Commutator relations in simple harmonic oscillator

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SUMMARY

The discussion centers on the commutator relations for the simple harmonic oscillator (SHO), specifically demonstrating that [a, \hat H] = \hbar\omega and [a^+, \hat H] = -\hbar\omega, where \hat H = \hbar\omega(a^+a - \frac{1}{2}). The user initially miscalculated the commutator [a, \hat H] but later confirmed the correct results through multiple independent sources, aligning with established literature. The key takeaway is the correct application of the commutation relations and the Hamiltonian in quantum mechanics.

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  • Understanding of quantum mechanics principles, particularly the simple harmonic oscillator.
  • Familiarity with commutation relations in quantum mechanics.
  • Knowledge of the Hamiltonian operator in quantum systems.
  • Proficiency in manipulating algebraic expressions involving operators.
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  • Study the derivation of the simple harmonic oscillator Hamiltonian in quantum mechanics.
  • Learn about the significance of commutation relations in quantum mechanics.
  • Explore the implications of the Heisenberg uncertainty principle in relation to the SHO.
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Students and professionals in quantum mechanics, physicists studying harmonic oscillators, and anyone interested in the mathematical foundations of quantum theory.

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Homework Statement



Show that, [a, \hat H] = \hbar\omega, [a^+, \hat H] = -\hbar\omega

Homework Equations

For the SHO Hamiltonian \hat H = \hbar\omega(a^+a - \frac{\ 1 }{2}) with [a^+, a] = 1

[a, b] = -[b, a]

The Attempt at a Solution



I have tried the following:

[a, \hat H] = a\hat H - \hat Ha = \hbar\omega ( (aa^+a - \frac{\ 1 }{2}a) - (a^+a - \frac{\ 1 }{2})a )<br /> = \hbar\omega (aa^+a -a^+aa) = \hbar\omega [a, a^+] a = - \hbar\omega a

And this is nothing like \hbar\omega I am supposed to get. Could anyone point out where I have gone wrong?
 
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I got that result too:

[a, \hat H] = \hbar \omega [a, a^\dagger a] = \hbar\omega\left( a^\dagger [a, a] + [a, a^\dagger] a \right) = \hbar\omega( 0 - a ) = - \hbar \omega a

which agrees perfectly well with the expressions given on Wikipedia.

So having three independent sources with the same result, I suppose that your question is incorrect?
 

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