In the canonical commutation relations, $$[\hat{x}, \hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x} = i \hbar $$ Both $\hat{x}$ and $\hat{p}$ are operators and so are $\hat{x}\hat{p}$ and $\hat{p}\hat{x}$. How does their difference turn out to be a constant? Shouldn't it also be an operator?
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4The RHS is missing the identity operator $I$ - which is usually omitted for notation convenience... – Tobias Fünke Sep 18 '23 at 14:40
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4$i\hbar$ has to be understood as $i\hbar I$... – Valter Moretti Sep 18 '23 at 14:40
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1The operators also must act on a wave function, which is why it is often discouraged to write the operators this way. This reads "the commutator of position and momentum acting on a wave function returns the wave function times a constant." – Matt Hanson Sep 18 '23 at 14:50
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Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis. – Frobenius Sep 18 '23 at 21:42
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The quantities that are not commuting in your example are the Hermitian operators associated with the position and momentum observables.
$$ \hat{x} = x $$
$$ \hat{p} = -i\hbar\frac{\partial}{\partial x} $$
By substituting these operators into the commutation relation, and remembering that they act on a dummy wavefunction $\psi(x)$ you get:
$$ [\hat{x},\hat{p}] \psi(x) = \hat{x}\hat{p} \psi(x) - \hat{p}\hat{x} \psi(x) = -i\hbar x \frac{\partial \psi}{\partial x} - -i\hbar \frac{\partial}{\partial x}\left( x \psi(x)\right) $$
This requires the product rule to evaluate the derivative in the 2nd term.
$$ [\hat{x},\hat{p}] \psi(x) = -i\hbar x \frac{\partial \psi}{\partial x} + i\hbar \psi + i\hbar x \frac{\partial \psi}{\partial x} = i\hbar \psi(x) $$
The resultant operator is then simply $[\hat{x},\hat{p}] = i\hbar$
Et voila.
Jamie Smith
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