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Source: Quantum Field Theory for the Gifted Amateur by Tom Lancaster, Stephen J. Blundell.

I am struggling to understand the logical step from the outline of the 'proof' in the footnote, to the fact that the probabilty density must look like eq. 6.12. Can anyone supply a supplemental text that walks through this more plainly? Moreover, I find my secondary source's derivation at a level above me as well.


6.2 Probability currents and densities

One of the reasons that Schr$\ddot{\rm o}$dinger wasn't happy with the Klein-Gordon equation after he'd derived it was that something rather nasty happens when you think about the flow of probability density. The probability of a particle being located somewhere depends on $\phi^{*}(x)\phi(x)$ and so if this quantity is time-dependent then particles must be sloshing around. The probability density $\rho$ and probability current density5 $\boldsymbol{j}$ obey a continuity equation \begin{equation} \dfrac{\mathrm d\rho}{\mathrm dt}\boldsymbol{+}\boldsymbol{\nabla \cdot}\boldsymbol{j}\boldsymbol{=}0, \tag{6.9}\label{6.9} \end{equation} which is more easily written in four-vector notation as \begin{equation} \partial_{\mu}j^{\mu}\boldsymbol{=}0. \tag{6.10}\label{6.10} \end{equation} If, as is usual in non-relativistic quantum mechanics,6 we take the spatial part to be \begin{equation} \boldsymbol{j}(x)\boldsymbol{=}\boldsymbol{-}\mathrm i\left[\phi^{*}(x)\boldsymbol{\nabla}\phi(x)\boldsymbol{-}\phi(x)\boldsymbol{\nabla}\phi^{*}(x)\right], \tag{6.11}\label{6.11} \end{equation} then, for eqn 6.10 to work,7 we require the probability density to look like8 \begin{equation} \rho(x)\boldsymbol{=}\mathrm i\left[\phi^{*}(x)\dfrac{\partial\phi(x)}{\partial t}\boldsymbol{-}\dfrac{\partial\phi^{*}(x)}{\partial t}\phi(x)\right]. \tag{6.12}\label{6.12} \end{equation} The resulting covariant probability current for the Klein–Gordon equation is then given by \begin{equation} j^{\mu}(x)\boldsymbol{=}\mathrm i\{\phi^{*}(x)\partial^{\mu}\phi(x)\boldsymbol{-}\left[\partial^{\mu}\phi^{*}(x)\right]\phi(x)\}, \tag{6.13}\label{6.13} \end{equation} which, as the notation suggests, is a four-vector. Substituting in our.............


$^7$ It will work, and you can prove it as follows. Take the Klein-Gordon equation (eqn 6.5) and premultiply it by $\phi^{*}(x)$. Then take the complex conjugate of eqn 6.5 and premultiply by $\phi(x)$. Subtracting these two results will give an equation of the form of eqn 6.9 with $\boldsymbol{j}$ and $\rho$ as given.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$


Secondary Source: Quantum Field Theory by Lewis H. Ryder.


