SHO ladder operators & some hamiltonian commutator relations

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SUMMARY

The discussion focuses on calculating specific commutators for the Simple Harmonic Oscillator (SHO) using ladder operators. The commutators [a_{-}, a_{-}a_{+}] and [a_{+}, a_{-}a_{+}] were evaluated, revealing that [a_{-}, a_{-}a_{+}] simplifies to a_{-} and [a_{+}, a_{-}a_{+}] equals zero. The user initially struggled with the inclusion of wavefunctions in their calculations but later recognized the utility of the product rule for commutators to streamline the process. The discussion concludes with the user successfully resolving their confusion regarding the commutators.

PREREQUISITES
  • Understanding of quantum mechanics and the Simple Harmonic Oscillator (SHO)
  • Familiarity with ladder operators a_{-} and a_{+}
  • Knowledge of commutation relations in quantum mechanics
  • Basic proficiency in manipulating wavefunctions and operators
NEXT STEPS
  • Study the derivation of commutation relations for quantum harmonic oscillators
  • Learn about the product rule for commutators in quantum mechanics
  • Explore the explicit forms of the Hamiltonian operator H, momentum operator p, and position operator x
  • Investigate the implications of commutators in quantum mechanics and their physical interpretations
USEFUL FOR

Students and researchers in quantum mechanics, particularly those studying the Simple Harmonic Oscillator and commutation relations, will benefit from this discussion.

Hakkinen
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Homework Statement


For the SHO, find these commutators to their simplest form:
[a_{-}, a_{-}a_{+}]
<br /> [a_{+},a_{-}a_{+}]<br />
<br /> [x,H]<br />
<br /> [p,H]<br />

Homework Equations


The Attempt at a Solution


I though this would be an easy problem but I am stuck on the first two parts. Here's what I did at first:
<br /> [a_{-}, a_{-}a_{+}]\psi = a_{-}(n+1)\psi_{n} - a_{-}n\psi_{n} = a_{-}\psi_{n}<br />
<br /> = \sqrt{n}\psi_{n-1}<br />

<br /> [a_{+}, a_{-}a_{+}]\psi = a_{+}(n+1)\psi_{n} - a_{-}a_{+}\sqrt{n+1}\psi_{n+1}<br />
<br /> = (n+1)^{3/2}\psi_{n+1} - (n+1)^{3/2}\psi_{n+1} = 0<br />Now what I am confused about is the \psi_{n-1} term in the first commutator. Surely there is a general form of the commutator without the test wavefunction? And I can't just drop this term and have root of n as the result. So did I do something wrong?

I tried the first part again using the explicit form of the ladder operators, in terms of H, p, x, with all of the constants. What I have gotten so far looks quite messy and involves [p,H] and [x,H], which I've yet to compute and are the last two parts of the problem... So it seems this route is not the easiest?

Any assistance is appreciated!
 
Last edited:
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Duh! The first answer is just a_{-}

And I didnt realize you could just use the "product rule" for commutators to simplify the algebra a bit!

All set now
 

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