next up previous contents
Next: The Harmonic Oscillator Up: Ladder Operators, Phonons and Previous: Ladder Operators, Phonons and   Contents

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

We define the two operators
$\displaystyle a:=\sqrt{\frac{m\omega}{2\hbar}}\hat{x}+\frac{i}{\sqrt{2m\hbar\om...
...r}:=\sqrt{\frac{m\omega}{2\hbar}}\hat{x}-\frac{i}{\sqrt{2m\hbar\omega}}\hat{p}.$     (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)
$\displaystyle [\hat{x},\hat{p}]=i\hbar\leadsto [a,a^{\dagger}] = 1.$     (264)

We define the number operator
$\displaystyle \hat{N}:=a^{\dagger}a$     (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$,
$\displaystyle \hat{N} \vert n\rangle= n \vert n\rangle.$     (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$.
$\displaystyle 0 \le \vert\vert a\vert n\rangle\vert\vert^2 = \langle n \vert a^...
...rangle
= \langle n \vert\hat{N}\vert n\rangle = n \langle n \vert n\rangle = n.$     (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$,...
$\displaystyle \hat{N}a$ $\displaystyle =$ $\displaystyle a^{\dagger}aa= \left( aa^ {\dagger}- [a,a^{\dagger}]\right) a =
\left( aa^ {\dagger}- 1\right) a =
a \left( \hat{N}- 1 \right)$  
$\displaystyle \leadsto \hat{N}\underline{a \vert n\rangle}$ $\displaystyle =$ $\displaystyle a \left( \hat{N}- 1 \right)\vert n\rangle = (n-1)\underline{a \vert n\rangle}$  
$\displaystyle \leadsto \hat{N}\underline{a^2 \vert n\rangle}$ $\displaystyle =$ $\displaystyle (\hat{N}a)a\vert n\rangle =
a \left( \hat{N}- 1 \right)a \vert n\rangle = a (n-1-1){a \vert n\rangle}
= (n-2) \underline{a^2 \vert n\rangle}$  
$\displaystyle ...$     (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$,
$\displaystyle a\vert n\rangle$ $\displaystyle =$ $\displaystyle C_n \vert n-1\rangle \leadsto n= \langle n a^{\dagger} a n \rangle
= \vert C_n\vert^2 \langle n-1 \vert n-1 \rangle =
\vert C_n\vert^2$  
$\displaystyle \leadsto a\vert n\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n} \vert n-1\rangle.$ (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:

$\displaystyle \hat{N}{a \vert n\rangle}$ $\displaystyle =$ $\displaystyle (n-1){a \vert n\rangle}$  
$\displaystyle \leadsto \hat{N}{a \vert 1\rangle}$ $\displaystyle =$ $\displaystyle 0\cdot {a \vert 1\rangle}$  

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)

$\displaystyle a^{\dagger}\vert n\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\vert n+1\rangle.$ (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 }$:
$\displaystyle \vert n\rangle = \frac{(a^{\dagger})^n}{\sqrt{n!}}\vert\rangle.$     (271)


next up previous contents
Next: The Harmonic Oscillator Up: Ladder Operators, Phonons and Previous: Ladder Operators, Phonons and   Contents
Tobias Brandes 2004-02-04