Show that [p^, x^] = ih/2π

Steps to Solve

1. Define the Operators

In quantum mechanics, the momentum operator $\hat{p}$ and the position operator $\hat{x}$ can be defined as follows:

• The momentum operator in one dimension is given by: $$ \hat{p} = -i\hbar \frac{d}{dx} $$
• The position operator is simply: $$ \hat{x} = x $$

2. Write the Commutator

The commutation relation we want to prove is defined as: $$ [\hat{x}, \hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x} $$

3. Calculate $\hat{x}\hat{p}$

To find $\hat{x}\hat{p}$ acting on a wave function $\psi(x)$: $$ \hat{x}\hat{p}\psi(x) = \hat{x}(-i\hbar \frac{d\psi}{dx}) = x(-i\hbar \frac{d\psi}{dx}) = -i\hbar x \frac{d\psi}{dx} $$

4. Calculate $\hat{p}\hat{x}$

Now, find $\hat{p}\hat{x}$ acting on the same wave function: $$ \hat{p}\hat{x}\psi(x) = \hat{p}(x\psi(x)) = -i\hbar \frac{d}{dx}(x\psi(x)) $$

Using the product rule of differentiation, we have: $$ \frac{d}{dx}(x\psi(x)) = \psi(x) + x\frac{d\psi}{dx} $$

Thus, $$ \hat{p}\hat{x}\psi(x) = -i\hbar \left( \psi(x) + x\frac{d\psi}{dx} \right) $$

5. Combine the Results

Now substitute the expressions we found back into the commutator: $$ [\hat{x}, \hat{p}]\psi(x) = \hat{x}\hat{p}\psi(x) - \hat{p}\hat{x}\psi(x) $$

This becomes: $$ [\hat{x}, \hat{p}]\psi(x) = (-i\hbar x \frac{d\psi}{dx}) - \left( -i\hbar \left( \psi(x) + x \frac{d\psi}{dx} \right) \right) $$

6. Simplify the Commutator

Now simplify the above expression: $$ [\hat{x}, \hat{p}]\psi(x) = -i\hbar x \frac{d\psi}{dx} + i\hbar \psi(x) + i\hbar x \frac{d\psi}{dx} $$ Canceling the terms, we have: $$ [\hat{x}, \hat{p}]\psi(x) = i\hbar \psi(x) $$

7. Final Conclusion

Thus, we conclude: $$ [\hat{x}, \hat{p}] = i\hbar $$