# Free Particles on a Line Segment (Quantum Mechanics)

1. Aug 5, 2008

### Domnu

Problem
At $$t = 0$$, $$10^5$$ noninteracting protons are known to be on a line segment $$10 \text{cm}$$ long. It is equally probable to find any proton at any point on this segment. How many protons remain on the segment at $$t = 10 \text{s}$$?

Attempt at Solution
Well, let the protons lie between $$x = -5$$ and $$x = 5$$. Now, we can model the initial state function to be

$$\psi(x, 0) = \begin{cases} \frac{1}{10}, & \mbox{if }|x|\le 5 \\ 0, & \mbox{if } |x| > 5 \end{cases}$$​

Our aim is to find

$$\int_{-\infty}^{\infty} |\psi(x, 10)|^2 dx$$​

Now, we have that

$$\psi(x, t) = \int_{-\infty}^{\infty} b(k) \curlyphi_k e^{-i \omega_k t} dx$$​

where

$$b(k) = \int_{-5}^{5} \psi(x, 0) \curlyphi_k^* dk$$ = \frac{1}{10\sqrt{2\pi}} \cdot \frac{2 \sin 5k}{k}​

and $$\curlyphi_k$$ represents the momentum eigenstate corresponding to wavenumber $$k$$. We try to find $$|\psi(x, t)|^2 = \psi(x, t) \cdot \psi^* (x, t)$$:

$$|\psi(x, t)|^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty b(k) b^* (k) \curlyphi_k \curlyphi_{k'}^* e^{-i \omega_k t + i \omega_k' t) dk dk'$$​

which is a rather formidable integral. However, note that $$\curlyphi_k \curlyphi_{k'}^* = \langle \curlyphi_{k'} | \curlyphi_{k} \rangle = \delta(k' - k)$$. This means that the entire integral breaks down to

$$|\psi(x, t)|^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty b(k) b^* (k) \curlyphi_k \curlyphi_{k'}^* e^{-i \omega_k t + i \omega_k' t) dk dk' = \int_{-\infty}^\infty b(k) b^*(k) dk$$​

since the integrand is zero whenever $$k \neq k'.$$. Substituting $$b(k)$$, we have

$$|\psi(x, t)|^2 = \int_{-\infty}^\infty b(k) b^*(k) dk = \int_{-\infty}^{\infty} \frac{1}{50 \pi} \frac{\sin^2 5k}{k^2} dk = \frac{1}{10}$$

Thus, we have that

$$10000\int_{-5}^{5} |\psi(x, t)|^2 dx = 10000 \cdot 10 \cdot \frac{1}{10} = 10000$$. But this is the same number of particles that were initially in the interval. Is this correct? Or have I made a mistake somewhere?

2. Aug 5, 2008

### nrqed

the notation is confusing. You used "curlyphi" in your latex code but it does not come out at all. I am assuming you mean curlyphi_k = $$e^{-i k x}$$, right?

You cannot do what you just did in your last line above and replace by a delta function in momentum space! This is only truw if one integrates over x

$$\int dx e^{ix(k-k')} \simeq \delta (k-k')$$

where I am not paying attention to the overall constant.

In your case you are not integrating over x so you can't replace by a delta function. You must carry out the k and k' integrals explicitly.

3. Aug 6, 2008

### Domnu

Ack.. here's the fixed version of my solution with Latex corrected... I used Mathematica terminology accidentally in my first post :tongue: If this is incorrect, how would I evaluate the integral? It seems extremely daunting... the integral is

$$\psi(x, t) = \frac{1}{100\pi^2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{\sin 5k}{k} \frac{\sin 5k'}{k'} e^{i [\hbar k'^2/2m - \hbar k^2/2m] t} e^{i (k-k') x} dk dk'$$

I tried evaluating the above using Mathematica, and it couldn't do it Anyways, here was my earlier solution:

At $$t = 0$$, $$10^5$$ noninteracting protons are known to be on a line segment $$10 \text{cm}$$ long. It is equally probable to find any proton at any point on this segment. How many protons remain on the segment at $$t = 10 \text{s}$$?

Attempt at Solution
Well, let the protons lie between $$x = -5$$ and $$x = 5$$. Now, we can model the initial state function to be

$$\psi(x, 0) = \begin{cases} \frac{1}{10}, & \mbox{if }|x|\le 5 \\ 0, & \mbox{if } |x| > 5 \end{cases}$$​

Our aim is to find

$$\int_{-5}^{5} |\psi(x, 10)|^2 dx$$​

Now, we have that

$$\psi(x, t) = \int_{-\infty}^{\infty} b(k) \varphi_k e^{-i \omega_k t} dx$$​

where

$$b(k) = \int_{-5}^{5} \psi(x, 0) \varphi_k^* dk = \frac{1}{10\sqrt{2\pi}} \cdot \frac{2 \sin 5k}{k}$$

and $$\varphi_k$$ represents the momentum eigenstate corresponding to wavenumber $$k$$. We try to find $$|\psi(x, t)|^2 = \psi(x, t) \cdot \psi^* (x, t)$$:

$$|\psi(x, t)|^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty b(k) b^* (k) \varphi_k \varphi_{k'}^* e^{-i \omega_k t + i \omega_k' t) dk dk'$$​

which is a rather formidable integral. However, note that $$\varphi_k \varphi_{k'}^* = \langle \varphi_{k'} | \varphi_{k} \rangle = \delta(k' - k)$$. This means that the entire integral breaks down to

$$|\psi(x, t)|^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty b(k) b^* (k) \varphi_k \varphi_{k'}^* e^{-i \omega_k t + i \omega_k' t) dk dk' = \int_{-\infty}^\infty b(k) b^*(k) dk$$​

since the integrand is zero whenever $$k \neq k'.$$. Substituting $$b(k)$$, we have

$$|\psi(x, t)|^2 = \int_{-\infty}^\infty b(k) b^*(k) dk = \int_{-\infty}^{\infty} \frac{1}{50 \pi} \frac{\sin^2 5k}{k^2} dk = \frac{1}{10}$$

Thus, we have that

$$10000\int_{-5}^{5} |\psi(x, t)|^2 dx = 10000 \cdot 10 \cdot \frac{1}{10} = 10000$$. But this is the same number of particles that were initially in the interval. Is this correct? Or have I made a mistake somewhere?

4. Aug 6, 2008

### nicksauce

A remark: Should the wave function initally be 1/sqrt(10) instead of 1/10? You need the integral of the square of the wave function to be 1 right?

5. Aug 6, 2008

### Domnu

Oh yea, that's right.. wow.. I'm making really careless errors. But that wouldn't affect the difficulty of evaluating the integral right? Everything (almost) is just off by a factor of sqrt(10)...

6. Aug 6, 2008

### nicksauce

I'm really not sure... I don't know enough quantum to understand most of the post (except for that part I mentioned).