Here I am talking about Harmonic Oscillators with Hamiltonian $$ H=\frac{1}{2m}(p^2+(m\omega x)^2), $$ with eigenstates $|1\rangle,|2\rangle,\ldots$

Many textbooks define the annihilation operator to be $$ A=\frac{m\omega x+ip}{\sqrt{2m\hbar\omega}}, $$ So $A|n\rangle=\sqrt{n}|n-1\rangle$.

The definition of $A$ look very complicated, so my question is, how can people possibly come up with it at the first place? Just imagine you are the first one who discover the operator $A$. How can you possible discover it?

More precisely, we want $A$ to have the property that, if $|\phi\rangle$ is an eigenstate, then so is $A|\phi\rangle$. Also, $A|\phi\rangle$ must not be a constant mutiple of $|\phi\rangle$. Can we obtain an expression of $A$ starting from this property?

  • $\begingroup$ Just to be clear, is $|\phi\rangle$ is an energy eigenstate? $\endgroup$ Commented Jun 29, 2019 at 23:32
  • $\begingroup$ @AlfredCentauri Yes $\endgroup$
    – Ma Joad
    Commented Jun 29, 2019 at 23:34
  • $\begingroup$ The number operator satisfies this property. Maybe you need something more than just $A|\phi\rangle$ is an energy eigenstate? $\endgroup$ Commented Jun 29, 2019 at 23:36
  • $\begingroup$ @AlfredCentauri No. The number operator $A^\dagger A$ gives the same eigenstate when it operates on an eigenstate. $A$ gives a different state. $\endgroup$
    – Ma Joad
    Commented Jun 30, 2019 at 2:46

1 Answer 1


It's not so bad if you just set all the constants to one, $m = \omega = \hbar = 1$. You can always recover these later by dimensional analysis.

Classically, we have the Hamiltonian $$H = \frac{p^2 + x^2}{2}.$$ Even without knowing quantum mechanics, you can think of phase space as the complex plane, with $x$ as the real axis and $p$ as the imaginary axis. That motivates looking at quantities like $x + i p$. In fact, this is useful because then you can factor the Hamiltonian as a "difference of squares". Defining $a = (x + i p)/\sqrt{2}$, we have the simple form $$H = \frac{(x + ip)(x - ip)}{2} = a^* a.$$ The variable $a$ is useful because it has a simple time evolution. We have $$\frac{dx}{dt} = p, \quad \frac{dp}{dt} = -x$$ which means that $$\frac{da}{dt} = \frac{p - i x}{\sqrt{2}} = - i a.$$ So the variable $a$ just rotates in a circle in the complex plane. It's all really nice, and doesn't require anything beyond high school algebra.

This all has nothing to do with quantum mechanics. Now, in the context of quantum mechanics, we find that $[a, a^\dagger] = 1$, which implies that $a$ and $a^\dagger$ behave like creation and annihilation operators. We also find that $H$ picks up an additional constant term. Finally, the fact that $a$ evolves by picking up a complex phase means that $[H, a] \propto a$, which implies $a$ shifts energy eigenvalues. However, just about all of these things are present in some modified form in the classical theory, where we're using nothing fancier than the complex plane.

  • 2
    $\begingroup$ Rather than setting all constants = 1, it's more common to work with "non-dimensionalized variables", that is, defining $x'= \sqrt{\dfrac{m\omega}{\hbar}}x $ and $p'=\dfrac{1}{\sqrt{m\hbar \omega}} p$. Those are variables with no dimensions, and you get the same hamiltonian, but now with the variables $x'$ and $p'$. Then you just un-do the variable switch. $\endgroup$
    – FGSUZ
    Commented Jun 29, 2019 at 23:48
  • 8
    $\begingroup$ @FGSUZ As long as dimensional analysis works, this is just another way of saying the same thing! $\endgroup$
    – knzhou
    Commented Jun 30, 2019 at 0:31
  • 1
    $\begingroup$ @kevin Yes, it's just a different way of saying the same thing - but for this particular question, it's a better way of saying it. The additional clarity can be clunky and unnecessary, but here its primary role is to add clarity. $\endgroup$ Commented Jun 30, 2019 at 6:28
  • $\begingroup$ @knzhou When things get complicated, recovering constants from dimensional analysis is burdensome. Much better using normalization. This is actually a long-standing fight between physicists and engineers. $\endgroup$ Commented Jun 30, 2019 at 7:44

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