$a^ta=n$ where $a^t$ is the raising operator. While doing the harmonic oscilaltor I encountered these. I could get that $n$ and Hamiltonian commute and if $|n\rangle$ the common eigenstate with eigenvalue $n$ then $a^t|n\rangle$ also eigen vector of operator n. From this they conclude that $a^t|n\rangle=c|n+1\rangle$ But how does this happen . Could someone explain the math? How do they know that $a^t$ raises the eigen value by 1. So far we only got $a^t|n\rangle$ eigenvector of $n$ with eigenvalue $n-1$. Dont tell me like raising operators are like that they raise it by one. I wanna know the clear cut reason.

  • $\begingroup$ I would recommend reading the section about the quantum harmonic oscillator in "Modern Quantum Mechanics" by J.J. Sakurai. $\endgroup$ Nov 29, 2020 at 8:57
  • $\begingroup$ Comment to the post (v4): 't' should be 'dagger'. $\endgroup$
    – Qmechanic
    Nov 29, 2020 at 9:37

3 Answers 3


To show this we need the following properties:

  1. The number operator is $\hat{n}=\hat{a}^{\dagger}\hat{a}$.
  2. The raising and lowering operators obey this commutation relation: $[\hat{a},\hat{a}^{\dagger}]=1$.

It then follows that:

$$ [\hat{n},\hat{a}^{\dagger}]=[\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}]=\hat{a}^{\dagger}[\hat{a},\hat{a}^{\dagger}]+[\hat{a}^{\dagger},\hat{a}^{\dagger}]\hat{a}=\hat{a}^{\dagger}, $$

where in the first equality we use the definition of $\hat{n}$, in the second the relation $[\hat{A}\hat{B},\hat{C}]=\hat{A}[\hat{B},\hat{C}]+[\hat{A},\hat{C}]\hat{B}$, and in the third the commutators $[\hat{a},\hat{a}^{\dagger}]=1$ and $[\hat{a}^{\dagger},\hat{a}^{\dagger}]=0$. This commutation we just derived implies that $\hat{n}\hat{a}^{\dagger}=\hat{a}^{\dagger}\hat{n}+\hat{a}^{\dagger}$.

We are now ready to turn to your question. The key step you are missing is to relize not only that $\hat{a}^{\dagger}|n\rangle$ is an eigenstate of the number operator $\hat{n}$, but that it is an eigenstate with eigenvalue $n+1$. To see this, consider:

$$ \hat{n}(\hat{a}^{\dagger}|n\rangle)=(\hat{a}^{\dagger}\hat{n}+\hat{a}^{\dagger})|n\rangle=(\hat{a}^{\dagger}n+\hat{a}^{\dagger})|n\rangle=(n+1)(\hat{a}^{\dagger}|n\rangle), $$

where in the first equality we use the relation we derived earlier for $\hat{n}\hat{a}^{\dagger}$, in the second we use the eigenvalue equation of the number operator $\hat{n}|n\rangle=n|n\rangle$, and in the third we simply re-organize the expresssion.

So what does this mean? The state $\hat{a}^{\dagger}|n\rangle$ is an eigenstate of the number operator $\hat{n}$ with eigenvalue $n+1$. We can write an eigenstate of the number operator with eigenvalue $n+1$ as $|n+1\rangle$, so this means that

$$ \hat{a}^{\dagger}|n\rangle\propto|n+1\rangle. $$

  • $\begingroup$ Why the last step holds ??? That was my question $\endgroup$
    – ramAN
    Nov 29, 2020 at 10:16
  • $\begingroup$ @ramAN: there are two points to the last step. The first is that we give the name $|m\rangle$ to an eigenstate of the number operator with eigenvalue $m$. Therefore, if $m=n+1$, we call the eigenstate $|n+1\rangle$. The second is that you have shown that $\hat{a}^{\dagger}|n\rangle$ is an eigenstate of the number operator with eigenvalue $n+1$. Put together, these two observations imply that $\hat{a}^{\dagger}|n\rangle$ and $|n+1\rangle$ must be the same state (up to a constant). Hence your result. $\endgroup$
    – ProfM
    Nov 29, 2020 at 10:46
  • $\begingroup$ Is it just labelling then?? $\endgroup$
    – ramAN
    Nov 29, 2020 at 13:23
  • 1
    $\begingroup$ @ramAN if you want to call it "labelling" that is fine, but the key is that they have the same "label" because they are representing the same state (up to a constant). $\endgroup$
    – ProfM
    Nov 29, 2020 at 14:28

Use $[a, a^\dagger] = 1$ here like this:

Since you agree that $a^\dagger|n \rangle$ is an eigenstate of $a^\dagger a$, it means

$$ a^\dagger a (a^\dagger |n \rangle) = \lambda a^\dagger |n \rangle\tag{1} $$ Now $a^\dagger a a^\dagger = a^\dagger(1+a^\dagger a)$, therefore (1) gives $\lambda = n+1$ by using $a^\dagger a|n \rangle = n |n \rangle$, which is the definiton.

Therefore, $a^\dagger |n\rangle$ is eigenstate of $N=a^\dagger a$ with eigenvalue of $n+1$, and must be therefore proportional to $|n+1\rangle$.


The operator $\hat n$ gives as eigenvalue the energy level of a given eigenstate. If you apply it to $a^\dagger|n\rangle$, you get $$\hat na^\dagger|n\rangle=a^\dagger a a^\dagger|n\rangle=a^\dagger (a^\dagger a+I)|n\rangle\\\quad =a^\dagger|n\rangle +a^\dagger (a^\dagger a)|n\rangle=a^\dagger|n\rangle + a^\dagger \hat n|n\rangle$$ Then $$\hat n a^\dagger |n\rangle= (n+1)a^\dagger|n\rangle$$ So if $\hat n|n\rangle=n|n\rangle$ and $\hat n a^\dagger|n\rangle= (n+1)a^\dagger|n\rangle$ it means that $a^\dagger|n\rangle\propto |n+1\rangle$


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