Griffiths uses an algebraic "brute force" technique to solve the harmonic oscillator. I'm somewhat confused regarding a few parts.

$$\frac{1}{2m}[p^2 + (m \omega x)^2] \psi = E \psi$$

$H = \frac{1}{2m}[p^2 + (m \omega x)^2]$

We are about to factor $H$, noting that the numerical equavilant is $u^2 + v^2 = (iu+v)(-iu+v)$. We now define $a_{\pm} = \frac{1}{\sqrt{2 \hbar m \omega }}(\pm ip+m \omega x)$

Skipping through the commutator, I don't understand how we can say: $a_+ a_ - = \frac{1}{\hbar \omega}H + \frac{1}{2}$

Looking at the original equation, we factored $[p^2 + (m \omega x)^2]$, so we can replace this with $a_+ a_ - $. Under this, couldn't we just say that $H = \frac{1}{2m} (a_+ a_ -) $

My second question has to do with the statement (this is a direct quote): Now, here comes the crucial step: I claim that if $\psi$ satisfies the Schrödinger equation with energy $E$, (that is: $H \psi = E \psi$), then $a_+ \psi$ satisfies the Schrödinger equation with energy $(E+ \hbar \omega)$

I don't understand why we're multiplying $\psi$ by a piece of Hamiltonian, and the by the Hamiltonian again.

  • $\begingroup$ For the second part: compute $[H,a_{+}]$. What do you get? Apply this commutator to the wavefunction. $\endgroup$ Oct 16, 2013 at 6:57
  • $\begingroup$ $[H,a_+] = Ha - aH$ Using a test function: $-\frac{\hbar}{i}\frac{d}{dx}(\frac{f(x)}{\sqrt{2 \hbar m \omega}} (\pm ip +m \omega x)) + \frac{1}{\sqrt{2\hbar m \omega}} (\pm i p +m \omega x) \frac{df}{dx}$ $\endgroup$
    – Astrum
    Oct 16, 2013 at 7:25

1 Answer 1


I think perhaps what you're missing is in the "skipping through the commutator" part. Do you understand where we get this equation (try computing it yourself, if not): $$a_{-}a_{+} = \frac{1}{2 \hbar m \omega}[p^{2} + (m\omega x)^{2}] - \frac{i}{2\hbar}[x,p]$$ Now, the canonical commutator, I'm sure you noticed (as it's boxed on the same page in Griffiths) is $[x,p] = ih$. Insert this into the above equation and note that we now have: $$a_{-}a_{+} = \frac{1}{2 \hbar m \omega}[p^{2} + (m\omega x)^{2}] + \frac{1}{2}$$ All you need to do from there recognize the first term as $\frac{1}{h\omega}H$.

Looking at the original equation, we factored $[p^{2}+(m\omega x)^{2}]$, so we can replace this with a+a−. Under this, couldn't we just say that $H=\frac{1}{2}(a_{+}a_{−})$

Careful here... remeber that $p$ and $x$ in this expression (and in the Hamiltonian generally) are operators, not scalars. This is why our "intuitive guesses" of $a_{\pm}$ are not exact factors of $[p^{2}+(m\omega x)^{2}]$, and why the canonical commutator above is important.

Edit: I just noticed that Griffiths does include this intermediate step in computing $a_{-}a_{+}$: $$a_{-}a_{+} = \frac{1}{2 \hbar m \omega}[p^{2} + (m\omega x)^{2}-im\omega(xp-px)]$$ Notice that if $x$ and $p$ were scalars, the rightmost term would be 0, and your intuition about $a_{-}$ and $a_{+}$ being "factors" would be correct. Once you realize they are operators, however, it's obvious that we need to substitute $[x,p] = xp-px = ih$.

  • $\begingroup$ I did understand the how/why of the canonical commutator, and I see that doing the multiplication out for $a_+ a_-$ gives you $a_+ a_ - = \frac{1}{\hbar \omega}H + \frac{1}{2}$. I see, so we used $a_{\pm}$ to find the exact factors of $H$, which involved the canonical commutator. $\endgroup$
    – Astrum
    Oct 16, 2013 at 7:34
  • $\begingroup$ We're actually not particularly concerned with the exact factors of $H$. We're primarily concerned with what the operators $a_{-}$ and $a_{+}$ do to the wavefunction $\psi$. $\endgroup$
    – RGMyr
    Oct 16, 2013 at 7:43
  • $\begingroup$ Now we can focus on the second of my questions. I understand why $H= \hbar \omega (a_+ a_- + \frac{1}{2})$ The reason behind the next step leaves me puzzled - $H(a_+ \psi)= \hbar \omega (a_+ a_- + \frac{1}{2})(a_+ \psi)$ The algebra that follows is clear, the motivation for $(a_+ psi)$ is not. $\endgroup$
    – Astrum
    Oct 16, 2013 at 7:47
  • $\begingroup$ The idea is to write the Hamiltonian in terms of these operators, and then show that for any solution of the Schrodinger EQ $\psi_{n}$, $a_{-}\psi_{n}$ and $a_{+}\psi_{n}$ are also solutions, and have the energy you mentioned in your post. $\endgroup$
    – RGMyr
    Oct 16, 2013 at 7:50
  • $\begingroup$ Sorry, typed that before I saw your specific question. Do you understand what is meant by an "eigenvalue equation?" (i.e. $H\psi = E\psi$). This is a ubiquitous concept in QM (and linear algebra in general), and it's very important to understand what that means. $\endgroup$
    – RGMyr
    Oct 16, 2013 at 7:52

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