Free particle and Heisenburg uncertainty principle

  • Thread starter KFC
  • Start date
  • #1
KFC
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4

Homework Statement


Assuming at time is zero, the wavefunction of a free particle is given as

[tex]
\Psi(x, 0) = \left\{
\begin{matrix}
0, \quad x<0\\
f(x), \quad x>0
\end{matrix}
\right.
[/tex]

where f(x) is integrable within [tex](0, \infty)[/tex]


Find the time evolution of [tex]\Psi(x, 0)[/tex]. Write down the steps how to show Heisenberg briefly.

2. The attempt at a solution
Since this is a free particle, we use the completeness of eigenstates in momentum space to do the problem.

[tex]\phi_p = \frac{1}{\sqrt{2\pi\hbar}}\exp(ipx/\hbar)[/tex]

The orthogonality is

[tex]\int_{-\infty}^{+\infty}\phi^*_p(x)\phi_{p'}(x)dx = \delta(p'-p)[/tex]

The energy of the particle is given by

[tex]E=\hbar\omega = \frac{p^2}{2m}[/tex]

Now, the initial state can be written in terms of eigenstates of momentum (or Fourier transform)

[tex]\Psi(x, 0) = \int c(p) \phi_p(x) dp[/tex]

multiply both side with [tex]\phi_{p'}(x)[/tex], take the integral and applying orthogonality, we find

[tex]c(p) = \int \Psi(x, 0)\phi^*_p(x)dx[/tex]

Hence , the time evolution of the initial state is just

[tex]
\begin{matrix}
\Psi(x, t) & = & \displaystyle{\int_{-\infty}^{+\infty} dp \left[\int \Psi(x, 0)\phi^*_p(x)dx\right]\phi_p(x) \exp(-i\omega t)} \\ \\
& = & \displaystyle{\int_{-\infty}^{+\infty} dp \left[\int \Psi(x, 0)\phi^*_p(x)\phi_p(x)dx\right] \exp(-i\omega t)} \\ \\
& = & \displaystyle{\frac{1}{2\pi\hbar} \int_{-\infty}^{+\infty} dp \exp(-i\omega t)\left[\int_0^\infty \Psi(x, 0)dx\right]} \\ \\
& = & \displaystyle{\frac{F(x)}{2\pi\hbar} \int_{-\infty}^{+\infty} dp \exp\left(-i\frac{p^2}{2m\hbar}t\right)}
\end{matrix}
[/tex]

where

[tex]
F(x) = \int_0^\infty f(x) dx
[/tex]


So far, I think my solution is quite correct, right? But I thought about it for long time and have no idea how to do the last integral!


For illustrating the Heisenberg principle based on the initial state, the steps could be concluded as

1) find [tex]\langle x^2 \rangle[/tex] and [tex]\langle x \rangle[/tex] in initial state [tex]\Psi(x, 0)[/tex], so the variance of x would be

[tex]\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}[/tex]

2) find [tex]\langle p^2 \rangle[/tex] and [tex]\langle p \rangle[/tex] in momentum state [tex]\phi(p)[/tex], so the variance of p would be

[tex]\Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}[/tex]

3) multiply [tex]\Delta x[/tex] and [tex]\Delta p[/tex] to find the relation.
 

Answers and Replies

  • #2
Avodyne
Science Advisor
1,396
88
No, sorry, you made a mistake in the first equation after "Hence". There are two different x's in this equation, the one being integrated over in brackets, and the one that is an argument of [itex]\Psi[/itex] on the left, and of [itex]\phi_p[/itex] on the right.
 
  • #3
KFC
488
4
No, sorry, you made a mistake in the first equation after "Hence". There are two different x's in this equation, the one being integrated over in brackets, and the one that is an argument of [itex]\Psi[/itex] on the left, and of [itex]\phi_p[/itex] on the right.
Oh, how careless I am ... so I know what's going on now.

What about the second part? Is that the right steps to prove Heisenberg uncertainty principle?
 
  • #4
Avodyne
Science Advisor
1,396
88
I guess. I actually don't think this is a very good question. Knowing that the initial wave function is zero for x<0 isn't very helpful for finding out the wave function at later times, so it's not clear to me what they're looking for in terms of the uncertainty principle.

I'm about to go offline for a week, so won't be able to reply further. Good luck!
 
  • #5
178
0
Well, in any case, the integral you were trying to do is commonly encountered in quantum mechanics. Try taking the Fourier transform of the Dirac delta function and see what happens.
 

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