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.

4 thoughts on “Klein-Gordon equation: probability density and current

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