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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.

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2 Answers 2

up vote 2 down vote accepted

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

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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
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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

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

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