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Free Particles on a Line Segment (Quantum Mechanics)

  1. Aug 5, 2008 #1
    Problem
    At [tex]t = 0[/tex], [tex]10^5[/tex] noninteracting protons are known to be on a line segment [tex]10 \text{cm}[/tex] long. It is equally probable to find any proton at any point on this segment. How many protons remain on the segment at [tex]t = 10 \text{s}[/tex]?

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

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

    Our aim is to find

    [tex]\int_{-\infty}^{\infty} |\psi(x, 10)|^2 dx[/tex]​

    Now, we have that

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

    where

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

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


    [tex]|\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'[/tex]​

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


    [tex]|\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[/tex]​

    since the integrand is zero whenever [tex]k \neq k'.[/tex]. Substituting [tex]b(k)[/tex], we have

    [tex]|\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}[/tex]

    Thus, we have that

    [tex]10000\int_{-5}^{5} |\psi(x, t)|^2 dx = 10000 \cdot 10 \cdot \frac{1}{10} = 10000[/tex]. 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. jcsd
  3. Aug 5, 2008 #2

    nrqed

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    Gold Member

    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 = [tex] e^{-i k x} [/tex], 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

    [tex] \int dx e^{ix(k-k')} \simeq \delta (k-k') [/tex]

    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.
     
  4. Aug 6, 2008 #3
    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

    [tex]\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'[/tex]

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

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

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

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

    Our aim is to find

    [tex]\int_{-5}^{5} |\psi(x, 10)|^2 dx[/tex]​

    Now, we have that

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

    where

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

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


    [tex]|\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'[/tex]​

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


    [tex]|\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[/tex]​

    since the integrand is zero whenever [tex]k \neq k'.[/tex]. Substituting [tex]b(k)[/tex], we have

    [tex]|\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}[/tex]

    Thus, we have that

    [tex]10000\int_{-5}^{5} |\psi(x, t)|^2 dx = 10000 \cdot 10 \cdot \frac{1}{10} = 10000[/tex]. But this is the same number of particles that were initially in the interval. Is this correct? Or have I made a mistake somewhere?
     
  5. Aug 6, 2008 #4

    nicksauce

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    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?
     
  6. Aug 6, 2008 #5
    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)...
     
  7. Aug 6, 2008 #6

    nicksauce

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