# Raising and lowering operators

$$a^ta=n$$ 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.

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

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

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

• Why the last step holds ??? That was my question Nov 29, 2020 at 10:16
• @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. Nov 29, 2020 at 10:46
• Is it just labelling then?? Nov 29, 2020 at 13:23
• @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). 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 $$a^\dagger|n\rangle\propto |n+1\rangle$$