# Probability current: a few examples

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.3, Exercises 5.3.2 – 5.3.4.

Here are a few examples of probability current.

Example 1 Suppose the wave function has the form

$\displaystyle \psi\left(\mathbf{r},t\right)=c\tilde{\psi}\left(\mathbf{r},t\right) \ \ \ \ \ (1)$

where ${c}$ is a complex constant and ${\tilde{\psi}\left(\mathbf{r},t\right)}$ is a real function of position and time. Then the probability current is

 $\displaystyle \mathbf{j}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}\left(\psi^*\nabla\psi-\psi\nabla\psi^*\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}\left(cc^*\left(\tilde{\psi}\nabla\tilde{\psi}\right)-\tilde{\psi}\nabla\tilde{\psi}\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

In particular, if ${\psi}$ itself is real, the probability current is always zero, so all the stationary states of systems like the harmonic oscillator and hydrogen atom that we’ve studied show no flow of probability, which is what we’d expect since they are, after all, stationary states.

Example 2 Now the wave function is

 $\displaystyle \psi_{\mathbf{p}}$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{i\mathbf{p}\cdot\mathbf{r}/\hbar} \ \ \ \ \ (5)$

where the momentum ${\mathbf{p}}$ is constant. In this case we have

 $\displaystyle \nabla\psi_{\mathbf{p}}$ $\displaystyle =$ $\displaystyle \frac{i}{\left(2\pi\hbar\right)^{3/2}\hbar}e^{i\mathbf{p}\cdot\mathbf{r}/\hbar}\mathbf{p}\ \ \ \ \ (6)$ $\displaystyle \nabla\psi_{\mathbf{p}}^*$ $\displaystyle =$ $\displaystyle \frac{-i}{\left(2\pi\hbar\right)^{3/2}\hbar}e^{-i\mathbf{p}\cdot\mathbf{r}/\hbar}\mathbf{p}\ \ \ \ \ (7)$ $\displaystyle \psi_{\mathbf{p}}^*$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{-i\mathbf{p}\cdot\mathbf{r}/\hbar} \ \ \ \ \ (8)$

This gives a probability current of

 $\displaystyle \mathbf{j}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}(\psi_{\mathbf{p}}^*\nabla\psi_{\mathbf{p}}-\psi_{\mathbf{p}}\nabla\psi_{\mathbf{p}}^*)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\hbar\right)^{3}2m}\left(\mathbf{p}+\mathbf{p}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\hbar\right)^{3}m}\mathbf{p} \ \ \ \ \ (11)$

The probability density is

$\displaystyle P=\psi_{\mathbf{p}}^*\psi_{\mathbf{p}}=\frac{1}{\left(2\pi\hbar\right)^{3}} \ \ \ \ \ (12)$

Thus the current can be written as

$\displaystyle \mathbf{j}=\frac{P}{m}\mathbf{p} \ \ \ \ \ (13)$

Classically, the momentum is ${\mathbf{p}=mv}$, so the current has the same form as ${\mathbf{j}=P\mathbf{v}}$. This is similar to the electromagnetic case where the electric current density ${\mathbf{J}=\rho\mathbf{v}}$ where ${\rho}$ is the charge density and ${\mathbf{v}}$ is the velocity of that charge. The probability density can be viewed as “probability” moving with velocity ${\mathbf{v}}$.

Example 3 Now consider a one-dimensional problem where the wave function consists of two oppositely-moving plane waves:

$\displaystyle \psi=Ae^{ipx/\hbar}+Be^{-ipx/\hbar} \ \ \ \ \ (14)$

In this case, we have

 $\displaystyle \frac{2mi}{\hbar}j$ $\displaystyle =$ $\displaystyle \psi^*\nabla\psi-\psi\nabla\psi^*\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(A^*e^{-ipx/\hbar}+B^*e^{ipx/\hbar}\right)\frac{ip}{\hbar}\left(Ae^{ipx/\hbar}+Be^{-ipx/\hbar}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(Ae^{ipx/\hbar}+Be^{-ipx/\hbar}\right)\frac{ip}{\hbar}\left(-A^*e^{-ipx/\hbar}+B^*e^{ipx/\hbar}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2ip}{\hbar}\left(\left|A\right|^{2}-\left|B\right|^{2}\right)\ \ \ \ \ (17)$ $\displaystyle j$ $\displaystyle =$ $\displaystyle \frac{p}{m}\left(\left|A\right|^{2}-\left|B\right|^{2}\right) \ \ \ \ \ (18)$

The probability current separates into two terms, one for each direction of momentum.

