Expectation value of kinetic energy operator

  • Thread starter docnet
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  • #1
docnet
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Homework Statement:
please see below
Relevant Equations:
please see below
Screen Shot 2021-02-11 at 3.30.14 PM.png




The expectation value of the kinetic energy operator in the ground state ##\psi_0## is given by
$$<\psi_0|\frac{\hat{p^2}}{2m}|\psi_0>$$
$$=<\psi_0|\frac{1}{2m}\Big(-i\sqrt{\frac{\hbar mw}{2}}(\hat{a}-\hat{a^{\dagger}})\Big)^2|\psi_0>$$
$$=\frac{-\hbar w}{4}<\psi_0|\hat{a}^2+-\hat{a}\hat{a^{\dagger}}-\hat{a^{\dagger}}\hat{a}+\hat{a^{\dagger}}^2|\psi_0>$$
$$=\frac{-\hbar w}{4}<\psi_0|\Big(\hat{a}^2|\psi_0>-\hat{a}\hat{a^{\dagger}}|\psi_0>-\hat{a^{\dagger}}\hat{a}|\psi_0>+\hat{a^{\dagger}}^2|\psi_0>\Big)$$
$$=\frac{-\hbar w}{4}<\psi_0|\Big(|-\psi_0>+\sqrt{2}|\psi_2>\Big)$$
$$=\frac{-\hbar w}{4}<-\psi_0|\psi_0>+<\psi_0|\psi_2>$$

$$=\frac{\hbar w}{4}$$
 

Answers and Replies

  • #2
anuttarasammyak
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So we know for ground state energy as well as excited states energy kinetic energy and potential energy are half and half. That coincide with classical mechanics.
 

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