1
$\begingroup$

David Bohm in his wonderful monograph Quantum Theory, in Section 4.6 discusses the difficulties one encounters in trying to develop a relativistic quantum mechanics. He starts from the relation \begin{equation} \hbar^2 \omega^2 = m^2 c^4 + \hbar^2 k^2 c^4 \end{equation} (which is equivalent to the classical relation $E^2=m^2 c^4 + p^2 c^2$), from which one derives (by proceeding as in Section 3.19) the second-order equation (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} Then he tries to define a probability function $P$ involving $\psi$ and its partial drivatives $\frac{\partial \psi}{ \partial t}$, $\frac{\partial \psi}{\partial x_i}$: \begin{equation} P(x,t)= \hbar^2 \left| \frac{\partial \psi}{ \partial t} \right|^2 + \hbar^2 c^2 \lvert \nabla \psi \rvert^2 + m^2 c^4 \lvert \psi \rvert^2, \end{equation} which can be seen to have an integral $\int P(\mathbf{x},t) d\mathbf{x}$ which is conserved over time. Anyway, Bohm says that this function does not give rise to a physically acceptable probability, since if we choose e.g. $\psi= \exp i \left( \frac{Et-\mathbf{p} \cdot \mathbf{x} } {\hbar} \right)$, we get \begin{equation} P(x,t)=E^2+p^2c^2+m^2c^4=2E^2, \end{equation} so that $P$ behaves likes the (4,4)-component of a rank-2 tensor. From this he concludes that under a Lorentz transformation the integral $\int P(\mathbf{x},t) d\mathbf{x}$ transforms like an energy, that is like the fourth component of a four-vector, so it is not invariant (for a proof of the last statement see my post Tensors and the Klein-Gordon Equation).

Bohm then states without proof that it is not possible to define any (reasonable) probability density function, by using the solution $\psi$ of the wave equation above and its partial derivatives, which is invariant under Lorentz transformation.

Does someone know some compelling reason why this is true?

$\endgroup$
  • 3
    $\begingroup$ Pretty much any introductory lecture notes on QFT will explain the problems involved in using this equation. If you Google Michael Luke, David Tong or M. Srednicki, the PDF versions of their lectures all cover this in the first three chapters. $\endgroup$ – user207480 Oct 5 '18 at 15:08
2
$\begingroup$

I have discovered that Kazemi, Hashamipour and Barati in their work Probability density of relativistic spinless particles suceeded in finding in the one-dimensional case a physically acceptable probability function for the Klein-Gordon equation. This probability function satisfies all the properties of a meaningful probability function, and in particular its integral is Lorentz invariant.

Anyhow, this probability function does not disprove Bohm's statement, since $P(x,t)$ does not depend on $\psi(x,t)$ and the values of the partial derivatives of $\psi$ computed in $(x,t)$, but it is a functional of $\psi$, that is it depends on the whole function $\psi$. So this probability function is not a counterexample to Bohm's statement.

Finally, I have found a very interesting work Uniqueness of conserved currents in quantum mechanics by Peter Holland, who shows that an essentially unique conserved four-vector current $\mathbf{J}$ exists for the Klein-Gordon equation, which has covariant components

\begin{equation} J_{\mu} = \frac{i \hbar}{2m} \left( \psi^{*} \partial_{\mu} \psi - \psi \partial_{\mu} \psi^{*} \right). \end{equation}

This current corresponds to the one usually defined for the Klein-Gordon equation. Its density component is $P=\frac{i \hbar}{2m} \left(\psi^{*} \frac{\partial \psi}{\partial t} - \psi \frac{\partial \psi^{*}}{\partial t} \right) $, we see that $P$ does not satisfy the property $P \geq 0$ so that Bohm's statement follows.

We must anyhow remark that Holland assumes in his proof that that $\mathbf{J}$ depends only on $\psi$ and its first derivatives, which is an assumption not explicitly made by Bohm, even though a perfectly plausible one (see the argument given by Holland in his work to justify the requirement that conserved currents should depend solely on the 'state variables').

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.