# Probability current with complex potential

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.3, Exercise 5.3.1.

Shakar’s derivation of the probability current in 3-d is similar to the one we reviewed earlier, so we don’t need to repeat it here. We can, however, look at a slight variant where the potential has a constant imaginary part, so that

$\displaystyle V\left(\mathbf{r}\right)=V_{r}\left(\mathbf{r}\right)-iV_{i} \ \ \ \ \ (1)$

where ${V_{r}\left(\mathbf{r}\right)}$ is a real function of position and ${V_{i}}$ is a real constant. A Hamiltonian containing such a complex potential is not Hermitian.

To see what effect this has on the total probability of finding a particle in all space, we can repeat the derivation of the probability current. From the Schrödinger equation and its complex conjugate, we have

 $\displaystyle i\hbar\frac{\partial\psi}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V_{r}\psi-iV_{i}\psi\ \ \ \ \ (2)$ $\displaystyle -i\hbar\frac{\partial\psi^*}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi^*+V_{r}\psi^*+iV_{i}\psi^* \ \ \ \ \ (3)$

Multiply the first equation by ${\psi^*}$ and the second by ${\psi}$ and subtract to get

$\displaystyle i\hbar\frac{\partial}{\partial t}\left(\psi\psi^*\right)=-\frac{\hbar^{2}}{2m}\left(\psi^*\nabla^{2}\psi-\psi\nabla^{2}\psi^*\right)-2iV_{i}\psi\psi^* \ \ \ \ \ (4)$

As in the case with a real potential, the first term on the RHS can be written as the divergence of a vector:

 $\displaystyle \mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}(\Psi^*\nabla\Psi-\Psi\nabla\Psi^*)\ \ \ \ \ (5)$ $\displaystyle \nabla\cdot\mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}\left(\psi^*\nabla^{2}\psi-\psi\nabla^{2}\psi^*\right)\ \ \ \ \ (6)$ $\displaystyle \frac{\partial}{\partial t}\left(\psi\psi^*\right)$ $\displaystyle =$ $\displaystyle -\nabla\cdot\mathbf{J}-\frac{2V_{i}}{\hbar}\psi\psi^* \ \ \ \ \ (7)$

If we define the total probability of finding the particle anywhere in space as

$\displaystyle P\equiv\int\psi^*\psi d^{3}\mathbf{r} \ \ \ \ \ (8)$

then we can integrate 4 over all space and use Gauss’s theorem to convert the volume integral of a divergence into a surface integral:

 $\displaystyle \frac{\partial}{\partial t}\left(\int\psi\psi^*d^{3}\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle -\int\nabla\cdot\mathbf{J}d^{3}\mathbf{r}-\frac{2V_{i}}{\hbar}\int\psi\psi^*d^{3}\mathbf{r}\ \ \ \ \ (9)$ $\displaystyle \frac{\partial P}{\partial t}$ $\displaystyle =$ $\displaystyle -\int_{S}\mathbf{J}\cdot d\mathbf{a}-\frac{2V_{i}}{\hbar}P \ \ \ \ \ (10)$

We make the usual assumption that the probability current ${\mathbf{J}}$ tends to zero at infinity fast enough for the first integral on the RHS to be zero, and we get

$\displaystyle \frac{\partial P}{\partial t}=-\frac{2V_{i}}{\hbar}P \ \ \ \ \ (11)$

This has the solution

$\displaystyle P\left(t\right)=P\left(0\right)e^{-2V_{i}t/\hbar} \ \ \ \ \ (12)$

That is, the probability of the particle existing decays exponentially. Although Shankar says that such a potential can be used to model a system where particles are absorbed, it’s not clear how realistic it is since the Hamiltonian isn’t hermitian, so technically the energies in such a system are not observables.

# Dirac equation: conserved probability current

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.13.

