# 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.

# Stress-energy tensor: conservation equations

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 20; Box 20.5.

We can express conservation of energy and momentum in terms of the stress-energy tensor. Recall that the physical meaning of the component ${T^{tt}}$ is the energy density.

To get the conservation laws, consider a small box with dimensions ${dx}$, ${dy}$ and ${dz}$, and restrict our attention to the case of ‘dust’, that is, a fluid containing particles that are all at rest relative to each other. In that case, the tensor has the form

$\displaystyle T^{ij}=\rho_{0}u^{i}u^{j} \ \ \ \ \ (1)$

where ${\rho_{0}}$ is the energy density of the dust in its own rest frame and ${u^{i}}$ is the four-velocity of the fluid as measured in some observer’s local inertial frame. Then the flow of energy into the box over, say, the left-hand face perpendicular to the ${x}$ axis at position ${x}$ in time ${dt}$ is the energy density multiplied by the velocity component in the ${x}$ direction ${v^{x}}$.

$\displaystyle dE_{x}=\left(T_{x}^{tt}v^{x}dt\right)dydz \ \ \ \ \ (2)$

We’ve multiplied by ${dydz}$ since this is the area of the face of the box through which the energy is flowing, and thus the total flow of energy is the density multiplied by the volume that crosses the box’s face, which is ${v^{x}dtdydz}$. The subscript ${x}$ on ${T_{x}^{tt}}$ means that the tensor is evaluated at position ${x}$. We have

$\displaystyle T^{tt}=\rho_{0}u^{t}u^{t} \ \ \ \ \ (3)$

and ${u^{t}v^{x}=\gamma v^{x}=u^{x}}$, so ${T^{tt}v^{x}=T^{tx}}$ using 1, and thus

$\displaystyle dE_{x}=\left(T_{x}^{tx}dt\right)dydz \ \ \ \ \ (4)$

Similarly, the energy flowing across the face at position ${x+dx}$ is then

$\displaystyle dE_{x+dx}=\left(T_{x+dx}^{tx}dt\right)dydz \ \ \ \ \ (5)$

Taking the difference of these two equations we get

$\displaystyle \left(T_{x+dx}^{tx}-T_{x}^{tx}\right)dtdydz=\partial_{x}T^{tx}dxdtdydz \ \ \ \ \ (6)$

We can write similar equations for the ${y}$ and ${z}$ directions:

 $\displaystyle \left(T_{y+dy}^{ty}-T_{y}^{ty}\right)dtdxdz$ $\displaystyle =$ $\displaystyle \partial_{y}T^{ty}dxdtdydz\ \ \ \ \ (7)$ $\displaystyle \left(T_{z+dz}^{tz}-T_{z}^{tz}\right)dtdydx$ $\displaystyle =$ $\displaystyle \partial_{z}T^{tz}dxdtdydz \ \ \ \ \ (8)$

Adding these up gives the net total change in energy within the boxes

 $\displaystyle dE$ $\displaystyle =$ $\displaystyle -\left(\partial_{x}T^{tx}+\partial_{y}T^{ty}+\partial_{z}T^{tz}\right)dxdydzdt \ \ \ \ \ (9)$

The minus sign occurs because if, say, ${\partial_{x}T^{tx}<0}$, this indicates that ${T_{x}^{tx}>T_{x+dx}^{tx}}$ so more energy flows in at position ${x}$ than flows out at position ${x+dx}$, resulting in ${dE>0}$.

The net change in energy due to its flow across the boundaries of the box must be reflected in the change of the energy within the box. The energy density is given by ${T^{tt}}$ so we must have

 $\displaystyle dE$ $\displaystyle =$ $\displaystyle \left(T_{t+dt}^{tt}-T_{t}^{tt}\right)dxdydz\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{t}T^{tt}dxdydzdt \ \ \ \ \ (11)$

The energy conservation law is then given by

 $\displaystyle \partial_{t}T^{tt}dxdydzdt$ $\displaystyle =$ $\displaystyle -\left(\partial_{x}T^{tx}+\partial_{y}T^{ty}+\partial_{z}T^{tz}\right)dxdydzdt \ \ \ \ \ (12)$

