# Derivation of current density in quantum mechanics

1. Apr 14, 2008

### Lojzek

I read some derivations of current density from the quantum equations of motion (like Scrödinger's and Klein-Gordon's). They derive an equation with the same form as continuity equation:

div(A)+dB/dt=0

Then they conclude that A=current density and B=density.

However there are non-zero vector fields that have zero divergence, so we could add any of them to A and the continuity equation would still be true. How can we know that A is the true expression for current density?

Last edited: Apr 14, 2008
2. Apr 15, 2008

### genneth

You don't really. In the case of Schroedinger, you can treat it as a classical complex field and apply Lagrangian methods. The field, being complex, has global phase symmetry, which by Noether's theorem identifies a pair of charge and currents. The charge is of the form $$\psi^* \psi$$ which we postulated is the probability density over position, and so we identify the current identified as the probability current.

3. Apr 15, 2008

### lbrits

The Lagrangian treatment is done here: http://www.physics.thetangentbundle.net/wiki/Quantum_mechanics/Lagrangian_formulation

To answer your question, one normally derives the continuity equation from saying $$\mathbf{J}$$ is a current density and $$\rho$$ is a density, and then from using flux arguments. So when you derive an equation like $$\boldsymbol\nabla\cdot\mathbf{J} + \tfrac{\partial\rho}{\partial t} = 0$$, you simply recognize it as the continuity equation and make the identification.

Note, the continuity equation isn't a constitutive equation. For example, in the heat equation, the continuity equation must be supplemented by $$\mathbf{J} \propto \boldsymbol\nabla T$$. In the case of the wavefunction, your constitutive equation is the thing you derive from the S.E.:

$$\mathbf{J} = \frac{\hbar}{2mi}\left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^*\right) = \frac\hbar m \mbox{Im}(\Psi^*\vec\nabla\Psi)=\mbox{Re}(\Psi^* \frac{\hbar}{im} \vec\nabla \Psi)$$

4. Jan 25, 2011

### TriKri

We have

$$\mathbf{\hat{p}}=m\mathbf{\hat{v}} \Leftrightarrow \mathbf{\hat{v}}=\frac{\mathbf{\hat{p}}}{m} = \frac{-i\hbar\mathbf{\nabla}}{m}=\frac{\hbar\mathbf{\nabla}}{im}$$

and if we assume that the wave function is normalized, we will get the expectation value of the particle's velocity by

$$\left<\mathbf{\hat{v}}\right> = \left<\Psi\left|\mathbf{\hat{v}}\right|\Psi\right> = \left<\Psi\left|\frac{\hbar}{im}\mathbf{\nabla}\right|\Psi\right> = \int\limits_{}^{}\Psi^*\frac{\hbar}{im}\mathbf{\nabla}\Psi\, dV.$$

where $$dV$$ is a small volume element. If we define

$$\mathbf{J_\mathbb{C}} = \Psi^*\frac{\hbar}{im}\mathbf{\nabla}\Psi$$

so that

$$\mathbf{J} = \mbox{Re}(\mathbf{J_\mathbb{C}})$$

we get the simple relation

$$\left<\mathbf{\hat{v}}\right> = \int\limits_{}^{}\mathbf{J_\mathbb{C}}\, dV.$$

I think this is a fundamental property that you want to have.

Let's continue with looking at how the expectation value of the position changes with time. It is well known that the time derivative of the expectation value of a time-independent operator $$\hat{A}$$ easily can be obtained using the Hamiltonian, namely

$$\frac{d}{dt}\left<\hat{A}\right> = (i\hbar)^{-1}\left<\left[\hat{A},\,\hat{H}\right]\right>$$

so in our case we have

$$\frac{d}{dt}\left<\hat{x}\right> = (i\hbar)^{-1}\left<\left[\hat{x},\,\hat{H}\right]\right> = (i\hbar)^{-1}\left<\left[\hat{x},\,\frac{\mathbf{\hat{p}}^2}{2m}+V(\mathbf{r},\,t)\right]\right>$$

