Tuesday, October 25, 2016

quantum mechanics - What's wrong with this derivation that $ihbar = 0$?


Let $\hat{x} = x$ and $\hat{p} = -i \hbar \frac {\partial} {\partial x}$ be the position and momentum operators, respectively, and $|\psi_p\rangle$ be the eigenfunction of $\hat{p}$ and therefore $$\hat{p} |\psi_p\rangle = p |\psi_p\rangle,$$ where $p$ is the eigenvalue of $\hat{p}$. Then, we have $$ [\hat{x},\hat{p}] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar.$$ From the above equation, denoting by $\langle\cdot\rangle$ an expectation value, we get, on the one hand $$\langle i\hbar\rangle = \langle\psi_p| i \hbar | \psi_p\rangle = i \hbar \langle \psi_p | \psi_p \rangle = i \hbar$$ and, on the other $$\langle [\hat{x},\hat{p}] \rangle = \langle\psi_p| (\hat{x}\hat{p} - \hat{p}\hat{x}) |\psi_p\rangle = \langle\psi_p|\hat{x} |\psi_p\rangle p - p\langle\psi_p|\hat{x} |\psi_p\rangle = 0$$ This suggests that $i \hbar = 0$. What went wrong?





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