Kinetic and potential energy of harmonic oscillator

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SUMMARY

The discussion focuses on calculating the kinetic and potential energy of eigenstates of the harmonic oscillator using ladder operator methods. The potential energy is derived as V = (1/2)(n + 1/2)ħω, which is confirmed to be correct. The kinetic energy is established to be equal to the potential energy in this context. Participants suggest using the virial theorem for independent verification of the calculations.

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  • Understanding of quantum mechanics concepts, specifically harmonic oscillators
  • Familiarity with ladder operator methods in quantum mechanics
  • Knowledge of the virial theorem and its application
  • Basic proficiency in manipulating quantum mechanical equations
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Students of quantum mechanics, physicists working with harmonic oscillators, and anyone interested in the mathematical foundations of energy calculations in quantum systems.

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Homework Statement


Use ladder operator methods to determine the kinetic and potential energy of eigenstates of the harmonic oscillator.

Homework Equations


H=\frac{p^2}{2m} + \frac{1}{2}m\omega x^2
x=\sqrt{\frac{\hbar}{2m \omega}}(a+a^{\dagger})

The Attempt at a Solution


So I squared x, and then substituted it in the expression for V:

\langle n | V | n \rangle

I ended up getting V=\frac{1}{2}(n+\frac{1}{2})\hbar \omega. This is one half of the energy expectation, so KE must be the same.

I just don't have any references with me to confirm if this is right. Can anyone tell me? Thanks in advance. I could write out what I got V as, if someone wants, but if the answer is right, there may not be much point.
 
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It is correct. You can verify your calculations independently using the virial theorem.
 

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