# Use the commutation relation between the position operator $\hat X$ and the momentum operator $\hat P_x$ to show the given equivalence relation

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 $$[\hat X^n, \hat P_x]=ihn\hat X^{n-1}$$

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 $$\frac12(\hat X^2 \hat P_x + \hat P_x \hat X^2) - \hat X\hat P_x\hat X = \frac12(\hat X^2 \hat P_x-\hat X\hat P_x\hat X)+\frac12(\hat P_x \hat X^2-\hat X\hat P_x\hat X) = \\ = \frac12 \hat X(\hat X\hat P_x - \hat P_x\hat X) + \frac12(\hat P_x \hat X-\hat X\hat P_x)\hat X =\\ = \frac12 \hat X * i\hbar + \frac12 (-i\hbar) *\hat X = 0$$

• Brilliant, thank you for your explanation. Feb 24 '20 at 19:04
• It's not necessary to act on some wavefunction $$xp_{x}-p_{x}x=i\hbar\:\: \Longrightarrow \left. \begin{cases} x\cdot \:\:\text{(from left)}\\ \cdot x \:\:\text{(from right)} \end{cases}\!\!\right\} \Longrightarrow \left. \begin{cases} x^2p_{x}-xp_{x}x=i\hbar x \\ xp_{x}x-p_{x}x^2=i\hbar x \end{cases}\!\!\right\} \Longrightarrow x^2p_{x}+p_{x}x^2=2xp_{x}x$$ Feb 25 '20 at 0:54