4
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
15
  • $\begingroup$ What I am asking if the operator transformation $\hat x \rightarrow \hat x '$ is unitary $\endgroup$ Aug 19, 2021 at 13:31
  • $\begingroup$ As I said, you did not change any operator: only the coordinates of its eigenvalues. You might be thinking about unitary canonical transformations, but you are barking up the very very wrong tree! The coordinate-free TDSE never changed. $\endgroup$ Aug 19, 2021 at 13:54
  • $\begingroup$ we are in two dimension with the position operators $\hat x$ and $\hat y $ isn't this transformation $\hat r=\sqrt{\hat x\hat x+\hat y\hat y}$ $\hat \varphi=\operatorname{atan} (\hat y, \hat x)$ a valid transformation? $\endgroup$ Aug 19, 2021 at 14:00
  • $\begingroup$ @ Cosmas Zachos I am not understanding you since in quantum mechanics coordinates are operators so changing coordinates should be changing operators $\endgroup$ Aug 19, 2021 at 14:13
  • $\begingroup$ To change to polar coordinates, you are not to change the operators in this coordinate transformation: you are merely changing the coordinates. Are you familiar with (1) and how the (1a) coordinate representation of it follows? Coordinates are not operators. Start with matrix mechanics which defines the theory. $\endgroup$ Aug 19, 2021 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.