Commutator relations in simple harmonic oscillator

In summary, the conversation discussed the proof of the commutation relations for the SHO Hamiltonian and how they lead to the expression [a, \hat H] = \hbar\omega, [a^+, \hat H] = -\hbar\omega. The attempted solution was also provided, but it was shown to be incorrect as it did not match the expected result. However, three independent sources were mentioned that confirmed the correct result.
  • #1
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Homework Statement



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

Homework Equations

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

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

The Attempt at a Solution



I have tried the following:

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

And this is nothing like [tex] \hbar\omega [/tex] I am supposed to get. Could anyone point out where I have gone wrong?
 
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  • #2
I got that result too:

[tex][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[/tex]

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?
 

Related to Commutator relations in simple harmonic oscillator

What is a commutator?

A commutator is an operator used in quantum mechanics to determine the order in which variables are applied to a function. It is denoted by [A, B] and is defined as the product of the two operators A and B minus the product of B and A.

How are commutator relations useful in the simple harmonic oscillator?

In the simple harmonic oscillator, commutator relations are used to determine the uncertainty in the position and momentum of a particle. This is because the commutator [x, p] is equal to iħ, where x is the position operator, p is the momentum operator, and ħ is the reduced Planck's constant. This relationship allows us to calculate the uncertainty principle in the simple harmonic oscillator.

What is the commutator relation for the position and momentum operators in the simple harmonic oscillator?

The commutator relation for the position and momentum operators in the simple harmonic oscillator is given by [x, p] = iħ, where x is the position operator, p is the momentum operator, and ħ is the reduced Planck's constant. This relationship is a fundamental property of quantum mechanics and is used to determine the uncertainty in the position and momentum of a particle.

How can commutator relations be extended to other systems?

Commutator relations in the simple harmonic oscillator can be extended to other systems by replacing the position and momentum operators with the corresponding operators for the system in question. For example, in a system with multiple particles, the position and momentum operators would be replaced with the operators for the positions and momenta of each particle. This allows for the calculation of uncertainty principles and other properties in a variety of quantum systems.

What is the physical significance of commutator relations in the simple harmonic oscillator?

The physical significance of commutator relations in the simple harmonic oscillator is that they provide a mathematical framework for understanding the uncertainty principle and other properties of quantum systems. They also allow for the calculation of important physical quantities, such as energy levels and transition probabilities, in the simple harmonic oscillator and other quantum systems.

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