# What is the physical dimension/unit of Probability current?

## Homework Statement

Question:
What is the physical dimension of Probability Current for a particle in 1 dimension? (Quantum Mechanics)

## Homework Equations

Quantum mechanical Probability Current:

## The Attempt at a Solution

I know the physical dimension of mass, that is kg. If I know every dimension, I can try some things and I can find the dimension of the observable. But now, I'm stuck. I guess that the wave function has no dimension, because it is very related to probability. But what's the case with $$\frac{\partial \Psi}{\partial x}$$?

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BvU
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Hello Feest,

m-1 I would dare to venture .... (on your last question)

haruspex
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2020 Award
Hello Feest,

m-1 I would dare to venture .... (on your last question)
But h has dimension of action. Taking Ψ to be dimensionless, j would appear to have dimension of velocity.

BvU
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Actually ##j## is a probability current density.

haruspex
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2020 Award
Actually ##j## is a probability current density.
Just realised you meant m as metres, not mass. The question is for dimension, so I would answer L-1 there.
But what do you think the dimension of j is, if not LT-1?

BvU
Homework Helper
With $${\partial \rho\over \partial t} + \nabla\cdot {\bf j} = 0$$ where ##\rho = \psi^*\psi ## and ##\int \rho \, d\tau = 1##, I would end up with L-2T-1

 And now I have to backtrack to 1 dimension as cleary stated in post #1 () ending up with T-1

BvU
Homework Helper
I guess that the wave function has no dimension, because it is very related to probability.
So here we must react also: It is related to probability in the sense that $$\int \psi^*\psi\,d\tau = 1$$where the integral is over all space. The ##1## is dimensionless: a genuine probability. So guess again !

EhrEnFeest
So here we must react also: It is related to probability in the sense that $$\int \psi^*\psi\,d\tau = 1$$where the integral is over all space. The ##1## is dimensionless: a genuine probability. So guess again !
This is a nice question to answer. It took me some time but I think:
∫ψ∗ψdτ=1 has no dimensions.
But as you integrate over space, and in this case this is one-dimensional, the dimension of ψ∗ψ is multiplied with [length]. And then ψ has a dimension of [(1/length)^½]?

BvU