I have been solving the harmonic oscillator problem in quantum mechanics using Algebraic Method and since then I am consulting the books of Tannoudji and Griffiths for that matter. While studying both of them I ham having a small confusion while defining the Hamiltonian operator Tannoudji used this expression $\frac{H}{\hbar w}$ for defining the Hamiltonian operator where $H$ is classical Hamiltonian while Griffiths have not used this expression at all. The confusion is how this expression from Tannoudji came across and why Griffiths did not use this expression at all.


1 Answer 1


It is always possible to multiply an operator by a scalar without changing the eigenvectors. If $|\psi_n\rangle$ is an eigenvectot of $H$ with eigenvalue $E_n$, then $|\psi_n\rangle$ will also be an eigenvalue of $k\cdot H$, for any value $k$:

$$ (kH)|\psi_n\rangle = k(H|\psi_n\rangle) = kE_n|\psi_n\rangle. $$

We see that the eigenvalue corresponding to the vector is scaled by a factor of $k$. Physically, this corresponds to changing the units the different quantities are measured in. In particular, the choice of dividing by $\hbar \omega$ means that the spectrum of the operator will be measured in units of $\hbar \omega$, and the eigenvalues of the harmonic oscillator will hence be $\left(n+\frac{1}{2}\right)$, with $n = 0, 1, 2,3\ldots$

  • $\begingroup$ Can you show me how defining the Hamiltonian operator in this way gives me energy eigen value in terms of ${\hbar w}$ $\endgroup$ May 26, 2014 at 8:22
  • $\begingroup$ You have swapped the words eigenvector ($|\psi\rangle$) and eigenvalue ($E_n$) in your explanation. $\endgroup$
    – Tom-Tom
    May 26, 2014 at 8:29
  • $\begingroup$ @V.Rossetto So I have - it's edited now! $\endgroup$
    – Sten
    May 26, 2014 at 8:30
  • $\begingroup$ @RoshanShrestha Which part is unclear? If $H' = \frac{H}{\hbar \omega}$, then an eigenvalue of $E_n$ for $H'$ corresponds to an eigenvalue of $\hbar \omega E_n$ for $H$. $\endgroup$
    – Sten
    May 26, 2014 at 8:33

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