Since this must be true for any choice of differentials, the energy conservation law is expressed in the compact form

$\displaystyle \partial_{j}T^{tj}=0 \ \ \ \ \ (13)$

We can do a similar argument for momentum. The component ${T^{tj}}$ (where ${j}$ is a spatial coordinate) is the density of the ${j}$ component of momentum and the components ${T^{ij}}$ are the rates of flow of the ${j}$ component of momentum in the ${i}$ direction, so the net change in momentum component ${j}$ due to differences in the flow rate at the boundaries of the box is

$\displaystyle dp^{j}=-\left(\partial_{x}T^{xj}+\partial_{y}T^{yj}+\partial_{z}T^{zj}\right)dxdydzdt \ \ \ \ \ (14)$

This must be equal to the net change of ${p^{j}}$ within the box over time ${dt}$, so

$\displaystyle dp^{j}=\partial_{t}T^{tj}dxdydzdt \ \ \ \ \ (15)$

and

$\displaystyle \partial_{i}T^{ij}=0 \ \ \ \ \ (16)$

This is therefore true for all four values of ${j}$ and represents conservation of energy and momentum, or just four-momentum.

We derived this formula for the special case of a local inertial frame (LIF). We’ve seen that the appropriate generalization of the gradient is the absolute gradient or covariant derivative, so the appropriate tensor equation for conservation of four-momentum is

$\displaystyle \boxed{\nabla_{i}T^{ij}=0} \ \ \ \ \ (17)$

In terms of Christoffel symbols, this is

$\displaystyle \nabla_{i}T^{ij}=\partial_{i}T^{ij}+\Gamma_{ik}^{i}T^{kj}+\Gamma_{ik}^{j}T^{ik}=0 \ \ \ \ \ (18)$

We can apply this equation to the more general case of a perfect fluid in general coordinates, where the tensor is

$\displaystyle T^{ij}=\left(\rho_{0}+P_{0}\right)u^{i}u^{j}+P_{0}g^{ij} \ \ \ \ \ (19)$

We can work out the covariant derivative in a LIF. In a LIF, all Christoffel symbols are zero so we get

 $\displaystyle \nabla_{i}T^{ij}$ $\displaystyle =$ $\displaystyle \partial_{i}T^{ij}\ \ \ \ \ (20)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle u^{i}u^{j}\partial_{i}\left(\rho_{0}+P_{0}\right)+\left(\rho_{0}+P_{0}\right)\left[u^{i}\partial_{i}u^{j}+u^{j}\partial_{i}u^{i}\right]+\eta^{ij}\partial_{i}P_{0} \ \ \ \ \ (21)$

The four-velocity always satisfies the relation ${\mathbf{u}\cdot\mathbf{u}=u_{j}u^{j}=-1}$ so we have

 $\displaystyle \partial_{i}\left(\mathbf{u}\cdot\mathbf{u}\right)$ $\displaystyle =$ $\displaystyle \partial_{i}\left(u^{j}u_{j}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{i}\left(\eta_{jk}u^{k}u^{j}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{jk}\left[u^{k}\partial_{i}u^{j}+u^{j}\partial_{i}u^{k}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle u_{j}\partial_{i}u^{j}+u_{k}\partial_{i}u^{k}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2u_{j}\partial_{i}u^{j}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (27)$

We can now multiply 21 by ${u_{j}}$ and use the above result to get

 $\displaystyle u^{i}u_{j}u^{j}\partial_{i}\left(\rho_{0}+P_{0}\right)+\left(\rho_{0}+P_{0}\right)\left[u^{i}u_{j}\partial_{i}u^{j}+u_{j}u^{j}\partial_{i}u^{i}\right]+\eta^{ij}u_{j}\partial_{i}P_{0}$ $\displaystyle =$ $\displaystyle -u^{i}\partial_{i}\left(\rho_{0}+P_{0}\right)-\left(\rho_{0}+P_{0}\right)\partial_{i}u^{i}+u^{i}\partial_{i}P_{0} \ \ \ \ \ (28)$ $\displaystyle =$ $\displaystyle 0$ $\displaystyle \left(\rho_{0}+P_{0}\right)\partial_{i}u^{i}+u^{i}\partial_{i}\rho_{0} \ \ \ \ \ (29)$ $\displaystyle =$ $\displaystyle 0$ $\displaystyle \partial_{i}\left(u^{i}\rho_{0}\right)+P_{0}\partial_{i}u^{i} \ \ \ \ \ (30)$ $\displaystyle =$ $\displaystyle 0$

