Consider the Klein-Gordon equation: \begin{equation} \frac{\partial^2 \psi}{\partial t^2} = c^2 \Delta \psi - \frac{m^2 c^4}{\hbar^2} \psi, \end{equation} and define for each one of its solutions $\psi$ the quantity:

\begin{equation} P(\mathbf{x},t)= \hbar^2 \frac{\partial \psi}{ \partial t} \frac{\partial \psi^{*}}{ \partial t} + \hbar^2 c^2 \nabla \psi \cdot \nabla \psi^{*} + m^2 c^4 \psi \psi^{*}, \end{equation} Let us adopt the convention in which the generic point of the Minkowski space-time is $(x,y,z,ct)$. In Section (4.6) of his wonderful treatise Quantum Theory Bohm states that under a Lorentz transformation

(i) $P(\mathbf{x},t)$ transforms as the (4,4)-coordinate a rank two tensor,

(ii) $\int P(\mathbf{x},t) d\mathbf{x}$ transforms as the fourth component of a four-vector.

Could someone give me a proof of these two statements, please?

NOTE (1). All that I know about the Klein-Gordon equation is that $\psi$ is invariant under Lorentz transformations, that is if $\psi(\mathbf{x},t)$ is a solution of the Klein-Gordon equation, then the new function $\phi(\mathbf{x'},t')$ obtained by replacing the equations of a Lorentz boost $(\mathbf{x},ct) \rightarrow (\mathbf{x'},ct')$ in $\psi$ is again a solution of the Klein-Gordon equation.

NOTE (2). Bohm justifies assertion (i) by considering the particular solution $\psi= \exp i \left( \frac{Et-\mathbf{p} \cdot \mathbf{x} } {\hbar} \right)$, for which we get \begin{equation} P=E^2+p^2c^2+m^2c^4=2E^2, \end{equation} so that $P$ transforms actually as the square of an energy.

  • 1
    $\begingroup$ Double-check your signs, then see if you can spot a tensor $A_{\mu\nu}$ for which $P=A_{tt}$. $\endgroup$
    – J.G.
    Commented Oct 16, 2018 at 18:04
  • $\begingroup$ Sorry, what do you mean? $\endgroup$ Commented Oct 16, 2018 at 20:20

1 Answer 1


Actually, the $P(\mathbf{x},t)$ you mention is the 4,4-component of the stress-tensor $T_{ik}$ of the Klein-Gordon (K-G) field. In the following instead I will use the metric tensor $\eta_{ik}=diag(1,-1,-1,-1)$ and identify $P(\mathbf{x},t)$ with the 0,0-component of $T_{ik}$. Bohm aparently uses the other metric $diag(-1,-1,-1,1)$ convention. Moreover $c=1=\hbar$ is assumed.

One starts best off from the Lagrange density of the complex K-G: (Double appearing indices is summed over, i.e. Einstein summation convention):

$$ L = \partial_i \phi^\star \partial^i \phi - m^2 \phi^\star \phi$$

For a complex field $\phi$ and $\phi^\star$ are considered as independent variables. The definition of the stress tensor is given by:

$$T_{ik}=\sum_{\varphi}\varphi_{,k}\frac{\partial L}{\partial \varphi^{,i}} -L\eta_{ik}$$

with $\varphi = (\phi, \phi^\star)$.

Upon inserting the expression for the Lagrange density of the K-G field in the stress tensor's definition we get:

$$T_{ik} = \partial_i\phi^\star \partial_k\phi + \partial_k\phi^\star \partial_i\phi -L\eta_{ik}$$

The tensor property of $T_{ik}$ is rather obvious, as the partial derivatives $\partial_i$ respectively $\partial_k$ transform like covariant vectors and $\eta_{ik}$ is also a tensor. This is in particular true for the $(i,k)=(0,0)$-component:

$$T_{00} = 2\frac{\partial\phi^\star}{\partial t}\frac{\partial\phi}{\partial t} -L=\frac{\partial\phi^\star}{\partial t}\frac{\partial\phi}{\partial t}+\nabla\phi^\star\nabla\phi + m^2\phi^\star \phi\equiv P(\mathbf{x},t) $$

The 4-momentum vector $P_i$ yields:

$$P_i = \int_{\partial\Omega} T_{ik} d\sigma^k$$

where $d\sigma^k$ is the vectorial hypersurface element which is parametrized by values $(u,v,w)$:

$$d\sigma_i =\epsilon_{ijkm}\frac{\partial x^j}{\partial u}\frac{\partial x^k}{\partial v}\frac{\partial x^m}{\partial w}du dv dw$$

If now the hypersurface $t=const$ is chosen, as parameters $(u,v,w)=(x,y,z)\equiv(x^1,x^2,x^3)$ can be used:

$$d\sigma_i =\epsilon_{ijkm}\delta^j_1\delta^k_2\delta^m_3 dx^1 dx^2 dx^3= \epsilon_{i,1,2,3}d^3x = (d^3x,\mathbf{0})$$

so we get for this particular hypersurface $t=const$:

$$P_i = \int_{t=const} T_{i0} d^3x $$

It can be shown that $P_i$ considered at another hypersurface $\partial\Omega$ which is Lorentz-transformed with respect to the original hypersurface $t=const$ has the same value if it is computed by the more general formula:

$$P_i = \int_{\partial\Omega} T_{ik} d\sigma^k$$

But due to the covariant way of writing it is clear that $P_i$ is a 4-vector (but this would not be longer true in curved space-time) and in particular

$$P_0 = \int_{t=const} T_{00} d^3x \equiv \int_{t=const}P(\mathbf{x},t) d^3x $$

the 0-component of the 4-vector $P_i$ (the momentum 4-vector), the energy of the K-G field.

  • $\begingroup$ Dear Frederic thank you very very ... much for having answered my question: I could never have done it by myself! I add here the only missing step needed to justify the claim that $P_i$ can be computed by the general formula you have given above. Let us note that $T_{ik}$ has zero divergence, that is $\partial^{k} T_{ik}=0$, as it is immediately checked by using the Klein-Gordon equation for $\phi$ and $\phi^{*}$. Then your claim follows from the general argument given in Møller, Theory of Relativity, $\S 63$ or Pauli, Theory of Relativity, $\S 21$. $\endgroup$ Commented Oct 17, 2018 at 9:37
  • $\begingroup$ You are welcome. The claim you refer to in your comment is it that $P_i$ has the same value if the hypersurface $t=const$ is Lorentz-transformed to another hypersurface $\partial\Omega$ when computed with the more general formula? $\endgroup$ Commented Oct 19, 2018 at 13:50
  • $\begingroup$ It is exactly that, Frederic. $\endgroup$ Commented Oct 19, 2018 at 15:17

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