# 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.

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