The Dirac equation is

$\displaystyle \left(i\gamma^{\mu}\partial_{\mu}-m\right)\left|\psi\right\rangle =0 \ \ \ \ \ (1)$

and the adjoint Dirac equation is

$\displaystyle i\partial_{\mu}\left\langle \bar{\psi}\right|\gamma^{\mu}+m\left\langle \bar{\psi}\right|=0 \ \ \ \ \ (2)$

where there are four solution vectors ${n=1,\ldots,4}$ and the adjoint solutions are given by

$\displaystyle \left\langle \bar{\psi}\right|=\left\langle \psi\right|\gamma^{0} \ \ \ \ \ (3)$

We’d like to find a conserved quantity analogous to that for the Klein-Gordon equation. We can do this as follows. First we multiply 1 on the left by the adjoint solutions:

$\displaystyle i\left\langle \bar{\psi}\left|\gamma^{\mu}\partial_{\mu}\right|\psi\right\rangle =m\left\langle \bar{\psi}\left|\psi\right.\right\rangle \ \ \ \ \ (4)$

Then, multiply 2 on the right by the original solutions:

$\displaystyle i\left(\partial_{\mu}\left\langle \bar{\psi}\right|\right)\gamma^{\mu}\left|\psi\right\rangle =-m\left\langle \bar{\psi}\left|\psi\right.\right\rangle \ \ \ \ \ (5)$

We’ve kept the bra in parentheses since the derivative applies only to it and not the ket portion. Adding these two equations gives

 $\displaystyle i\left\langle \bar{\psi}\left|\gamma^{\mu}\partial_{\mu}\right|\psi\right\rangle +i\left(\partial_{\mu}\left\langle \bar{\psi}\right|\right)\gamma^{\mu}\left|\psi\right\rangle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle i\partial_{\mu}\left\langle \bar{\psi}\left|\gamma^{\mu}\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

where we’ve used the product rule to combine the two derivatives, so that the ${\partial_{\mu}}$ in the last line does apply to the full bracket. Note also that in the bracket in the last line, there is no integration over space, since all we’ve done is multiply the adjoint solution into the regular solution.

By the way, you might think that this result is trivial, since spacetime enters into ${\left|\psi\right\rangle }$ only in the form ${e^{\pm ipx}}$ and therefore into ${\left\langle \bar{\psi}\right|}$ in the form ${e^{\mp ipx}}$, so it would seem that the bracket ${\left\langle \bar{\psi}\left|\gamma^{\mu}\right|\psi\right\rangle }$ has no dependence on ${x}$ so obviously its derivative must be zero. However, the ${\left|\psi\right\rangle }$ here can refer to a sum of states with different momenta ${\mathbf{p}}$, so the result isn’t quite as trivial as it looks.

We can therefore define a conserved current ${j^{\mu}}$ as

$\displaystyle j^{\mu}\equiv\left\langle \bar{\psi}\left|\gamma^{\mu}\right|\psi\right\rangle \ \ \ \ \ (8)$

so that

$\displaystyle \partial_{\mu}j^{\mu}=0 \ \ \ \ \ (9)$

Again, remember that there is no integration over space in the definition of ${j^{\mu}}$.

In Klauber’s equations 4-36 and 4-37, he shows that if we consider a single particle in the state

$\displaystyle \left|\psi\right\rangle =\sum_{r,\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}C_{r}\left(\mathbf{p}\right)u_{r}e^{-ipx} \ \ \ \ \ (10)$

(remember that ${r}$ ranges over the four solutions and ${\mathbf{p}}$ over all possible discrete momenta) and integrate ${\rho\equiv j^{0}}$ over space, we get the condition

$\displaystyle \sum_{r,\mathbf{p}}\left|C_{r}\left(\mathbf{p}\right)\right|^{2}=1 \ \ \ \ \ (11)$

Thus we can interpret ${\left|C_{r}\left(\mathbf{p}\right)\right|^{2}}$ as the probability of finding the particle in state ${r}$ with momentum ${\mathbf{p}}$.

# Klein-Gordon equation: probability density and current

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 3, Problem 3.3.

In non-relativistic quantum mechanics governed by the Schrödinger equation, the probability density is given by

$\displaystyle \rho=\Psi^{\dagger}\Psi \ \ \ \ \ (1)$

and the probability current is given by (generalizing our earlier result to 3-d and using natural units):

$\displaystyle \mathbf{J}=\frac{i}{2m}\left(\Psi\nabla\Psi^{\dagger}-\Psi^{\dagger}\nabla\Psi\right) \ \ \ \ \ (2)$

The continuity equation for probability is then

$\displaystyle \frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{J}=0 \ \ \ \ \ (3)$

We’ll now look at how these results appear in relativistic quantum mechanics, using the Klein-Gordon equation:

