QHO
As a warm-up, consider the QHO in one dimension.
\[\begin{split}
\begin{gather*}
\hat{H}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega ^2\hat{x}^2 \\
\left[ \hat{x},\hat{p} \right] =i\hbar \\
\end{gather*}
\end{split}\]
in position basis, \(\hat{p}=-i\hbar \partial _x\). Check
\[\begin{split}
\begin{align*}
\left[ \hat{x},\hat{p} \right] f\left( x \right) &=-i\hbar x\partial _xf+i\hbar \partial _x\left( xf \right) \\
&=-i\hbar x\partial _xf+i\hbar f+i\hbar x\partial _xf\\
&=i\hbar f
\end{align*}
\end{split}\]
In position basis, the eigenvalues satisfy
\[ -\frac{\hbar ^2}{2m}\partial _{x}^{2}\phi \left( x \right) +\frac{1}{2}m\omega ^2x^2\phi \left( x \right) =E\phi \left( x \right) \]
We can, however, solve it algebraically (without going to the position basis). First, let’s adopt some dimensionless coordinates. We know \([\hat{H}]=[\hbar\omega]\); write
\[ \hat{H}=\frac{\hbar \omega}{2}\left( \frac{m\omega}{\hbar}\hat{x}^2+\frac{1}{m\hbar \omega}\hat{p}^2 \right) \]
Define
\[ \hat{X}=\sqrt{\frac{m\omega}{\hbar}}\hat{x},\quad \hat{P}=\frac{1}{\sqrt{m\hbar \omega}}\hat{p} \]
Then
\[ \left[ \hat{X},\hat{P} \right] =\sqrt{\frac{m\omega}{\hbar}}\frac{1}{\sqrt{m\hbar \omega}}\left[ \hat{x},\hat{p} \right] =i \]
Let’s now define
\[ \hat{a}^{\dagger}=\frac{1}{\sqrt{2}}\left( \hat{X}-i\hat{P} \right) ,\quad \hat{a}=\frac{1}{\sqrt{2}}\left( \hat{X}+i\hat{P} \right) \]
which gives
\[\begin{split}
\begin{gather*}
\hat{a}^{\dagger}\hat{a}=\frac{1}{2}\left( \hat{X}-i\hat{P} \right) \left( \hat{X}+i\hat{P} \right) =\frac{1}{2}\left( \hat{X}^2+\hat{P}^2+i\left[ \hat{X},\hat{P} \right] \right) \\
\hat{H}=\frac{\hbar \omega}{2}\left( \hat{X}^2+\hat{P}^2 \right) =\hbar \omega \left( \hat{a}^{\dagger}\hat{a}+\frac{1}{2} \right)
\end{gather*}
\end{split}\]
This gives a convenient way to construct the spectrum. First, check
\[ \left[ \hat{a},\hat{a}^{\dagger} \right] =\frac{1}{2}\left[ \hat{X}+i\hat{P},\hat{X}-i\hat{P} \right] =\frac{1}{2}i\left[ \hat{P},\hat{X} \right] -\frac{1}{2}i\left[ \hat{X},\hat{P} \right] =1 \]
So, if \(\hat{H}|E\rangle =E|E\rangle\), we have
\[\begin{split}
\begin{align*}
\hat{H}\left( \hat{a}^{\dagger}|E\rangle \right) &=\hbar \omega \left( \hat{a}^{\dagger}\hat{a}+\frac{1}{2} \right) \hat{a}^{\dagger}|E\rangle \\
&=\hbar \omega \left( \hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}+\frac{1}{2}\hat{a}^{\dagger} \right) |E\rangle \\
&=\hbar \omega \left( \hat{a}^{\dagger 2}\hat{a}+\hat{a}^{\dagger}+\frac{1}{2}\hat{a}^{\dagger} \right) |E\rangle \\
&=\hat{a}^{\dagger}\left( \hat{H}+\hbar \omega \right) |E\rangle \\
&=\hat{a}^{\dagger}\left( E+\hbar \omega \right) |E\rangle \\
&=\left( E+\hbar \omega \right) \left( \hat{a}^{\dagger}|E\rangle \right)
\end{align*}
\end{split}\]
which means \(\hat{a}^{\dagger}|E\rangle\) is another eigenstate with energy \(E+\hbar\omega\). This relates the different eigenstates. We just need to find the ground state. Since
\[ \langle \phi |\hat{a}^{\dagger}\hat{a}|\phi \rangle =\left\| \hat{a}|\phi \rangle \right\| ^2\ge 0,\quad \forall |\phi \rangle \]
If \(\hat{a}\) has a null vector, then it will be the ground state. Let’s just posit such a state exist, i.e.,
\[ \exists |0\rangle \quad \mathrm{s}.\mathrm{t}.\quad \hat{a}|0\rangle =0\]
Then the eigen spectrum is given by
\[ \left\{ |n\rangle ;E_n=\hbar \omega \left( n+\frac{1}{2} \right) \right\} \]
We can also find the matrix elements of \(\hat{a},\hat{a}\) in this eigen basis. Recall
\[ |n+1\rangle =\mathcal{N} \hat{a}^{\dagger}|n\rangle \]
where \(\mathcal{N}\in\mathbb{R}^+\) is the normalization factor. Also, from the preceding discussion we have
\[ \hat{a}^{\dagger}\hat{a}|n\rangle =n|n\rangle \]
The normalization is then
\[\begin{split}
\begin{align*}
1=\langle n+1|n+1\rangle &=\mathcal{N} ^2\langle n|\hat{a}\hat{a}^{\dagger}|n\rangle \\
&=\mathcal{N} ^2\langle n|\left( \hat{a}^{\dagger}\hat{a}+1 \right) |n\rangle \\
&=\left( n+1 \right) \mathcal{N} ^2
\end{align*}
\end{split}\]
\[ \hat{a}^{\dagger}|n\rangle =\sqrt{n+1}|n+1\rangle \]
similar argument gives \(\hat{a}|n\rangle =\sqrt{n}|n-1\rangle\). In a matrix picture.
