(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

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# Homework Help: Free particle and Heisenburg uncertainty principle

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