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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?

Qmechanic
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TaeNyFan
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1 Answers1

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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.

Hans Moleman
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  • 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? – TaeNyFan May 08 '19 at 14:18
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    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). – Hans Moleman May 08 '19 at 14:21