Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 1 down vote favorite

When considering the 'Heisenberg' picture of the harmonic oscillator, I've come across the step: $$\begin{align} \left\langle n\left|(\hat{q_H}\hat{H}-\hat{H}\hat{q_H})\right|k\right\rangle &= (E_k-E_n)q_{nk} \end{align}$$

where $\hat{q_H}$ is a position operator. I realise that this must be a pretty simple step for someone who is familiar with Dirac notation, but I'm not sure how the author has unlocked the $\hat{H}$ from the $\hat{H}\hat{q_H}$?

Any comments greatly appreciated as always.

share|improve this question

2 Answers

up vote 2 down vote accepted

Hamiltonian can act on the left or on the right state vector.

share|improve this answer
Hey, thanks again. Somewhat surprisingly, I'm sure that this is the first time I've heard that :s. What is the motivation for an operator acting 'backwards'? – TCTopCat Apr 30 '12 at 10:14
1  
In a scalar product $(\psi_1,\hat{H}\psi_2)$ there are two state vectors. There is a definition of a conjugate operator: it is another operator and it acts on the other state vector, like this: $(\hat{H}^+\psi_1,\psi_2)=(\psi_1,\hat{H}\psi_2)$. The conjugate operator $\hat{H}^+$ is not obliged to be the same as $\hat{H}$, it may even not exists at all. But for a Hamiltonian the conjugate operator coincides with its the original Hamiltonian and this gives the eigenvalue of the original Hamiltonian. – Vladimir Kalitvianski Apr 30 '12 at 12:53
Thank you very much, I've used the adjoint of operators before, but I wasn't sure about the rigour of applying the operation to parts of a two operator product like $\hat{H}\hat{q_h}$. – TCTopCat Apr 30 '12 at 13:57
up vote 1 down vote

$$\begin{align} \left\langle n\left|\hat{q_H}\hat{H}-\hat{H}\hat{q_H}\right|k\right\rangle &= \left\langle n\left|\hat{q_H}\hat{H}\right|k\right\rangle -\left\langle n\left|\hat{H}\hat{q_H}\right|k\right\rangle\\ &= E_k\left\langle n\left|\hat{q_H}\right|k\right\rangle -E_n\left\langle n\left|\hat{q_H}\right|k\right\rangle\\ &= (E_k-E_n)\left\langle n\left|\hat{q_H}\right|k\right\rangle\\ &\equiv (E_k-E_n)q_{nk} \end{align}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.