2
$\begingroup$

Consider a commutator acting on a 1D wavefunction: $$[\frac{\hbar}{i} \frac{d}{dx},x]\psi(x)=(\frac{\hbar}{i} \frac{d}{dx}x-x\frac{\hbar}{i} \frac{d}{dx})\psi(x).$$

Now does this mean

  1. $\frac{\hbar}{i} (\frac{d}{dx}x) \psi(x)-x\frac{\hbar}{i} \frac{d}{dx} \psi(x)$ or
  2. $\frac{\hbar}{i} (\frac{d}{dx}x \psi(x))-x\frac{\hbar}{i} \frac{d}{dx} \psi(x)?$

In the first cast $\frac{d}{dx}$ only acts on $x$. In the second case $\frac{d}{dx}$ acts on $x\psi (x)$. Which is correct?

$\endgroup$
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/55773/2451 and links therein. $\endgroup$ – Qmechanic May 8 at 14:09
  • 1
    $\begingroup$ Note on formatting : if you need larger brackets try using "\left(" and "\right)" and similar for other bracket types. $\endgroup$ – StephenG May 8 at 16:36
3
$\begingroup$

The only sensible interpretation is the second one: any operator like $P$ or $X$ acts on whatever is to its right. For instance in linear algebra, if we have two matrices/operators $A,B$ and a vector $v$, then $ABv$ really means $A(B(v))$ and likewise $$[A,B]v = ABv - BAv = A(B(v)) - B(A(v)).$$ In your case, you can check this easily. One the one hand we have the famous commutator $$[-i\hbar \frac{d}{d x}, x] = -i \hbar.$$ Interpretation (1) is not consistent with the above formula; interpretation (2) is.

$\endgroup$
  • $\begingroup$ Regarding your matrices/vector argument, we can also represent wavefunctions/vectors as a Nx1 matrix, and by the associative property of matrices, $A(Bv)=(AB)v$. So $A$ can only act on $B$ instead of $Bv$. How does one resolve this problem? $\endgroup$ – TaeNyFan May 8 at 14:18
  • 1
    $\begingroup$ It's a moot point since matrix multiplication is associative; in this case $(AB)(v) = A(B(v))$. In general you shouldn't think of matrix multiplication as an operator "acting on" something. Operators (matrices) can only act on states (vectors). $\endgroup$ – Hans Moleman May 8 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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