I was studying the algebric method of solving the wavefunction for a harmonic oscillator from Quantum Mechanics by Zettili and the part where he finally brings the wavefunction derived from analytic method equal to one he got from ladder operator method (Page 246) , he writes

At this level, we can show that the wave function (4.165) derived from the algebraic method is similar to the one obtained from the first method (4.118). To see this, we simply need to use the following operator identity:

$$\displaystyle e^{\frac{-x^2}{2}}(x-\frac{d}{dx})e^{\frac{x^2}{2}}= -\frac{d}{dx}$$ $$e^{\frac{-x^2}{2 x_{0}^{2}}}(x- x_0^2 \frac{d}{dx})e^{\frac{x^2}{2 x_{0}^{2}}}=-x^2_0\frac{d}{dx}$$

I want to know what identity is this cause if you solve this (LHS) you get zero as an answer.

Here is the image link

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    $\begingroup$ Just compute the left hand side! $\endgroup$ Aug 23 '20 at 15:11
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    $\begingroup$ You should apply both sides to generic function... $\endgroup$ Aug 23 '20 at 15:22
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    $\begingroup$ you're not taking the derivative correctly then. Maybe you need to be reminded that both the LHS and RHS act on an arbitrary function $g(x)$? $\endgroup$ Aug 23 '20 at 15:22
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    $\begingroup$ the problem is that you were simply differentiating the LHS, but you have to use test function. $\endgroup$
    – crabNebula
    Aug 23 '20 at 15:25
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    $\begingroup$ Do you understand the commutator $[\frac{d}{dx},x ]=1$ evaluated by the chain rule? $\endgroup$ Aug 23 '20 at 15:38

Here's a golden rule whenever you're dealing with differential operators: always use a test function $f(x)$.

If you call the operator on the left hand side $\hat{O}$, then what you want to show is that for any generic function $f(x)$,

$$\hat{O} f(x) = -\frac{\text{d}f}{\text{d}x}.$$

In other words, you need to show that

$$e^{-x^2/2} \left(x - \frac{\text{d}}{\text{d}x} \right)\left( e^{x^2/2} f\right) = -\frac{\text{d}f}{\text{d}x}.$$

I leave it to you to complete the proof.

  • $\begingroup$ Got it. Identities always need a test function beside them. $\endgroup$ Aug 23 '20 at 15:53

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