$\displaystyle \frac{\partial^{2}\phi}{\partial t^{2}}=\left(\nabla^{2}-\mu^{2}\right)\phi=0 \ \ \ \ \ (4)$

We can multiply this equation by ${\phi^{\dagger}}$ and then subtract the hermitian conjugate of the result from the original equation to get

 $\displaystyle \phi^{\dagger}\frac{\partial^{2}\phi}{\partial t^{2}}-\phi\frac{\partial^{2}\phi^{\dagger}}{\partial t^{2}}$ $\displaystyle =$ $\displaystyle \phi^{\dagger}\left(\nabla^{2}-\mu^{2}\right)\phi-\phi\left(\nabla^{2}-\mu^{2}\right)\phi^{\dagger}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \phi^{\dagger}\nabla^{2}\phi-\phi\nabla^{2}\phi^{\dagger} \ \ \ \ \ (6)$

The LHS can be written as

$\displaystyle \phi^{\dagger}\frac{\partial^{2}\phi}{\partial t^{2}}-\phi\frac{\partial^{2}\phi^{\dagger}}{\partial t^{2}}=\frac{\partial}{\partial t}\left(\phi^{\dagger}\frac{\partial\phi}{\partial t}-\phi\frac{\partial\phi^{\dagger}}{\partial t}\right) \ \ \ \ \ (7)$

(use the product rule on the RHS and cancel terms).

The RHS of 6 can be written as (use the product rule again):

$\displaystyle \phi^{\dagger}\nabla^{2}\phi-\phi\nabla^{2}\phi^{\dagger}=\nabla\cdot\left(\phi^{\dagger}\nabla\phi-\phi\nabla\phi^{\dagger}\right) \ \ \ \ \ (8)$

We can write this as a continuity equation for the Klein-Gordon equation, with the following definitions:

 $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle i\left(\phi^{\dagger}\frac{\partial\phi}{\partial t}-\phi\frac{\partial\phi^{\dagger}}{\partial t}\right)\ \ \ \ \ (9)$ $\displaystyle \mathbf{j}$ $\displaystyle =$ $\displaystyle -i\left(\phi^{\dagger}\nabla\phi-\phi\nabla\phi^{\dagger}\right) \ \ \ \ \ (10)$

[The extra ${i}$ is introduced to make ${\rho}$ and ${\mathbf{j}}$ real. Note that the factor within the parentheses in both expressions is a complex quantity minus its complex conjugate, which always gives a pure imaginary term. Thus multiplying by ${i}$ ensures the result is real.]

Then

$\displaystyle \frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{j}=0 \ \ \ \ \ (11)$

We can put this in 4-vector form if we use (for some 3-vector ${\mathbf{A}}$):

$\displaystyle \nabla\cdot\mathbf{A}=-\partial^{i}a_{i} \ \ \ \ \ (12)$

where the implied sum over ${i}$ is from ${i=1}$ to ${i=3}$ (spatial coordinates), and the minus sign appears because we’ve raised the index on ${\partial^{i}}$. If we define

$\displaystyle j_{i}=i\left(\phi^{\dagger}\partial_{i}\phi-\phi\partial_{i}\phi^{\dagger}\right) \ \ \ \ \ (13)$

(that is, the negative of 10), then ${\nabla\cdot\mathbf{j}=\partial^{i}j_{i}}$. To make ${j_{\mu}}$ into a 4-vector, we add ${j_{0}=\rho}$ and we get

$\displaystyle \frac{\partial j_{0}}{\partial t}+\partial^{i}j_{i}=\partial^{\mu}j_{\mu}=0 \ \ \ \ \ (14)$

[Note that my definition of ${j_{i}}$ is the negative of the middle term in Klauber’s equation 3-21, although raising the ${i}$ index agrees with the last term in 3-21. I can’t see how his middle and last equations for ${j_{i}}$ and ${j^{i}}$ can both be right, since raising the ${i}$ in the middle equation for ${j_{i}}$ merely raises the ${\phi_{,i}}$ to ${\phi^{,i}}$ without changing the sign.]

