# Homework Help: Units of j

1. Aug 2, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
What are the units of the probability flux j?

$$\vec j = \frac{\hbar}{2mi}\left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^*\right)$$

2. Relevant equations

3. The attempt at a solution

The units of psi are probability amplitude and the units of del(psi) are probability amplitude per unit area, so I would say the units are psi* del psi are probability density per unit area?

2. Aug 2, 2007

### Irid

Think about normalization equation for units of Psi.
Also, the del operator is a fancy sort of derivative (d/dx). It reduces dimension by one power of length, since x has the units of lenght.

3. Aug 2, 2007

### nrqed

First off, psi is a probability amplitude but we can't say that the units are "probability amplitude". This does not quite make sense. Its' like saying that the units of "m" are "mass". Mass is the name of the quantity, it does not give its dimensions. Dimensions are expressible in terms of length, time and mass units (meter, second and kilogram in MKSA). So the first question is: what are the units of psi? The answer is provided by noticing that a normalized psi obeys $\int dV | \psi|^2 = 1$. That should give you the units of psi.

Not per unita area, per unit length. $\frac{\partial }{\partial x}$ for example introduced a factor fo 1/meter in the dimensions of the quantity differentiated.

4. Aug 2, 2007

### ehrenfest

Are they sqrt(1/dV) then? So, then del * psi would have units 1/dV^2? That does not seem right.

5. Aug 3, 2007

### Dick

del(psi) will have units of sqrt(1/V)/L. del is a derivative wrt length L.

6. Aug 3, 2007

### ehrenfest

So psi* del psi has units 1/V/L? How does this translate into probability density per unit length which I thought nrqed implied were the correct units of j?

7. Aug 3, 2007

### olgranpappy

he didn't imply that. anyways, a probability is a pure number... so the dimension of "probability density" is just 1/L^3. thus a "probability density per unit length" would have dimensions of 1/L^4 which is the wrong units for j.

Perhaps what he was implying was that the units of psi*del psi are 1/L^4 which is correct. Now, what are the units of the rest of the things in your equation...

8. Aug 3, 2007

### ehrenfest

Planck's constant is J * s, mass is kg, so we have

kg * m^2 / s^2 * 1/kg * 1/ m^4 = 1/m^2/s^2

where m is meters. So how do you put that in words: length * probability density per second per second?

9. Aug 3, 2007

### olgranpappy

you left off a factor of seconds... you said "Joule seconds" but you only put in the "joule" part.

10. Aug 3, 2007

### ehrenfest

Sorry.

kg * m^2 / s * 1/kg * 1/ m^4 = 1/m^2/s

11. Aug 3, 2007

### olgranpappy

thats right
$$\frac{1}{TL^2}$$

12. Aug 3, 2007

### nrqed

Yes, psi has units of q/sqrt(volume), so the units of $\psi$ are $\frac{1}{L^{3/2}}$ where L stands for length.

Therefore, the units of del psi are $\frac{1}{L^{5/2}}$, so the units of psi del psi are $\frac{1}{L^4}$ as someone else pointed out.

Last edited: Aug 3, 2007