Free Particles on a Line Segment (Quantum Mechanics)

[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?

[nrqed]

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


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.

[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 :frown: 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?

[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?

[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)...

[nicksauce]

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)...

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