# Quantum physics - probability density,

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1. Jan 27, 2015

### Jillds

1. The problem statement, all variables and given/known data
consider a particle at an interval $[-L/2, L/2]$, described by the wave function $\psi (x,t)= \frac{1}{\sqrt{L}}e^{i(kx-wt)}$
a) Calculate the probability density $\rho (x,t)$ and the current density $j(x,t)$ of the particle
b) How can you express $j(x,t)$ as a function of $\rho (x,t)$ and the velocity $v$?
c) What is de probability to find the particle at the intervals $[-L/2, L/2]$, $[-L/2, 0]$ and $[0, L/4]$

(note: not sure whether 'current density' is the proper translation of the term I have in Dutch for j(x,t).

2. Relevant equations
$\rho (x,t) = \psi^* (x,t) \psi (x,t)$
$j(x,t) = - \frac{i \hbar}{2m} \big[ \psi^* (x,t) \frac{\partial}{\partial x}\psi (x,t) - \psi (x,t) \frac{\partial}{\partial x}\psi^* (x,t) \big]$
$P(\Omega)= \int_{\Omega} |\psi (x,t)|^2 dx$
no clue on the relevant equation for b and c

3. The attempt at a solution
a) $\rho (x,t) = \frac{1}{\sqrt{L}} e^{-i(kx-wt)} . \frac{1}{\sqrt{L}}e^{i(kx-wt)} = \frac{1}{L} e^{-i(kx-wt)+i(kx-wt)} = \frac{1}{L} e^0 = \frac{1}{L}$
$j(x,t) = - \frac{i \hbar}{2m} \big[ \frac{1}{\sqrt{L}} e^{-i(kx-wt)} \frac{\partial}{\partial x}\frac{1}{\sqrt{L}}e^{i(kx-wt)} - \frac{1}{\sqrt{L}} e^{i(kx-wt)} \frac{\partial}{\partial x}\frac{1}{\sqrt{L}}e^{-i(kx-wt)} \big]$
$= - \frac{i \hbar}{2m} \big[ \frac{1}{L} e^{-i(kx-wt)} e^{i(kx-wt)} ik - \frac{1}{L} e^{-i(kx-wt)} e^{i(kx-wt)} (-ik) \big]$
$= - \frac{i \hbar}{2m} \big[ \frac{ik}{L} + \frac{ik}{L} \big]$
$= - \frac{i \hbar}{m} \frac{ik}{L} = \frac{\hbar k}{m L}$

b) I was not sure how to interprete the question, express first $j(x,t)$ as a function of $\rho (x,t)$ AND afterwards in relation to velocity, OR all at the same time?

I have suggested for myself at least that $j(x,t) = \frac{\hbar k}{m} \rho (x,t)$
I have no idea how the velocity comes into that.

c) For the last part I have

$\frac{1}{L} \int_{-L/2}^{L/2} dx = \frac{x}{L}|_{-L/2}^{L/2} = 2L/2L =1$
Since the second interval is half of the first interval the probability should be 1/2, and checking the integral that is what I got. Same logic applies for the third given interval, which is a 4th of the first, and the probability is 1/4.

Can someone please review what I did solve, and whether I applied the correct logic or proper techniques. And can someone help me out with the velocity part?

2. Jan 27, 2015

### Staff: Mentor

There is just one equation that combines them and has a nice expression. Your wave function describes a moving wave. What is its (phase) velocity?

(a) and (c) look fine.

3. Jan 27, 2015

### Jillds

I know there a continuity equation: $\frac{\partial}{\partial t} \rho(x,t) + \frac{\partial}{\partial x} j(x,t) = 0$. Is that the one you mean?

ETA: correction on the continuity equation per mfb's following comment

Last edited: Jan 27, 2015
4. Jan 27, 2015

### Staff: Mentor

That continuity equation does not look right (edit: the edited version is correct now), but that is not what I meant.
What is the phase velocity v of your wave? It is the same as in the classical case.

Last edited: Jan 28, 2015
5. Jan 27, 2015

### Jillds

phase velocity : $v_p = \frac{\lambda}{T} = \lambda f = \frac{\omega}{k}$
group velocity : $v_g = \frac{\partial \omega}{\partial k}$

6. Jan 27, 2015

### Jillds

I know I could rewrite the wave in its classic notation as $\psi (x,t) = \frac{1}{\sqrt{L}} sin (kx - wt)$
I know the phase velocity is $v_p = \lambda f = \frac{\omega}{k}$
But I don't yet see how this relates to the probability density and j(x,t)... Help very much appreciated.

7. Jan 28, 2015

### Jillds

Found the solution:
De Broglie: $p = \hbar k$
Classical mechanics: $p=mv$

Hence, $j(x,t) = \frac{\hbar k}{mL} = \frac{p}{mL} = \frac{v}{L} = v \rho(x,t)$

So, I started out correct by replacing the $\frac{1}{L}$ with the $\rho$, but failed to recognize De Broglie relation and to apply Classical mechanics in that.