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Suppose we have a variable $x$ and a probability density $\rho(x)$. The pushforward of this density under a bijective function $y = f(x)$ is given by \begin{equation*} \rho'(y) = \frac{\rho(f^{-1}(y))}{|Df(f^{-1}(y))|} \end{equation*} where $Df$ is the Jacobian of $f$. My question is: is there an analogous transformation law for the wavefunction?

As a start, immediately we can see that since $|\psi(x)|^2$ is a probability measure, the transformation law given above fixes the form of $|\psi'(y)|^2$ as $|\psi(f^{-1}(y))|^2|Df^{-1}(y)|$. However, this does not fix the form of the transformation law for the phase. Is there an obvious way to get the form of both the modulus transformation law and the phase transformation?

Qmechanic
  • 201,751

1 Answers1

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TL;DR: As the overall phase of the wavefunction is not physical, OP's question has a non-unique answer that ultimately comes down to a choice of convention. Within a given class of situations we often choose the simplest possible consistent convention.

Examples:

  1. Normally the inner product $$\langle \phi|\psi\rangle~=~\int \! d^nx~\rho(x)~ \phi^{\ast}(x)\psi(x) $$ is (possibly implicitly) defined with a density measure $\rho(x)$ (which is often chosen to be $1$ in some preferred coordinate system), while the wavefunction $\psi(x)$ is a scalar, $$\begin{align} \psi\quad\longrightarrow\quad \psi^{\prime}~=~&\psi, \cr \rho\quad\longrightarrow\quad \rho^{\prime}~=~&\frac{\rho}{|J|}, \quad J~:=~\det\frac{\partial x^{\prime}}{\partial x}, \cr \end{align} $$ under a $t$-independent coordinate transformation $$x\quad\longrightarrow\quad x^{\prime}~=~f(x)$$ of space.

  2. Sometimes the inner product $$\langle \phi|\psi\rangle~=~\int \! d^nx~ \phi^{\ast}(x)\psi(x) $$ is defined without a density, and the wavefunction $\psi(x)$ is a half-density/semi-density $$\psi\quad\longrightarrow\quad \psi^{\prime}~=~\frac{\psi}{\sqrt{|J|}}. $$

  3. In the holomorphic/coherent state phase space over a complex manifold, it is natural to introduce holomorphic (antiholomorhic) transformation rules for holomorphic (antiholomorhic) wavefunctions, respectively. This will produce a phase factor, cf. OP's question.

  4. For $t$-dependent coordinate transformations of space, the situation is more complicated, e.g. this Phys.SE post.

Qmechanic
  • 201,751
  • This makes a reasonable amount of sense. I want to clarify a point-- while the phase portion of the wavefunction is defined only up to an overall phase factor, phase differences between different points are quite physical. So the transformation should be unique, up to that global factor (contingent on which of cases 1-4 you are working in). – Panopticon May 24 '23 at 16:43
  • I guess the message is just that there is no "funny business" with the phase and it just transforms like an ordinary function as opposed to a measure-like object. – Panopticon May 24 '23 at 16:45