Transformation of wavefunction

Question

While learning QM, I was wondering how would the wavefunction of a particle, suppose charged particle, look for different observers moving with respect to each other.

To begin with, let the electric potential be a function of space and time denoted by $V(x,t)$.

This is the electric potential according to an observer suppose $A$ but for another observer suppose $B$ who is moving with $v$ velocity with respect to $A$ the potential will be $V_B(x_B,t)=V( x_B-vt,t)$ assuming that $A,B$ were at the same point when $t=0$.

Let $\psi(x,t)$ be the wavefunction of a charged particle (with unit charge) according to $A$ while for $B$ it is $\psi_B(x_B,t)$.

$x_A$ and $x_B$ are the $x$ coordinates for $A,B$ respectively.

It is obvious that $\psi_B(x_B,t)=\psi(x_B-vt,t)$ because $x$ at $t$ and $x_B-vt$ at $t$ denotes the same point.

So this is the transformation of wave function.

If so then both wavefunctions should satisfy Schrödinger equation in their respective frames.

We assume $\psi(x,t)$ satisfies time dependent schrodinger equation in $A$.

Notation: If $f(v_1,v_2)$ is a function of 2 variables then $f^1(v_1,v_2)$ is the partial derivative function of $f$ with respect to first variable and $f^2(v_1,v_2)$ is to second variable . Eg:- let $f(x,y)=\sin(x+2y)$ then $f^1(x,y)=\cos(x+2y)$ and $f^2(x,y)=2\cos(x+2y)]$. We know :

$$\frac{\partial\psi(a,b)}{\partial c}=\frac{\partial\psi(a,b)}{\partial a}\frac{\partial a}{\partial c}+\frac{\partial \psi(a,b)}{\partial b}\frac{\partial b}{\partial c}.$$

So we have $$\frac{i\hbar \partial \psi(x_B-vt,t)}{\partial t}=i\hbar (-v\psi^1(x_B-vt,t)+\psi^2(x_B-vt,t))$$

And also:

$$\frac{-\hbar^2\partial^2\psi(x_B-vt,t)}{2m\partial x_B^2 }+V(x_B-vt,t)\psi(x_B-vt,t)=\frac{-\hbar^2\psi^11(x_B-vt,t)}{2m}+V(x_B-vt,t)\psi(x_B-vt,t).$$

Thus if the transformation is correct, $$i\hbar\psi^2(x,t)-vi\hbar\psi^1(x,t)=V(x,t)\psi(x,t)-\frac{\hbar^2}{2m}\psi^{11}(x,t)$$ for all $x,t$. The wave function satisfies time-dependent Schrödinger equation in $A$ frame also so this implies $$vi\hbar\psi^1(x,t)=0$$ for all $x,t$ which is not true. This implies the transformation is wrong. This is extremely shocking result because if this transformation is wrong then the 2 observers will have different probability of finding the particle at the same point. Does that mean that the Schrödinger equation is wrong because it is not compatible with the fact of same wavefunction at same point irrespective of the observer's frame of reference?

Answer 1

It is obvious that $\psi_B(x_B,t)=\psi(x_B-vt,t)$ because $x$ at $t$ and $x_B-vt$ at $t$ denotes the same point.

This is where your mistake starts. Wavefunction is not an observable, so you can't make the assumption that its value will be the same in a different frame of reference. You can only say this about the probability density. And indeed, for a free particle with definite momentum the transformation law for the wavefunction will introduce a phase factor$^\dagger$:

$$\Psi(\vec r,t)=\Psi'(\vec r-\vec vt,\,t)\exp\left[\frac{i}{\hbar}\left(m\vec v\cdot \vec r-\frac{mv^2}2 t\right)\right],\tag1$$

where $\Psi$ is the wavefunction in frame $K$ and $\Psi'$ is in frame $K'$, and the $K'$ is moving with velocity $\vec v$ with respect to $K$.

For a general wavefunction note that it can be expanded in the basis of plane waves, so $(1)$ will still hold for each component of this expansion.

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_{$^\dagger$For derivation see the problem after chapter 17 "Schrödinger's equation" in Landau, Lifshitz "Quantum Mechanics. Non-relativistic theory."}