Since $$\hat{x}$$ commutes with $$V(\mathbf{r},t)$$ and also with $$\hat{p}_y$$ and $$\hat{p}_z$$, we have

$$\frac{d}{dt}\left<\hat{x}\right> = \frac{1}{2mi\hbar}\left<\left[\hat{x},\,\hat{p}_x^2}\right]\right> = \frac{1}{2mi\hbar}\left<\left[\hat{x},\,\hat{p}_x}\right]\hat{p}_x + \hat{p}_x\left[\hat{x},\,\hat{p}_x}\right]\right>$$

We know that $$\left[\hat{x},\hat{p}_x\right]=i\hbar$$, so we get

$$\frac{d}{dt}\left<\hat{x}\right> = \frac{1}{2mi\hbar}\left<i\hbar\hat{p}_x + \hat{p}_x i\hbar\right> = \frac{\left<\hat{p}_x\right>}{m}$$

Since we will get the same result with the expectation values of y and z, we obtain

$$\frac{d}{dt}\left<\mathbf{\hat{r}}\right> = \frac{\left<\mathbf{\hat{p}}\right>}{m} = \left<\mathbf{\hat{v}}\right>$$

Using the result from above, we get

$$\frac{d}{dt}\left<\mathbf{\hat{r}}\right> = \int\limits_{}^{}\mathbf{J_\mathbb{C}}\, dV.$$

Or, if we would so like

$$\frac{d}{dt}\left<\mathbf{\hat{r}}\right> = \left<\mathbf{\hat{v}}\right> = \int\limits_{}^{}\mathbf{J_\mathbb{C}}\, dV.$$

5. May 5, 2011

### andremelzi

How can I show this?

6. May 6, 2011

7. May 6, 2011

### andremelzi

My doubt is more basic. How do I show the equalities?

8. Nov 20, 2011

### preposterous

9. Nov 13, 2012

### tstin

If z is a complex number, then $$z=z_{1}+iz_{2}$$ and its complex conjugate is $$z^*=z_{1}-iz_{2}$$ Then it is easy to prove $$z+z^*=z_{1}+iz_{2}+z_{1}-iz_{2}=2z_{1}$$ and from this is is obvious that $$z_{1}=\mbox{Re}\left(z\right)=\frac{z+z^*}{2}$$

Similarly, $$z_{2}=\mbox{Im}\left(z\right)=\frac{z-z^*}{2}$$

Now consider a complex-valued function $$\psi$$ and some operator whose result is also complex valued: $$\hat{O}\psi$$. Now let our complex number z be $$z=\psi^*\left(\hat{O}\psi\right)$$. Then from the above identity, $$\frac{1}{2}\left[\psi^*\left(\hat{O}\psi\right)+\psi\left(\hat{O} \psi \right)^*\right]=\mbox{Re}\left[\psi^*\left(\hat{O}\psi\right)\right]$$

Now we substitute the general operator for the quantum mechanical momentum operator, $$\hat{O}=\frac{\hbar}{i}\nabla$$

Then we find $$\hat{O}\psi=\frac{\hbar}{i}\nabla\psi$$ and $$\left(\hat{O}\psi\right)^*=-\frac{\hbar}{i}\nabla\psi^*$$

Then it follows that $$\mbox{Re}\left[\psi^*\left(\hat{O}\psi\right)\right]=\mbox{Re}\left[\psi^*\frac{\hbar}{i}\nabla\psi\right]=\frac{1}{2}\left[\psi^*\frac{\hbar}{i}\nabla\psi-\psi\frac{\hbar}{i}\nabla\psi^*\right]=\frac{\hbar}{2i}\left[\psi^*\nabla\psi-\psi\nabla\psi^*\right]$$

Now if $$z=\left(\psi^*\nabla\psi\right)$$

then $$z^*=\psi\nabla\psi^*$$

So we get the second identity (which has a typo in the previous post): $$\frac{\hbar}{im}\mbox{Im}\left[\psi^*\nabla\psi\right]=\frac{\hbar}{2mi}\left[\psi^*\nabla\psi-\psi\nabla\psi^*\right]$$

Put it all together and you get the equality
$$\frac{\hbar}{2mi}\left[\psi^*\nabla\psi-\psi\nabla\psi^*\right]=\frac{\hbar}{im}\mbox{Im}\left[\psi^*\nabla\psi\right]=\mbox{Re}\left[\psi^*\frac{\hbar}{im}\nabla\psi\right]$$