I'm reading simple harmonic oscillator problem in "Modern Quantum Mechanics" by J.J. Sakurai.
The approach is by defining the annihilation ($a^{t}$) and creation ($a$) operators, then a number operator is defined as the product between these operators $N=a^{t}a$. Also, an energy eigenket of $N$ is denoted by its eigenvalue $n$, that is, $$N|n\rangle=n|n\rangle.$$ Then we can find $$Na|n\rangle=(n-1)a|n\rangle$$ and the book says this implies $$a|n\rangle=c|n-1\rangle,$$ where $c$ is a numerical constant, the problem is I can't understand why this is true. I'm new with Dirac's notation, maybe that is the problem.


the problem is I can't understand why this is true

From your equation 1, we get that:

$$Nc|n-1\rangle = (n-1)c|n-1\rangle$$

Your equation 2 is:

$$Na|n\rangle = (n-1)a|n\rangle$$

Thus, it must be that

$$a|n\rangle = c|n-1\rangle $$

To be clear, $a$ is an operator that, given a ket, returns a ket so we could write something like

$$|m\rangle = a|n\rangle$$

and then your equation 2 becomes

$$N|m\rangle = (n-1)|m\rangle$$

which implies that $|m\rangle$ is proportional to $|n-1\rangle$, i.e.,

$$|m\rangle = c|n-1\rangle$$


This becomes more apparent with some parenthesis:

$$ N(a|n\rangle)=(n-1)(a|n\rangle) $$

Since $N$ yields an eigenvalue of $(n-1)$ when operating on the state $a|n\rangle$, then $a|n\rangle$ must be an eigenstate of $N$ with an eigenvalue of $(n-1)$, aka $|n-1\rangle$ (a overall phase factor/numerical constant $c$ could also be included in this state, which would be cancelled from both sides).


Your question is why the eigenvalue-$n-1$ eigenspace has dimension $1$. Since repeated application of $a$ can reduce the eigenvalue until it is $0$, while repeated application of $a^\dagger$ can increase the eigenvalue, each eigenspace has the same dimension. You can easily prove this dimension is $1$ in the $n=0$ case.

  • $\begingroup$ So, if eigenspace hasn't dimension 1 the equality doesn't hold, does it? $\endgroup$ – Alberto Navarro Jun 17 '18 at 20:06
  • $\begingroup$ @AlbertoNavarro But it does have dimension $1$. Can you prove that for the eigenvalue-$0$ eigenspace? $\endgroup$ – J.G. Jun 17 '18 at 20:09

protected by Qmechanic Jun 18 '18 at 2:56

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