...where the Schr$\ddot{\rm o}$dinger equation and its complex conjugate have been used. What are the corresponding expressions for the Klein-Gordon equation? To be properly relativistic, $\rho$ should not, as in (2.18), transform as a scalar, but as the time component of a 4-vector, whose space component is $\mathbf{j}$, given by (2.19). Then $\rho$ is given by \begin{equation} \rho(x)\boldsymbol{=}\dfrac{\mathrm i \hbar}{2m}\left(\phi^{*}\dfrac{\partial\phi}{\partial t}\boldsymbol{-}\phi\dfrac{\partial\phi^{*}}{\partial t}\right) \tag{2.20}\label{2.20} \end{equation} and with \begin{equation} j^{\mu}\boldsymbol{=}\left(\rho,\mathbf{j}\right)\boldsymbol{=}\dfrac{\mathrm i \hbar}{m}\phi^{*}\left(\overset{\boldsymbol{\leftrightarrow}}{\partial_{0}},\overset{\boldsymbol{\leftrightarrow}}{\boldsymbol{\nabla}}\right)\phi \boldsymbol{=}\dfrac{\mathrm i \hbar}{m}\phi^{*}\overset{\boldsymbol{\leftrightarrow}}{\partial^{\mu}}\,\phi \tag{2.21}\label{2.21} \end{equation} where \begin{equation} A\overset{\boldsymbol{\leftrightarrow}}{\partial^{\mu}}B\stackrel{\text{def}}{\boldsymbol{=}}\tfrac12\left[A\partial^{\mu}B\boldsymbol{-}(\partial^{\mu}A)B\right], \tag{2.22}\label{2.221} \end{equation} and we have used (2.9), we have the continuity equation \begin{equation} \partial_{\mu}j^{\mu}\boldsymbol{=}\dfrac{\mathrm i \hbar}{2m}\left(\phi^{*}\square \,\phi\boldsymbol{-}\phi\,\square\,\phi^{*}\right)\boldsymbol{=}0, \tag{2.23}\label{2.3} \end{equation} since $\phi^{*}$ also obeys the Klein-Gordon equation. Then $\rho$ and $\mathbf{j}$ are the probability density and current we want. But this immediately presents a problem, because $\rho$, given by equation (2.20), unlike expression (2.18) for the...

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  • $\begingroup$ Do you understand (6.11)? (6.12) is just the 0-th (temporal) component of the tensor equation (6.11). $\endgroup$ – Solenodon Paradoxus Feb 5 at 4:06
  • $\begingroup$ The current density can be easily derived from Noether's Theorem. If you are not aware of Noether's Theorem you can read about it on Wikipedia. Using four-vector form you can directly arrive on (6.13) . From there both (6.11) and (6.12) can be derived. $\endgroup$ – user215742 Feb 5 at 7:09
  • $\begingroup$ Great comments all! @SolenodonParadoxus I don't believe that is correct, j^mu is the 4 vector — where rho is the zeroth entry (temporal), and /vec{j} is(are) the 1st, 2nd, and 3rd entries (spacial). But your comment does make the Ryder piece clearer for me! Thank you! $\endgroup$ – Lopey Tall Feb 5 at 18:57
  • $\begingroup$ @user215742 I had not thought of that! I will work on that direction now, thank you! $\endgroup$ – Lopey Tall Feb 5 at 19:02
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The Schr$\ddot{\rm o}$dinger equation is non-relativistic and for a free particle is derived from the Hamiltonian \begin{equation} H\boldsymbol{=} \dfrac{p^2}{2m} \tag{K-01}\label{eqK-01} \end{equation} by the transcription \begin{equation} H\boldsymbol{\longrightarrow} i\hbar\dfrac{\partial}{\partial t}\quad \text{and}\quad \mathbf{p}\boldsymbol{\longrightarrow} \boldsymbol{-}i\hbar\boldsymbol{\nabla} \tag{K-02}\label{eqK-02} \end{equation} so that \begin{equation} i\hbar \dfrac{\partial \psi}{\partial t}\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\psi\boldsymbol{=} 0 \tag{K-03}\label{eqK-03} \end{equation} For a first try to derive a relativistic quantum mechanical equation we make use of the property that according to the theory of special relativity the total energy $\;E\;$ and momenta $\;(p_x,p_y,p_z)\;$ transform as components of a contravariant four-vector \begin{equation} p^\mu\boldsymbol{=}\left(p^0,p^1,p^2,p^3\right)\boldsymbol{=}\left(\dfrac{E}{c},p_x,p_y,p_z\right) \tag{K-04}\label{eqK-04} \end{equation} of invariant length \begin{equation} \sum\limits_{\mu\boldsymbol{=}0}^{3}p_{\mu} p^{\mu}\boldsymbol{\equiv}p_{\mu} p^{\mu}\boldsymbol{=}\dfrac{E^2}{c^2}\boldsymbol{-}\mathbf{p}\boldsymbol{\cdot}\mathbf{p}\boldsymbol{\equiv}m^2c^2\tag{K-05}\label{eqK-05} \end{equation} where $\;m\;$ is the rest mass of the particle and $\;c\;$ the velocity of light in vacuum.