The curious thing about the Klein-Gordon equation is that its probability density ${\rho}$ in 9 need not be positive, depending on the values of ${\phi}$ and its time derivative. To see how this can affect the physical meaning of the equation, consider the general plane wave solution to the Klein-Gordon equation

$\displaystyle \phi=\sum_{\mathbf{k}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(A_{\mathbf{k}}e^{-ikx}+B_{\mathbf{k}}^{\dagger}e^{ikx}\right) \ \ \ \ \ (15)$

Klauber explores this starting with his equation 3-24, where he takes a test solution in which all ${B_{\mathbf{k}}^{\dagger}=0}$ and shows that ${\int\rho\;d^{3}x=\sum_{\mathbf{k}}\left|A_{\mathbf{k}}\right|^{2}=1}$ so that in this case, the total probability of finding the system in some state is +1 as it should be. Let’s see what happens if we take all ${A_{\mathbf{k}}=0}$. In that case, 9 becomes

 $\displaystyle \rho$ $\displaystyle =$ $\displaystyle i\left[\sum_{\mathbf{k}}\frac{B_{\mathbf{k}}}{\sqrt{2\omega_{\mathbf{k}}V}}e^{-ikx}\right]\left[\sum_{\mathbf{k}^{\prime}}\frac{i\omega_{\mathbf{k}^{\prime}}B_{\mathbf{k}^{\prime}}^{\dagger}}{\sqrt{2\omega_{\mathbf{k}^{\prime}}V}}e^{ik^{\prime}x}\right]-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle i\left[\sum_{\mathbf{k}^{\prime}}\frac{B_{\mathbf{k}^{\prime}}^{\dagger}}{\sqrt{2\omega_{\mathbf{k}^{\prime}}V}}e^{ik^{\prime}x}\right]\left[\sum_{\mathbf{k}}\frac{-i\omega_{\mathbf{k}}B_{\mathbf{k}}}{\sqrt{2\omega_{\mathbf{k}}V}}e^{-ikx}\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left[\sum_{\mathbf{k}}\frac{B_{\mathbf{k}}}{\sqrt{2\omega_{\mathbf{k}}V}}e^{-ikx}\right]\left[\sum_{\mathbf{k}^{\prime}}\frac{\omega_{\mathbf{k}^{\prime}}B_{\mathbf{k}^{\prime}}^{\dagger}}{\sqrt{2\omega_{\mathbf{k}^{\prime}}V}}e^{ik^{\prime}x}\right]-\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\sum_{\mathbf{k}^{\prime}}\frac{B_{\mathbf{k}^{\prime}}^{\dagger}}{\sqrt{2\omega_{\mathbf{k}^{\prime}}V}}e^{ik^{\prime}x}\right]\left[\sum_{\mathbf{k}}\frac{\omega_{\mathbf{k}}B_{\mathbf{k}}}{\sqrt{2\omega_{\mathbf{k}}V}}e^{-ikx}\right] \ \ \ \ \ (18)$

We now wish to calculate ${\int\rho\;d^{3}x}$. We can use the orthonormality of solutions to do the integral. We have

$\displaystyle \frac{1}{V}\int e^{i\left(k^{\prime}-k\right)x}d^{3}x=\delta_{\mathbf{k},\mathbf{k}^{\prime}} \ \ \ \ \ (19)$

We get

 $\displaystyle -\int\left[\sum_{\mathbf{k}}\frac{B_{\mathbf{k}}}{\sqrt{2\omega_{\mathbf{k}}V}}e^{-ikx}\right]\left[\sum_{\mathbf{k}^{\prime}}\frac{\omega_{\mathbf{k}^{\prime}}B_{\mathbf{k}^{\prime}}^{\dagger}}{\sqrt{2\omega_{\mathbf{k}^{\prime}}V}}e^{ik^{\prime}x}\right]d^{3}x$ $\displaystyle =$ $\displaystyle -\sum_{\mathbf{k}}\frac{\left|B_{\mathbf{k}}\right|^{2}}{2}\ \ \ \ \ (20)$ $\displaystyle -\int\left[\sum_{\mathbf{k}^{\prime}}\frac{B_{\mathbf{k}^{\prime}}^{\dagger}}{\sqrt{2\omega_{\mathbf{k}^{\prime}}V}}e^{ik^{\prime}x}\right]\left[\sum_{\mathbf{k}}\frac{\omega_{\mathbf{k}}B_{\mathbf{k}}}{\sqrt{2\omega_{\mathbf{k}}V}}e^{-ikx}\right]d^{3}x$ $\displaystyle =$ $\displaystyle -\sum_{\mathbf{k}}\frac{\left|B_{\mathbf{k}}\right|^{2}}{2}\ \ \ \ \ (21)$ $\displaystyle \int\rho\;d^{3}x$ $\displaystyle =$ $\displaystyle -\sum_{\mathbf{k}}\left|B_{\mathbf{k}}\right|^{2} \ \ \ \ \ (22)$

Thus the total probability of finding the system in one of the state ${\mathbf{k}}$ is negative.

