- #1
Domnu
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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
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?
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]
\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?