Following this it is natural to take as the Hamiltonian of a relativistic free particle \begin{equation} H\boldsymbol{=}\sqrt{p^{2}c^2\boldsymbol{+}m^2c^4} \tag{K-06}\label{eqK-06} \end{equation} and to write for a relativistic quantum analogue of \eqref{eqK-03} \begin{equation} i\hbar \dfrac{\partial \psi}{\partial t}\boldsymbol{=}\sqrt{\boldsymbol{-}\hbar^2c^2 \nabla^{2}\boldsymbol{+}m^2c^4}\,\psi \tag{K-07}\label{eqK-07} \end{equation} Facing the problem of interpreting the square root operator on the right in eq. \eqref{eqK-07} we simplify mathematics by removing this square root operator, so that \begin{equation} \left[\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\boldsymbol{-}\nabla^{2}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\right]\psi\boldsymbol{=}0 \tag{K-08}\label{eqK-08} \end{equation} or recognized as the classical wave equation \begin{equation} \left[\square\boldsymbol{+}\left(\dfrac{mc}{\hbar}\right)^2\right]\psi\boldsymbol{=}0 \tag{K-09}\label{eqK-09} \end{equation} where(1) \begin{equation} \square\boldsymbol{\equiv}\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\boldsymbol{-}\nabla^{2}\boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\dfrac{\partial}{\partial x^\mu} \tag{K-10}\label{eqK-10} \end{equation}

Equation \eqref{eqK-09} is the Klein-Gordon equation for a free particle. With its complex conjugate we have

\begin{align} & \dfrac{1}{c^2}\dfrac{\partial^2 \psi\hphantom{^{\boldsymbol{*}}}}{\partial t^2}\boldsymbol{-}\nabla^{2}\psi\hphantom{^{\boldsymbol{*}}}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\psi\hphantom{^{\boldsymbol{*}}}\boldsymbol{=} 0 \tag{K-11.1}\label{eqK-11.1}\\ &\dfrac{1}{c^2}\dfrac{\partial^2 \psi^{\boldsymbol{*}}}{\partial t^2}\boldsymbol{-}\nabla^{2}\psi^{\boldsymbol{*}}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\psi^{\boldsymbol{*}}\boldsymbol{=} 0 \tag{K-11.2}\label{eqK-11.2} \end{align} Multiplying them by $\;\psi^{\boldsymbol{*}},\psi\;$ respectively and subtracting side by side we have(2) \begin{align} \dfrac{1}{c^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial^2 \psi}{\partial t^2}\boldsymbol{-}\psi\dfrac{\partial^2 \psi^{\boldsymbol{*}}}{\partial t^2}\right)\boldsymbol{-}\left(\psi^{\boldsymbol{*}}\nabla^{2}\psi\boldsymbol{-}\psi\nabla^{2}\psi^{\boldsymbol{*}}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)&\boldsymbol{=} 0\quad \boldsymbol{\Longrightarrow} \nonumber\\ \dfrac{1}{c^2}\dfrac{\partial}{\partial t}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\boldsymbol{+}\boldsymbol{\nabla \cdot}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)&\boldsymbol{=} 0 \tag{K-12}\label{eqK-12} \end{align} We multiply above equation by $\;i\hbar/2m\;$ in order to have real quantities on one hand and on the other hand to have an identical expression for the probability current density vector as that one from the Schr$\ddot{\rm o}$dinger equation
\begin{equation} \dfrac{\partial}{\partial t}\left[\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\right]\boldsymbol{+}\boldsymbol{\nabla \cdot}\left[\dfrac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)\right]\boldsymbol{=} 0 \tag{K-13}\label{eqK-13} \end{equation} so \begin{equation} \dfrac{\partial \varrho}{\partial t}\boldsymbol{+}\boldsymbol{\nabla \cdot}\boldsymbol{S}\boldsymbol{=} 0 \tag{K-14}\label{eqK-14} \end{equation} where \begin{equation} \boxed{\:\:\varrho\boldsymbol{\equiv}\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\:\:}\quad \text{and} \quad \boxed{\:\:\boldsymbol{S}\boldsymbol{\equiv}\dfrac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)\:\:} \tag{K-15}\label{eqK-15} \end{equation} We would like to interpret $\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)$ as a probability density $\varrho$. However, this is impossible, since it is not a positive definite expression.