This last equation is known as the equation of continuity. Note that it is valid only in a LIF, since the derivative isn’t covariant.

Now we can multiply 29 by ${u^{j}}$and subtract it from 21:

 $\displaystyle u^{i}u^{j}\partial_{i}P_{0}+\left(\rho_{0}+P_{0}\right)u^{i}\partial_{i}u^{j}+\eta^{ij}\partial_{i}P_{0}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (31)$ $\displaystyle \left(\rho_{0}+P_{0}\right)u^{i}\partial_{i}u^{j}$ $\displaystyle =$ $\displaystyle -\left(u^{i}u^{j}+\eta^{ij}\right)\partial_{i}P_{0} \ \ \ \ \ (32)$

This is the equation of motion, also valid in a LIF.

In the non-relativistic limit, the density in any LIF will be the same, as will the pressure. Also, ${P_{0}\ll\rho_{0}}$ so we can approximate 30 by

$\displaystyle \partial_{i}\left(u^{i}\rho_{0}\right)\approx0 \ \ \ \ \ (33)$

Using ${\mathbf{u}\approx\left[1,v^{x},v^{y},v^{z}\right]}$ this becomes

$\displaystyle \partial_{t}\rho_{0}=-\vec{\nabla}\cdot\left(\rho_{0}\mathbf{v}\right) \ \ \ \ \ (34)$

where the arrow above ${\vec{\nabla}}$ indicates this is the 3-d gradient, not the covariant derivative. This is the Newtonian equation of continuity for a perfect fluid, which expresses conservation of mass.

We can also approximate 32 by neglecting any products of velocity components, since ${v^{i}v^{j}\ll1}$ if both ${i}$ and ${j}$ are spatial coordinates. The LHS becomes

 $\displaystyle \left(\rho_{0}+P_{0}\right)u^{i}\partial_{i}u^{j}$ $\displaystyle \approx$ $\displaystyle \rho_{0}u^{i}\partial_{i}u^{j}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho_{0}\left[\partial_{t}v^{j}+\left(\vec{\mathbf{v}}\cdot\vec{\nabla}\right)v^{j}\right] \ \ \ \ \ (36)$

The term with ${j=t}$ drops out, since ${u^{t}=1}$ and its derivatives are all zero. We can combine the three spatial coordinates into a single vector expression:

$\displaystyle \rho_{0}\left[\partial_{t}\vec{\mathbf{v}}+\left(\vec{\mathbf{v}}\cdot\vec{\nabla}\right)\vec{\mathbf{v}}\right] \ \ \ \ \ (37)$

The RHS is

$\displaystyle -\left(u^{i}u^{j}+\eta^{ij}\right)\partial_{i}P_{0} \ \ \ \ \ (38)$

If ${j=t}$, the ${i=t}$ term in the sum is zero because ${u^{t}u^{t}+\eta^{ij}=+1-1=0}$. If we ignore all other terms that are second order or higher in ${v}$ and/or ${P_{0}}$, we are left with only ${-\eta^{ij}\partial_{i}P_{0}}$. Looking at the 3 terms with ${j}$ being a spatial coordinate, this is ${-\vec{\nabla}P_{0}}$ so we get the approximation

$\displaystyle \rho_{0}\left[\partial_{t}\vec{\mathbf{v}}+\left(\vec{\mathbf{v}}\cdot\vec{\nabla}\right)\vec{\mathbf{v}}\right]=-\vec{\nabla}P_{0} \ \ \ \ \ (39)$

which is Euler’s equation of motion for a perfect fluid.