What kind of function will represent this initial state?

3. Apr 27, 2008

### IHateMayonnaise

I don't quite understand what you are implying...the particle-in-a-box wavefunction takes the form of

$$\Psi(x)=\sqrt{\frac{2}{L}}Sin\left(\frac{n\pi x}{L}\right)$$

this way it satisfies the boundary conditions at x=0 and at x=L.

4. Apr 27, 2008

### G01

Yes, that is the correct wave function for the eigenstates of the particle. The particle is not in an eigenstate though, it is localized at L/2. This means we need a function where the probability of measurement for the particles position is 1 for x = L/2 and zero for every other point.

Thus, we need the area under x=L/2 to be 1 and 0 over every other point.

Do you know of any "function" that has the property of having and area of 1 inside of an infinitesimal silver above one point? (I put function in quotes because it isn't really a function in the rigorous sense. This should be a clue as to what "function" your looking for.)

5. Apr 27, 2008

### IHateMayonnaise

Why, yes of course! The so-called Kronecker's Delta!

Just so that I am sure that this is clicking correctly, let me summarize:

We can say that the wave function for the many possible eigenstates of a particle confined to an infinite potential well are given by:

$$\Psi(x)=\sqrt{\frac{2}{L}}Sin\left(\frac{n\pi x}{L}\right)$$

But, when a measurement is made, the wavefunction is no longer in the land of eigenstates (meaning, subsequent to a measurement the measured value is a definite number, i.e. it is localized), so the wavefunction collapses down to something along the lines of:

$$\Psi(x)=\delta(x-L/2)$$

So that if x=L/2 (which it is, since this is where the particle is localized), $$\delta(x-L/2)=\delta(0)=1$$.

Assuming that this is all correct, I will not attempt to solve the original prompt, which asks to show tha tin a subsequent measurement, the particle will be in any odd eigenstate with equal probability. Thus, I will compute essentially the same thing that I attempted before, but with our new delta-dependent wavefunction:

$$\left|a_{2n+1}\right|^2=<\Psi(x),\Psi_{2n+1}(x)>^2 =\left<\delta(x-L/2),\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)\right >^2 =\left [\int_0^L \delta(x-L/2)^*\left(\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)\right)dx\right]^2$$

And if we recall the defining property of the Dirac Delta Function, which says

$$\int_{-\infty}^{\infty}\delta(x-y)f(x)dx=f(y)$$

So, applying this to the above equation, we have:

$$\left [\int_0^L \delta(x-L/2)^*\left(\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)\right)dx\right]^2=\left(\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi \left(\frac{L}{2}\right)}{L}\right)\right)^2=\left(\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi }{2}\right)\right)^2=\frac{2}{L}$$

where this value holds for all odd values of n. Yay or nay?

Last edited: Apr 27, 2008
6. Apr 27, 2008

### G01

Your work and reasoning looks good.

Just to make one point clear, you said "Kronecker Delta" in the beginning of the post and "Dirac Delta" near the end. Throughout this problem you are using a Dirac Delta, not a Kronecker Delta. Also for the Dirac:

$$\delta (0)=\infty$$

You used the delta correctly under the integral sign, so this point shouldn't affect your result.

Anyway, your work still looks good, I just want to make sure you understand that you are using a Dirac delta here.

Good Job!

P.S. (That delta sure made that integral easier huh? )

Last edited: Apr 27, 2008