# Finding the expression for probability density (the Klein Gordon equation)

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 Schrödinger 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 $$$$\dfrac{\mathrm d\rho}{\mathrm dt}+\boldsymbol{\nabla} \cdot \boldsymbol{j}=0, \tag{6.9}\label{6.9}$$$$ which is more easily written in four-vector notation as $$$$\partial_{\mu}j^{\mu}=0. \tag{6.10}\label{6.10}$$$$ If, as is usual in non-relativistic quantum mechanics,6 we take the spatial part to be $$$$\boldsymbol{j}(x)=-\mathrm i\left[\phi^{*}(x)\boldsymbol{\nabla}\phi(x)-\phi(x)\boldsymbol{\nabla}\phi^{*}(x)\right], \tag{6.11}\label{6.11}$$$$ then, for eqn 6.10 to work,7 we require the probability density to look like8 $$$$\rho(x)\boldsymbol{=}\mathrm i\left[\phi^{*}(x)\dfrac{\partial\phi(x)}{\partial t}\boldsymbol{-}\dfrac{\partial\phi^{*}(x)}{\partial t}\phi(x)\right]. \tag{6.12}\label{6.12}$$$$ The resulting covariant probability current for the Klein–Gordon equation is then given by $$$$j^{\mu}(x)=\mathrm i\{\phi^{*}(x)\partial^{\mu}\phi(x)-\left[\partial^{\mu}\phi^{*}(x)\right]\phi(x)\}, \tag{6.13}\label{6.13}$$$$ 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 $$$$\rho(x)=\dfrac{\mathrm i \hbar}{2m}\left(\phi^{*}\dfrac{\partial\phi}{\partial t}-\phi\dfrac{\partial\phi^{*}}{\partial t}\right) \tag{2.20}\label{2.20}$$$$ and with $$$$j^{\mu}=\left(\rho,\mathbf{j}\right)=\dfrac{\mathrm i \hbar}{m}\phi^{*}\left(\overset{\leftrightarrow}{\partial_{0}},\overset{\leftrightarrow}{\boldsymbol{\nabla}}\right)\phi =\dfrac{\mathrm i \hbar}{m}\phi^{*}\overset{\leftrightarrow}{\partial^{\mu}}\,\phi \tag{2.21}\label{2.21}$$$$ where $$$$A\overset{\leftrightarrow}{\partial^{\mu}}B\stackrel{\text{def}}{=}\tfrac12\left[A\partial^{\mu}B\boldsymbol{-}(\partial^{\mu}A)B\right], \tag{2.22}\label{2.221}$$$$ and we have used (2.9), we have the continuity equation $$$$\partial_{\mu}j^{\mu}=\dfrac{\mathrm i \hbar}{2m}\left(\phi^{*}\square \,\phi-\phi\,\square\,\phi^{*}\right)=0, \tag{2.23}\label{2.3}$$$$ since $$\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...

• Do you understand (6.11)? (6.12) is just the 0-th (temporal) component of the tensor equation (6.11). Feb 5, 2019 at 4:06
• 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. Feb 5, 2019 at 7:09
• 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! Feb 5, 2019 at 18:57
• @user215742 I had not thought of that! I will work on that direction now, thank you! Feb 5, 2019 at 19:02

The Schr$$\ddot{\rm o}$$dinger equation is non-relativistic and for a free particle is derived from the Hamiltonian $$$$H\boldsymbol{=} \dfrac{p^2}{2m} \tag{K-01}\label{eqK-01}$$$$ by the transcription $$$$H\boldsymbol{\longrightarrow} i\hbar\dfrac{\partial}{\partial t}\quad \text{and}\quad \mathbf{p}\boldsymbol{\longrightarrow} \boldsymbol{-}i\hbar\boldsymbol{\nabla} \tag{K-02}\label{eqK-02}$$$$ so that $$$$i\hbar \dfrac{\partial \psi}{\partial t}\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\psi\boldsymbol{=} 0 \tag{K-03}\label{eqK-03}$$$$ For a first try to derive a relativistic quantum mechanical equation we make use of the property that according to the theory of special relativity the total energy $$\;E\;$$ and momenta $$\;(p_x,p_y,p_z)\;$$ transform as components of a contravariant four-vector $$$$p^\mu\boldsymbol{=}\left(p^0,p^1,p^2,p^3\right)\boldsymbol{=}\left(\dfrac{E}{c},p_x,p_y,p_z\right) \tag{K-04}\label{eqK-04}$$$$ of invariant length $$$$\sum\limits_{\mu\boldsymbol{=}0}^{3}p_{\mu} p^{\mu}\boldsymbol{\equiv}p_{\mu} p^{\mu}\boldsymbol{=}\dfrac{E^2}{c^2}\boldsymbol{-}\mathbf{p}\boldsymbol{\cdot}\mathbf{p}\boldsymbol{\equiv}m^2c^2\tag{K-05}\label{eqK-05}$$$$ where $$\;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 $$$$H\boldsymbol{=}\sqrt{p^{2}c^2\boldsymbol{+}m^2c^4} \tag{K-06}\label{eqK-06}$$$$ and to write for a relativistic quantum analogue of \eqref{eqK-03} $$$$i\hbar \dfrac{\partial \psi}{\partial t}\boldsymbol{=}\sqrt{\boldsymbol{-}\hbar^2c^2 \nabla^{2}\boldsymbol{+}m^2c^4}\,\psi \tag{K-07}\label{eqK-07}$$$$ Facing the problem of interpreting the square root operator on the right in eq. \eqref{eqK-07} we simplify mathematics by removing this square root operator, so that $$$$\left[\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\boldsymbol{-}\nabla^{2}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\right]\psi\boldsymbol{=}0 \tag{K-08}\label{eqK-08}$$$$ or recognized as the classical wave equation $$$$\left[\square\boldsymbol{+}\left(\dfrac{mc}{\hbar}\right)^2\right]\psi\boldsymbol{=}0 \tag{K-09}\label{eqK-09}$$$$ where(1) $$$$\square\boldsymbol{\equiv}\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\boldsymbol{-}\nabla^{2}\boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\dfrac{\partial}{\partial x^\mu} \tag{K-10}\label{eqK-10}$$$$

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
$$$$\dfrac{\partial}{\partial t}\left[\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\right]\boldsymbol{+}\boldsymbol{\nabla \cdot}\left[\dfrac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)\right]\boldsymbol{=} 0 \tag{K-13}\label{eqK-13}$$$$ so $$$$\dfrac{\partial \varrho}{\partial t}\boldsymbol{+}\boldsymbol{\nabla \cdot}\boldsymbol{S}\boldsymbol{=} 0 \tag{K-14}\label{eqK-14}$$$$ where $$$$\boxed{\:\:\varrho\boldsymbol{\equiv}\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\:\:}\quad \text{and} \quad \boxed{\:\:\boldsymbol{S}\boldsymbol{\equiv}\dfrac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)\:\:} \tag{K-15}\label{eqK-15}$$$$ We would like to interpret $$\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 $$$$\boldsymbol{\nabla \cdot}\left(\psi\mathbf{a}\right)\boldsymbol{=}\mathbf{a}\boldsymbol{\cdot}\boldsymbol{\nabla}\psi\boldsymbol{+}\psi\boldsymbol{\nabla \cdot}\mathbf{a} \nonumber$$$$

• @Lorenzo B. : Thanks for your attention. I reject your editing. I want my post to be as it is. Feb 23 at 13:50

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.