(1) We define \begin{align} \blacktriangleright x^\mu\boldsymbol{=}\left(ct,\mathbf{x}\right)&\blacktriangleright \nabla^\mu\boldsymbol{=}\partial^\mu\boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\boldsymbol{=}\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\boldsymbol{-}\boldsymbol{\nabla}\right) \nonumber\\ &\blacktriangleright \nabla_\mu\boldsymbol{=}\partial_\mu\boldsymbol{=}\dfrac{\partial}{\partial x^\mu}\boldsymbol{=}\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\boldsymbol{+}\boldsymbol{\nabla}\right)\blacktriangleright\square \boldsymbol{=}\nabla^\mu\nabla_\mu \boldsymbol{=}\partial^\mu\partial_\mu \boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\dfrac{\partial}{\partial x^\mu} \nonumber \end{align}

(2) If $\;\psi\;$ and $\;\mathbf{a}\;$ are scalar and vector functions in $\;\mathbb{R}^{3}$ then \begin{equation} \boldsymbol{\nabla \cdot}\left(\psi\mathbf{a}\right)\boldsymbol{=}\mathbf{a}\boldsymbol{\cdot}\boldsymbol{\nabla}\psi\boldsymbol{+}\psi\boldsymbol{\nabla \cdot}\mathbf{a} \nonumber \end{equation}

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You start off as described in the footnote 7 (We assume the validity of the Klein-Gordon equation for $\phi$ and $\phi^\ast$):

$$0 = -i\phi^{\ast} (\Box +m^2)\phi +i \phi(\Box +m^2)\phi^{\ast} = i \left[\phi^{\ast}\partial_\mu\partial^\mu \phi - \phi\partial_\mu\partial^\mu\phi^{\ast}\right] =i \left[ \partial_\mu\phi^{\ast} \partial^\mu \phi +\phi^{\ast}\partial_\mu\partial^\mu\phi - \partial_\mu\phi\partial^\mu\phi^{\ast} - \phi\partial_\mu \partial^\mu \phi^{\ast}\right] = i\left[\partial_\mu(\phi^{\ast}\partial^{\mu}\phi - \phi\partial^{\mu}\phi^{\ast})\right] = \partial_\mu j^{\mu}$$

where we used the definition $$j^\mu = i[\phi^{\ast}\partial^\mu\phi - \phi\partial^\mu\phi^{\ast}]$$ and $$\Box =-\partial_\mu\partial^\mu$$ Then with $\mu=(0,i)$ and $(i=1,2,3)$ $$\partial^i =-\nabla$$ you get the relation you wanted to proof (using $j^\mu =(\rho, \bf{j})$ as $j^\mu$ is a 4-vector ):

$$\bf{j} = -i[\phi^{\ast}\nabla\phi - \phi\nabla\phi^{\ast}]$$ respectively $$j^0\equiv \rho = i\left[ \phi^{\ast}\frac{\partial\phi}{\partial t} - \phi\frac{\partial\phi^{\ast}}{\partial t} \right]$$

As the found $j^{\mu}$ fulfills the continuity equation $0=\partial_\mu j^{\mu}$ it is the current density for the Klein-Gordon field $\phi$. It can of course also be found by using the Noether theorem.

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