# Probability current in 3-d

Required math: calculus

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.41.

We’ve seen the concept of the probability current in one dimension. A generalization to 3-d gives the vector

$\displaystyle \mathbf{J}=\frac{i\hbar}{2m}(\Psi\nabla\Psi^*-\Psi^*\nabla\Psi) \ \ \ \ \ (1)$

We can use the product rule from vector calculus to get

 $\displaystyle \nabla\cdot\mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2m}(|\nabla\Psi|^{2}+\Psi\nabla^{2}\Psi^*-|\nabla\Psi|^{2}-\Psi^*\nabla^{2}\Psi)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2m}(\Psi\nabla^{2}\Psi^*-\Psi^*\nabla^{2}\Psi)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Psi\left(-\frac{\partial\Psi^*}{\partial t}+\frac{i}{\hbar}\Psi^*V\right)-\Psi^*\left(\frac{\partial\Psi}{\partial t}+\frac{i}{\hbar}\Psi V\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial}{\partial t}|\Psi|^{2} \ \ \ \ \ (5)$

where in the third line we used the Schrodinger equation rearranged to:

$\displaystyle \frac{i\hbar}{2m}\nabla^{2}\Psi=\frac{\partial\Psi}{\partial t}+\frac{i}{\hbar}\Psi V \ \ \ \ \ (6)$

Again generalizing the 1-d case, if we integrate the outward component of ${\mathbf{J}}$ over the surface area of a volume, this gives us the net rate at which probability ‘flows’ across the surface, and should, therefore, be equal to the rate at which probability changes within the volume. From the divergence theorem, we know that

 $\displaystyle \int_{A}\mathbf{J}\cdot d\mathbf{a}$ $\displaystyle =$ $\displaystyle \int_{V}\nabla\cdot\mathbf{J}d^{3}\mathbf{r}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{d}{dt}\int_{V}\left|\Psi\right|^{2}d^{3}\mathbf{r} \ \ \ \ \ (8)$

Since ${\left|\Psi\right|^{2}}$ is the probability density, this equation expresses exactly this idea.

As an example, we can look at the ${\psi_{211}}$ state of the hydrogen atom. We worked out this function earlier, so we quote the result here:

 $\displaystyle \psi_{211}$ $\displaystyle =$ $\displaystyle -\frac{1}{8a^{2}\sqrt{\pi a}}re^{-r/2a}\sin\theta e^{i\phi} \ \ \ \ \ (9)$

The gradient in spherical coordinates is

$\displaystyle \nabla=\hat{\mathbf{r}}\frac{\partial}{\partial r}+\frac{1}{r}\hat{\mathbf{\theta}}\frac{\partial}{\partial\theta}+\frac{1}{r\sin\theta}\hat{\mathbf{\phi}}\frac{\partial}{\partial\phi} \ \ \ \ \ (10)$

so we can calculate ${\nabla\psi_{211}}$:

$\displaystyle \nabla\psi_{211}=\frac{-1}{8a^{2}\sqrt{\pi a}}e^{-r/2a}e^{i\phi}\left[\left(1-\frac{r}{2a}\right)\sin\theta\hat{\mathbf{r}}+\cos\theta\hat{\mathbf{\theta}}+i\hat{\mathbf{\phi}}\right] \ \ \ \ \ (11)$

We can now calculate ${\mathbf{J}}$ from 1. The contributions along the ${\hat{\mathbf{r}}}$ and ${\hat{\mathbf{\theta}}}$ directions cancel out, so we are left with

 $\displaystyle \mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2m}\left[-\frac{1}{8a^{2}\sqrt{\pi a}}re^{-r/2a}\sin\theta e^{i\phi}\frac{i}{8a^{2}\sqrt{\pi a}}e^{-r/2a}e^{-i\phi}-\frac{1}{8a^{2}\sqrt{\pi a}}re^{-r/2a}\sin\theta e^{-i\phi}\frac{i}{8a^{2}\sqrt{\pi a}}e^{-r/2a}e^{i\phi}\right]\hat{\mathbf{\phi}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{64\pi ma^{5}}re^{-r/a}\sin\theta\hat{\mathbf{\phi}} \ \ \ \ \ (13)$

From 1, we see that the dimensions of ${\mathbf{J}}$ are those of velocity, since ${\hbar}$ has dimensions of ${\mbox{energy}\times\mbox{time}=\mbox{mass}\times\mbox{length}^{2}\times\mbox{time}^{-1}}$ so ${\frac{\hbar}{m}\nabla}$ has dimensions of ${\mbox{length}\times\mbox{time}^{-1}}$. The quantity ${m\mathbf{J}}$ therefore has dimensions of momentum, and can be interpreted as the flow of mass. This makes sense, since the probability density is the probability density of finding the particle within a given infintesimal volume, and the particle consists of its mass.

