The Ladder Operators $a$ and $a^{\dagger }$

We define the two operators

$$a:=\sqrt{\frac{m\omega}{2\hbar}}\hat{x}+\frac{i}{\sqrt{2m\hbar\om... ...r}:=\sqrt{\frac{m\omega}{2\hbar}}\hat{x}-\frac{i}{\sqrt{2m\hbar\omega}}\hat{p}.$$ (263)

You have showed in the problems that if two operators $A$ and $B$ are hermitian, $A=A^{\dagger}$, $B=B^{\dagger}$ the linear combination $C:=A+iB$ is not hermitian but $C^{\dagger}=A-iB$ (remember the analogy to a complex number $z=x+iy, z^*=x-iy$). We know that $\hat{x}$ and $\hat{p}$ are hermitian, therefore $a^{\dagger }$ (`a dagger') is the hermitian conjugate of $a$. From the commutator of $\hat{x}$ and $\hat{p}$ we easily find (see the problems)

$$[\hat{x},\hat{p}]=i\hbar\leadsto [a,a^{\dagger}] = 1.$$ (264)

We define the number operator

$$\hat{N}:=a^{\dagger}a$$ (265)

which is a hermitian operator because $\hat{N}^{\dagger}=(a^{\dagger}a)^{\dagger}= a^{\dagger}(a^{\dagger})^{\dagger}=\hat{N}$. The eigenvalues of $\hat{N}$ must be real therefore. We denote the eigenvalues of $\hat{N}$ by $n$ and show that the $n$ are non-negative integers: First, we denote the corresponding (normalized) eigenvectors of $\hat{N}$ by $\vert n\rangle$,

$$\hat{N} \vert n\rangle= n \vert n\rangle.$$ (266)

STEP 1: We show $n\ge 0$: remember the scalar product of two states $\vert\psi\rangle$ and $\vert\phi\rangle$ is denoted as $\langle \phi\vert \psi \rangle$.

$$0 \le \vert\vert a\vert n\rangle\vert\vert^2 = \langle n \vert a^... ...rangle = \langle n \vert\hat{N}\vert n\rangle = n \langle n \vert n\rangle = n.$$ (267)

STEP 2: We step down the ladder: if $\vert n\rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $n$, then $a\vert n\rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $n-1$, $a^2\vert n\rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $n-2$, $a^3\vert n\rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $n-3$,...

$$\hat{N}a = a^{\dagger}aa= \left( aa^ {\dagger}- [a,a^{\dagger}]\right) a = \left( aa^ {\dagger}- 1\right) a = a \left( \hat{N}- 1 \right)$$

$$\leadsto \hat{N}\underline{a \vert n\rangle} = a \left( \hat{N}- 1 \right)\vert n\rangle = (n-1)\underline{a \vert n\rangle}$$

$$\leadsto \hat{N}\underline{a^2 \vert n\rangle} = (\hat{N}a)a\vert n\rangle = a \left( \hat{N}- 1 \right)a \vert n\rangle = a (n-1-1){a \vert n\rangle} = (n-2) \underline{a^2 \vert n\rangle}$$

$$...$$ (268)

The state $a\vert n\rangle$ is an eigenstate of $\hat{N}$ with eigenvalue $n-1$ and therefore must be proportional to $\vert n-1\rangle$,

$$a\vert n\rangle = C_n \vert n-1\rangle \leadsto n= \langle n a^{\dagger} a n \rangle = \vert C_n\vert^2 \langle n-1 \vert n-1 \rangle = \vert C_n\vert^2$$

$$\leadsto a\vert n\rangle = \sqrt{n} \vert n-1\rangle.$$ (269)

The operator $a$ takes us from one eigenstate with eigenvalue $n$ to a lower eigenstate with eigenvalue $n$

STEP 3: We show that $n$ must be an integer, and the only possile eigenstates of $\hat{N}$ are $\vert\rangle$, $\vert 1\rangle$, $\vert 2\rangle$, ...

If we step down the ladder to lower and lower eigenvalues, we eventually would come to negative eigenvalues which can't be because all eigenvalues of $\hat{N}$ must be non-negative! The lowest possible eigenstate is $a\vert 1\rangle=\vert\rangle$ with eigenvalue 0:

$$\hat{N}{a \vert n\rangle} = (n-1){a \vert n\rangle}$$

$$\leadsto \hat{N}{a \vert 1\rangle} = 0\cdot {a \vert 1\rangle}$$

For any $n$ with $0<n<1$, the eigenvalue equation $\hat{N}{a \vert n\rangle} = (n-1){a \vert n\rangle}$ can only be true if $a \vert n\rangle=0$ is the zero-vector. It then becomes the trivial equation $0=0$ that contains no contradictions. But $a\vert n\rangle$ cannot be the zero-vector because the norm of $a\vert n\rangle$ is $\Vert a \vert n\rangle \Vert= \sqrt{n}>0$. Therefore, $0<n<1$ leads to a contradition. In the same way, there can't be values of $n$ with $1<n<2$ (application of $a$ leads us to the case $0<n<1$ which is already excluded. As a result, n is an integer.

Step 4: The normalized state $a^{\dagger}\vert n\rangle$ is (the proof is left for the problems)

$$a^{\dagger}\vert n\rangle = \sqrt{n+1}\vert n+1\rangle.$$ (270)

Therefore, $a^{\dagger }$ takes us up the ladder from one eigenstate $\vert n\rangle$ to the next higher $\vert n+1\rangle$. All the normalized eigenstates $\vert n\rangle$ can be created from the ground state $\vert\rangle$ by successive application of the ladder operator $a^{\dagger }$:

$$\vert n\rangle = \frac{(a^{\dagger})^n}{\sqrt{n!}}\vert\rangle.$$ (271)