Diagonalizing a (dimensionless) Hamiltonian

Thread starter Cocoleia
Start date Sep 25, 2019
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hamiltonian

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The discussion centers on the process of diagonalizing a dimensionless Hamiltonian, with a specific focus on handling an additional term. Participants highlight the standard approach to diagonalization and express confusion regarding the incorporation of this extra term. The conversation suggests that expressing the position operator x in terms of the creation and annihilation operators a and a† may be necessary for simplification. Clarification on this method is sought to ensure proper diagonalization. Understanding this relationship is crucial for successfully diagonalizing the Hamiltonian.

Sep 25, 2019

Cocoleia

Homework Statement: Diagonalize using creation / annihilation operator methods

Relevant Equations: a = 1/sqrt2 (y+dy)
a- = 1/sqrt2(y-dy)

I am given this Hamiltonian:

And asked to diagonalize.
I understand how we do such a Hamiltonian:

But I don't understand how to deal with the extra term in my given Hamiltonian. Usually we use

To get

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Sep 26, 2019

DrClaude

Mentor

Isn't it simply that you need to express ##x## in terms of ##a## and ##a^\dagger##?

Thread 'Eigenvalues of a "unusual" Hamiltonian of a harmonic oscillator'

for d), I am a bit confused. I have two trains of thoughts here any thoughts on which answer is correct, and why the other one is incorrect? Both seem like valid solutions to me. Or is the question ambiguous? thanks

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