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The Schrödinger equation is $$i \hbar \frac{\partial}{\partial t} \Psi(\mathbf{r}, t)=-\frac{\hbar^{2}}{2 m} \nabla^{2} \Psi(\mathbf{r}, t)+V(\mathbf{r}) \Psi(\mathbf{r}, t).$$

Under the coordinate transformation $r\rightarrow r'$ the Schrödinger equation goes to $$i \hbar \frac{\partial}{\partial t} \Psi'(\mathbf{r'}, t)=-\frac{\hbar^{2}}{2 m} \nabla'^{2} \Psi'(\mathbf{r'}, t)+V(\mathbf{r'}) \Psi(\mathbf{r'}, t). \tag 2$$

Now, since the position $r$ operator and the momentum operator are independent, coordinate transformations should not affect the momentum operator $p$. But, as we saw in $(2)$, the so-called coordinate transformations affect the momentum operator.

My question is: are coordinate transformations in quantum mechanics unitary transformations of the position operator?

Cosmas Zachos
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amilton moreira
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1 Answers1

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Operators in QM don't depend on coordinates, so changing coordinates does not affect them in the least; which is why your (2) is off. Recall how you handle Schroedinger's equation in polar coordinates. Stick to one dimension for simplicity.

The proper, operator, TDSE is $$ \left (-i\hbar \partial_t + \frac{\hat p ^2}{2m} +V(\hat x) \right ) |\Psi(t)\rangle =0 \tag{1}, $$ which, dotted by $\langle x|$, yields your first TDSE in coordinate space, $$ i \hbar \partial_t\langle x|\Psi(t)\rangle= \langle x| \frac{\hat p ^2}{2m} |\Psi(t)\rangle + \langle x|V(\hat x)|\Psi(t)\rangle \tag{1a}\\ i \hbar \partial_t \Psi( x, t)=-\tfrac{\hbar^{2}}{2 m} \partial_x ^{2} \Psi( x, t)+V(x) \Psi( x, t).$$

Under a coordinate transformation $ x\rightarrow x'= x'(x)$, this PDE is equivalent to $$i \hbar \partial_t \Psi( x', t)=-\tfrac{\hbar^{2}}{2 m} \partial_{x'} ^{2} \Psi( x', t)+V(x') \Psi( x', t) ~. \tag 2$$ The first term (kinetic) on the r.h.side can also be rewritten as $$ -\tfrac{\hbar^{2}}{2 m} \partial_{x'} ^{2} \Psi( x', t) = -\tfrac{\hbar^{2}}{2 m} \left (\frac{1}{\partial_x x'(x)} ~ \partial_{x} \right )^{2} \Psi( x'(x), t) , \tag{3} $$ on which you might practice with a simple example.

But this is exactly what you would have gotten by dotting (1) by $\langle x'(x)|$.

Edit as per comments, which have taken a left turn and point to a very different question, on canonical transformations. For the 2D polar coordinate question asked, however, the change of variable $(x,y) \to (r=\sqrt{x^2+y^2}, \phi =\arctan(y,x))$, (3) goes to the generalization $$ -\tfrac{\hbar^{2}}{2 m}(\partial_{x} ^{2}+ \partial_{y} ^{2})= -\tfrac{\hbar^{2}}{2 m} \left (\partial_r^2+\frac{1}{r}\partial_r + \frac{1}{r^2}\partial_\phi^2 \right )^{2} , $$ and you have a plain change of coordinates.

As stated above, $$ \hat x |r,\phi\rangle = r\cos\phi ~|r,\phi\rangle, ... \hbox{etc.} $$ To be sure, you could define formal operators such as $$ \hat r\equiv \sqrt{\hat x ^2 + \hat y ^2}, \implies ~~ \hat r |r,\phi\rangle= r|r,\phi\rangle, $$ but, as indicated in the linked answer, operators of angular variables (skipped here) are beset with formal problems revolving around hermiticity, so some wise bypass them.

Cosmas Zachos
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