From ${\mathbf{L}=\mathbf{r}\times\mathbf{p}}$, we can therefore write the (orbital, ignoring spin) angular momentum as

$\displaystyle \mathbf{L}=\int_{V}\mathbf{r}\times\left(m\mathbf{J}\right)d^{3}\mathbf{r} \ \ \ \ \ (14)$

Since ${\mathbf{J}}$ is parallel to ${\hat{\mathbf{\phi}}}$ and ${\hat{\mathbf{r}}\times\hat{\mathbf{\phi}}=-\hat{\mathbf{\theta}}}$, we have for the hydrogen atom above:

$\displaystyle \mathbf{r}\times\mathbf{J}=\frac{-\hbar}{64\pi ma^{5}}e^{-r/a}r^{2}\sin\theta\hat{\mathbf{\theta}} \ \ \ \ \ (15)$

If we want to find ${L_{z}}$, we need the component of ${\mathbf{r}\times\mathbf{J}}$ along the unit vector ${\hat{\mathbf{k}}}$. The component of ${\hat{\mathbf{\theta}}}$ along ${\hat{\mathbf{k}}}$ is ${-\sin\theta\hat{\mathbf{k}}}$ so to find ${L_{z}}$ we must do the integral

 $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle m\int|\mathbf{r}\times\mathbf{J}|\sin\theta d^{3}\mathbf{r}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{64\pi a^{5}}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}r^{4}e^{-r/a}\sin^{3}\theta drd\theta d\phi\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar \ \ \ \ \ (18)$

where we used Maple to do the integral. This is correct for the state ${\psi_{211}}$ since for quantum number (not mass) ${m=+1}$ we would expect ${L_{z}=+\hbar}$.

# Finite step potential – scattering

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.34.

A variant of the finite square well is the finite step, which has the potential

$\displaystyle V(x)=\begin{cases} 0 & x\le0\\ V_{0} & x>0 \end{cases} \ \ \ \ \ (1)$

where ${V_{0}}$ is a positive constant energy.

There are two distinct cases here:

1. Energy below the barrier: ${0\le E\le V_{0}}$
2. Energy greater than the barrier: ${E>V_{0}}$

We’ll consider first the case where ${0\le E\le V_{0}}$.

In this case, the Schrödinger equation for ${x>0}$ is

 $\displaystyle -\frac{\hbar^{2}}{2m}\psi''+V_{0}\psi$ $\displaystyle =$ $\displaystyle E\psi\ \ \ \ \ (2)$ $\displaystyle \psi''$ $\displaystyle =$ $\displaystyle \mu^{2}\psi \ \ \ \ \ (3)$

where

$\displaystyle \mu=\frac{\sqrt{2m(V_{0}-E)}}{\hbar} \ \ \ \ \ (4)$

This has solution

$\displaystyle \psi(x)=Ce^{\mu x}+De^{-\mu x} \ \ \ \ \ (5)$

To keep the solution finite as ${x\rightarrow\infty}$ we must have ${C=0}$ so the solution is an exponentially decaying wave function:

$\displaystyle \psi(x)=De^{-\mu x} \ \ \ \ \ (6)$

To the left of the barrier, the Schrödinger equation is

$\displaystyle \psi''=-\frac{2mE}{\hbar^{2}}\psi \ \ \ \ \ (7)$

Assuming particles coming in from the left, we have

$\displaystyle \psi(x)=Ae^{ikx}+Be^{-ikx} \ \ \ \ \ (8)$

where

$\displaystyle k=\frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (9)$

Since the potential is finite everywhere, both ${\psi}$ and ${\psi'}$ are continuous everywhere, which gives us two boundary conditions at ${x=0}$.

