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The eigenstates for the infinite potential well described by the Hamiltonian $$ \hat{H}= \frac{\hat{P}^2}{2m} + V(x),\, V(x) = \begin{cases} 0, & \text{if $x <0$} \\ \infty, & \text{if $x>L$} \end{cases} $$ are $$\Psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$

Now, if two operators $\hat{A},\hat{B}$ commute, they have eigenstates in common. $\hat{H}$ and $\hat{P}$ commute but they do not have simultaneous eigenstates. Why?

user1830663
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1 Answers1

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Hamiltonian does not commute with momentum operator .It only commutes if potential is zero. For more details check https://www.quora.com/When-does-the-Hamiltonian-H-commute-with-the-momentum-P-How-can-I-prove-that-H-P-0

Rishabh Jain
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