Commutators of the Position and Momentum Operators | Quantum Science Philippines

## Commutators of the Position and Momentum Operators

By: Levine T. Poralan, MS Physics, MSU-IIT

Show that

(a) $[x, p] = i\hbar$

(b) $[\hat{X}, \hat{P}] = i$

Solution:

(a) Given that

$p = p_{x} = -i\hbar\frac{\partial}{\partial x}$

we let an arbitrary function

$\Psi(x).$

Now solving for the commutator,

$\begin{array}{lll} [x, p]\Psi(x)&=&[xp - px]\Psi(x)\\\\&=&-xi\hbar\frac{\partial}{\partial x}\Psi(x) + i\hbar\frac{\partial}{\partial x} (x) \Psi(x)\\\\&=&-xi\hbar\frac{\partial}{\partial x}\Psi(x) + xi\hbar\frac{\partial}{\partial x}\Psi(x) + i\hbar\frac{\partial x}{\partial x}\\\\\end{array}$

The first and second terms cancel and we are left with

$\begin{array}{lll} [x, p]\Psi(x)&=&i\hbar\Psi(x)\end{array}$

Therefore,

$[x, p] = i\hbar.$

(b) Introducing dimensionless observable $\hat{X}$ and $\hat{P}$ ,

$\hat{X} = \sqrt{\frac{m\omega}{\hbar}} x$

$\hat{P} = \frac{1}{\sqrt{m\hbar\omega}} p$

where

$p = -i\hbar\frac{\partial}{\partial x}.$

Now,

$\begin{array}{lll} [\hat{X}, \hat{P}] \Psi(x) &=& (\hat{X}\hat{P} - \hat{P}\hat{X}) \Psi(x)\\\\&=& \sqrt{\frac{m\omega}{\hbar}}x (\frac{-i\hbar}{\sqrt{m\hbar\omega}}\frac{\partial}{\partial x}) \Psi(x) + (\frac{i\hbar}{\sqrt{m\hbar\omega}}\frac{\partial}{\partial x})(\sqrt{\frac{m\omega}{\hbar}} x ) \Psi(x)\\\\&=& \frac{-i\hbar}{\hbar}x\frac{\partial}{\partial x}\Psi(x) + \frac{i\hbar}{\hbar}\frac{\partial}{\partial x} (x(\Psi(x))\\\\&=& -ix\frac{\partial \Psi(x)}{\partial x} + ix\frac{\partial \Psi(x)}{\partial x} + i\Psi(x) \end{array}$

The first and second terms cancel and we are left with

$\begin{array}{lll} [\hat{X}, \hat{Y}]\Psi(x) &=& i\Psi(x)\end{array}$

Therefore,

$[\hat{X}, \hat{P}] = i.$