Question

Show that xp-px=ih for the ground state wave function of the quantum harmonic oscillator.

Answer 1

Consider $\hat A = \hat x$ and Bˆ = $\hat p = −ih\dfrac{d}{dx}$ ,

Then we have $\hat A\hat Bf(x) = \hat x\hat pf (x)$

We can of course also construct another new operator : $\hat p\hat x$

Then, by definition of the operator product, $\hat p\hat xf (x)$

means that $\hat x$ is first operating on f(x) and then $\hat p$ is operating on the function $\hat xf (x)$ .

Compare the results of operating with the products $\hat p\hat x \text{ and } \hat x\hat \ p \text{ on } f(x)$

$(\hat x\hat p− \hat p\hat x)f(x) = −ih(x\dfrac{d(f(x))}{dx} - \dfrac{d}{dx}(xf(x)))$  

and hence by the product rule of differentiation:

$(\hat x\hat p− \hat p\hat x)f(x) = ihf(x)$

and since this must hold for any differentiable function f(x), we can write this as an operator equation:

$\hat x\hat p− \hat p\hat x = ih$

Thus, we have shown that the operator product of $\hat x$ and $\hat p$ is non-commuting.