 $\displaystyle A+B$ $\displaystyle =$ $\displaystyle D\ \ \ \ \ (10)$ $\displaystyle ik\left(A-B\right)$ $\displaystyle =$ $\displaystyle -\mu D \ \ \ \ \ (11)$

This has solution

 $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{ik+\mu}{ik-\mu}A\ \ \ \ \ (12)$ $\displaystyle D$ $\displaystyle =$ $\displaystyle \frac{2ik}{ik-\mu}A \ \ \ \ \ (13)$

The reflection coefficient is

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\left|B\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (15)$

That is, the probability of an incoming particle being reflected is 1. This is because the wave function for ${x>0}$ is exponentially decaying, so the probability of a particle reaching infinity is zero, thus no particles can be transmitted.

For ${E>V_{0}}$ the Schrödinger equation for ${x>0}$ is

 $\displaystyle -\frac{\hbar^{2}}{2m}\psi''+V_{0}\psi$ $\displaystyle =$ $\displaystyle E\psi\ \ \ \ \ (16)$ $\displaystyle \psi''$ $\displaystyle =$ $\displaystyle -\kappa^{2}\psi \ \ \ \ \ (17)$

where

$\displaystyle \kappa=\frac{\sqrt{2m(E-V_{0})}}{\hbar} \ \ \ \ \ (18)$

We now get travelling wave solutions instead of exponentially decaying ones:

$\displaystyle \psi(x)=Ce^{i\kappa x}+De^{-i\kappa x} \ \ \ \ \ (19)$

Assuming incoming particles arrive only from the left, we can set ${D=0}$. Applying the boundary conditions, we get

 $\displaystyle A+B$ $\displaystyle =$ $\displaystyle C\ \ \ \ \ (20)$ $\displaystyle ik\left(A-B\right)$ $\displaystyle =$ $\displaystyle i\kappa C \ \ \ \ \ (21)$

with solutions

 $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{k-\kappa}{k+\kappa}A\ \ \ \ \ (22)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{2k}{k+\kappa}A \ \ \ \ \ (23)$

In this case, the reflection coefficient is

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\left|B\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{k-\kappa}{k+\kappa}\right)^{2} \ \ \ \ \ (25)$

Substituting the expressions for ${k}$ and ${\kappa}$ we get

$\displaystyle R=\left[\frac{E-\sqrt{E\left(E-V_{0}\right)}}{E+\sqrt{E\left(E-V_{0}\right)}}\right]^{2} \ \ \ \ \ (26)$

From this we can get the transmission coefficient

 $\displaystyle T$ $\displaystyle =$ $\displaystyle 1-R\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4E^{3/2}\sqrt{E-V_{0}}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}} \ \ \ \ \ (28)$

Note that this is not equal to ${\left|C\right|^{2}/\left|A\right|^{2}=4E^{2}/\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}}$. Have we done something wrong?

The answer lies in the fact that the wave for ${x>0}$ is not the same as the wave for ${x<0}$, since the net energy on the right is ${E-V_{0}}$ while on the left it is just ${E}$. One way of looking at it is in terms of the probability current for the free particle. The probability current must be conserved; this is just a way of saying that particles cannot vanish into, nor arise from, thin air. Since the probability current for a free particle with stationary state

$\displaystyle \Psi(x,t)=Ae^{ikx}e^{-i\hbar k^{2}t/2m} \ \ \ \ \ (29)$

is

$\displaystyle J=\frac{\hbar k}{m}\left|A\right|^{2} \ \ \ \ \ (30)$

the conservation law implies, for the case of the step potential

$\displaystyle \frac{\hbar k}{m}\left[\left|A\right|^{2}-\left|B\right|^{2}\right]=\frac{\hbar\kappa}{m}\left|C\right|^{2} \ \ \ \ \ (31)$

That is, the influx of particles from the left minus the reflected beam must equal the transmitted beam. Dividing through by ${\frac{\hbar k}{m}\left|A\right|^{2}}$ we get

$\displaystyle \frac{\left|B\right|^{2}}{\left|A\right|^{2}}+\frac{\kappa}{k}\frac{\left|C\right|^{2}}{\left|A\right|^{2}}=1 \ \ \ \ \ (32)$

The first term is the reflection coefficient we calculated in 26. The second term is the transmission coefficient, which works out to

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E-V_{0}}{E}}\frac{\left|C\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E-V_{0}}{E}}\frac{4E^{2}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4E^{3/2}\sqrt{E-V_{0}}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}} \ \ \ \ \ (35)$

which is what we got earlier.