$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$

Wavefunctions and Relativity

^{First created in December 2018}

Ref: Lesson 3: Representing Complex Things

A stationary particle observed from a moving observer.

$p_x=0,~~p_x$ known exactly, $x$ completely unknown, Prob($x$)=const, Energy=$mc^2$.

$\Large\Psi(x,t)=Ae^{-i\omega t}=Ae^{-i{E\over\hbar}t}.$

Lorentz transformation: $\Large t=\gamma\left(t'+{vx'\over c^2}\right)$ where $\gamma={1\over\sqrt{1-{v^2\over c^2}}}\ge 1$, the Lorentz factor.

$\displaystyle \Psi(x',t') =Ae^{-i{E\over\hbar}\gamma\left(t'+{vx'\over c^2}\right)} =Ae^{-i\gamma{E\over\hbar}t'-i{\gamma mvx'\over\hbar}} =Ae^{i(kx'-\omega t')},\text{ where } k=-{\gamma mv\over\hbar}={p_x'\over\hbar},\text{ and }\omega={\gamma E\over\hbar}={E'\over\hbar}.$

Please note that this is not a "proof" of $E=\omega\hbar$ and $p=k\hbar$. It is to show that Lorentz transformation is consistent with wavefunctions.

This means the moving observer will measure energy and momentum based on $\Psi(x',t')$.

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$\displaystyle E=mc^2,~ E^2 =(mc^2)^2 =(\gamma m_0c^2)^2 ={c^2\over{c^2-v^2}}m_0^2c^4\\ ={(c^2-v^2)+v^2\over{c^2-v^2}}m_0^2c^4 =m_0^2c^4+{v^2\over{c^2-v^2}}m_0^2c^4 =m_0^2c^4+v^2m_0^2\gamma^2c^2 .$

As $p=mv=\gamma m_0v,~~\boxed{\large E^2=m_0^2c^4+p^2c^2.}~~~~ (\omega\hbar)^2=m_0^2c^4+(k\hbar)^2c^2.$

Recall spacetime interval $(\Delta s)^2=(\Delta x)^2+(\Delta y)^2+(\Delta z)^2+(\Delta\tau)^2,$ where $\tau\equiv ict.$ So $c=-i\tau/t.$

$(\omega\hbar)^2=m_0^2c^4-(k\hbar)^2\tau^2/t^2,~~ (\omega\hbar)^2t^2=m_0^2c^4t^2-(k\hbar)^2\tau^2,~~ (\omega\hbar t)^2=(m_0c^2t)^2-(k\hbar\tau)^2 .$

$\large\boxed{ (Et)^2+(p\tau)^2=(E_0t)^2} ,$ where $\tau=ict$ and $E_0=m_0c^2$. i.e. the rest mass.

WORKING

$(\omega\hbar)^2=m_0^2c^4+(k\hbar)^2c^2,~~ \omega^2={m_0^2c^4\over\hbar^2}+k^2c^2 .$

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From the stationary observer, $p=0$ and $E^2=m_0^2c^4.$

From the moving observer, $p'>0$ and $E'^2=m_0^2c^4+p'^2c^2.~~ \omega^2={m_0^2c^4\over\hbar^2}+k^2c^2 .$

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Let us revisit $\large t =\gamma\left(t'+{vx'\over c^2}\right) =\gamma t'+{\gamma vx'\over c^2} .~~$ The first term $\gamma t'$ gives rise to the energy, and the second term ${\gamma vx'\over c^2}$ momentum.