# I'm off by sqrt(2) and it's driving me crazy! (expanded particle in 1D box)

## Homework Statement

I believe this is a pretty standard question. A particle is initially in the ground state of a box of length L. Suddenly, the box expands symmetrically to twice its size, without disturbing the wavefunction. Show that the probability of finding the particle in the ground state of the new box is (8/(3Pi))^2.

## Homework Equations

We're using the box defined (initially) from -L/2.. L/2, and the potential is infinite everywhere outside the box. The ground state wavefunction should be sqrt(2/L)sin(Pi*x/L).

## The Attempt at a Solution

First I determined how we would renormalize the wavefunction in the new box, which extends from -L to L. So, I integrated $$\int_{-L}^{L} A^2 sin^2\left(\frac{n \pi x}{2L}\right) dx$$ and got the new wavefunctions to be $$\sqrt{\frac{1}{L}} sin{\frac{n \pi x}{2L}}$$.

Now if I understand correctly, the probabilty of finding the particle in the new ground state is simply the square of the projection of this new state vector onto the old ground state vector. So I then integrated $$\int_{-L}^{L} \sqrt{\frac{2}{L}} sin\left(\frac{\pi x}{L}\right) \sqrt{\frac{1}{L}} sin\left(\frac{\pi x}{2L}\right) dx = \frac{8 \sqrt{2}}{3 \pi}$$.

If you square this, it's correct except for the square root of two (which makes my prob. off by a factor of 2!).

The extra $$\sqrt{2}$$ is driving me crazy!! What am I doing wrong?

Related Advanced Physics Homework Help News on Phys.org
If it helps, I can see that using the normalization factor sqrt(1/L) for both wavefunctions would produce the correct answer, but I don't understand why I'd tack the new normalization onto my old wavefunction. If someone can explain this I'd be much obliged...

gabbagabbahey
Homework Helper
Gold Member
Now if I understand correctly, the probabilty of finding the particle in the new ground state is simply the square of the projection of this new state vector onto the old ground state vector. So I then integrated $$\int_{-L}^{L} \sqrt{\frac{2}{L}} sin\left(\frac{\pi x}{L}\right) \sqrt{\frac{1}{L}} sin\left(\frac{\pi x}{2L}\right) dx = \frac{8 \sqrt{2}}{3 \pi}$$.
Do you really want to integrate this from $-L$ to $L$?....Isn't the particle in the state

$$\psi(x)=\begin{array}{lr} \sqrt{\frac{2}{L}} sin\left(\frac{\pi x}{L}\right) &, |x|\leq\frac{L}{2} \\ 0 &, |x|>\frac{L}{2}\end{array}$$

I suppose that makes more sense, since the initial state isn't defined past L/2.

The only problem is that that integral would then give me 4/3*pi, which is still off by a factor of 2!

gabbagabbahey
Homework Helper
Gold Member
Isn't the ground state [itex]\sqrt{\frac{2}{L}}\cos\left(\frac{\pi x}{L}\right)[/tex]?

Aha!!!

I always forget that the sine solution goes with the even integers (I tend to associate sine to odd things).

That totally solves my problem; thank you!! :-D