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I am attempting to prove the following relation

$\frac 1 2$$(\hat X^2 \hat P_x+\hat P_x \hat X^2)$ = $\hat X \hat P_x \hat X$

My attempt: $\hat X=x$ , $\hat P_x=-ih\frac d {dx}$

I commuted the commutator relation: $[\hat X,\hat P_x] = ih$

I'm unsure how I could use the commutator relation here.

Using the given values one should obtain: $\frac 1 2$$(x^2*-ih\frac d {dx}--ih\frac d {dx}*x^2)$ = $x*ih\frac d {dx}*x$

Which seems trivial, and this yields: $\frac 1 2$$(x^2*-ih\frac d {dx}-2xih)$=$-xih$

Which could be written as $\frac 1 2$$(x^2*-ih\frac d {dx})$ = $0$.

I have clearly gone wrong here and I'm not sure where. I should really apply the commutator relation somewhere or perhaps use the argument that $[\hat X^n, \hat P_x]=ihn\hat X^{n-1}$

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1 Answer 1

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It's not true that $$-i\hbar \frac{d}{dx} * x^2 = -2i\hbar x$$ That is because $*$ should be understood as multiplication of operators, and not as evaluating one operator on another.

The proper way to calculate this product is to act on some wavefunction:

$$ (\hat P_x \hat X^2 \psi)(x) = -i\hbar \frac{d}{dx} (x^2\psi(x)) = -i\hbar(2x\psi(x) + x^2\frac{d}{dx}\psi(x)) = ((-2i\hbar \hat X + \hat X^2\hat P_x)\psi)(x)$$

Since it works for any function $\psi$, it means that $$ \hat P_x \hat X^2 = -2i\hbar \hat X + \hat X^2\hat P_x $$

Similarly $$ \hat X\hat P_x \hat X = -i\hbar \hat X + \hat X^2\hat P_x $$

It is also possible to prove the desired relation without using any representation of the operators. We have $$ \frac12(\hat X^2 \hat P_x + \hat P_x \hat X^2) - \hat X\hat P_x\hat X = \frac12(\hat X^2 \hat P_x-\hat X\hat P_x\hat X)+\frac12(\hat P_x \hat X^2-\hat X\hat P_x\hat X) = \\ = \frac12 \hat X(\hat X\hat P_x - \hat P_x\hat X) + \frac12(\hat P_x \hat X-\hat X\hat P_x)\hat X =\\ = \frac12 \hat X * i\hbar + \frac12 (-i\hbar) *\hat X = 0$$

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  • $\begingroup$ Brilliant, thank you for your explanation. $\endgroup$
    – user254780
    Feb 24, 2020 at 19:04
  • $\begingroup$ It's not necessary to act on some wavefunction $$ xp_{x}-p_{x}x=i\hbar\:\: \Longrightarrow \left. \begin{cases} x\cdot \:\:\text{(from left)}\\ \cdot x \:\:\text{(from right)} \end{cases}\!\!\right\} \Longrightarrow \left. \begin{cases} x^2p_{x}-xp_{x}x=i\hbar x \\ xp_{x}x-p_{x}x^2=i\hbar x \end{cases}\!\!\right\} \Longrightarrow x^2p_{x}+p_{x}x^2=2xp_{x}x $$ $\endgroup$
    – Frobenius
    Feb 25, 2020 at 0:54

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