\[\begin{split} \hat{a}^\dagger=\left[ \begin{matrix}
0& & & & \\
1& 0& & & \\
& \sqrt{2}& 0& & \\
& & \sqrt{3}& 0& \\
& & & \ddots& \ddots\\
\end{matrix} \right] ,\quad \hat{a}=\left[ \begin{matrix}
0& 1& & & \\
& 0& \sqrt{2}& & \\
& & 0& \sqrt{3}& \\
& & & 0& \ddots\\
& & & & \ddots\\
\end{matrix} \right] \end{split}\]
This is a “number” basis. We call it the Fock space. It’s separable (but infinite dimensional). We are only left with showing \(\hat{a}|0\rangle =0\) {TODO-UNKNOWN-WORD} solution. To do so, let’s go back to the position basis. Recall
\[ \hat{a}=\frac{1}{\sqrt{2}}\left( \hat{X}+i\hat{P} \right) =\frac{1}{\sqrt{2}}\left( X+i\left( -i\frac{\partial}{\partial X} \right) \right) =\frac{1}{\sqrt{2}}\left( X+\frac{\partial}{\partial X} \right) \]
\[ \hat{a}|\phi _0\rangle =0 \]
\[ \left( X+\frac{\partial}{\partial X} \right) \phi _0\left( X \right) =0 \]
\[ \int{\frac{d\phi _0\left( X \right)}{\phi _0\left( X \right)}}=-\int{XdX}\]
\[ \phi _0\left( X \right) =\mathcal{N} e^{-X^2/2}\]
Which is a Gaussian. Recall \(X=\sqrt{\frac{m\omega}{\hbar}}x\),
\[ \phi _0\left( x \right) =\mathcal{N} e^{-m\omega x^2/\left( 2\hbar \right)} \]
To get the normalization,
\[\begin{split}
\begin{align*}
1&=\int_{-\infty}^{\infty}{\left| \phi _0\left( x \right) \right|^2 dx}=\mathcal{N} ^2\int_{-\infty}^{\infty}{e^{-m\omega x^2/\hbar}dx}\\
&=\mathcal{N} ^2\sqrt{\frac{\hbar}{2m\omega}}\int_{-\infty}^{\infty}{e^{-m\omega x^2/\hbar}\exp \left( -\left( \sqrt{\frac{\hbar}{2m\omega}}x \right) ^2/2 \right) d\left( \sqrt{\frac{\hbar}{2m\omega}}x \right)}\\
&=\mathcal{N} ^2\sqrt{\frac{\hbar}{2m\omega}}\sqrt{2\pi}
\end{align*}
\end{split}\]
\[ \mathcal{N} =\left( \frac{m\omega}{\pi \hbar} \right) ^{1/4} \]
\[ \phi _0\left( x \right) =\left( \frac{m\omega}{\pi \hbar} \right) ^{1/4}e^{-m\omega x^2/\left( 2\hbar \right)} \]
Side note: Gaussian integration
\[ \int_{-\infty}^{\infty}{e^{-ax^2/2}dx}=\sqrt{\frac{2\pi}{a}},\quad a>0 \]
A standard discussion will next introduce the real-space wave function of the excited states through the Hermite polynomial. Let’s try to avoid that!
coherent state
We have seen that the creation and annihilation operators provides a simple way to analyze the QHO. Let’s now take these as the “coordinates” for our problem. Recall position basis \(\hat{x}|x\rangle =x|x\rangle\). Similarly, we consider the eigenstates of the form
\[ \hat{a}|\alpha \rangle =\alpha |\alpha \rangle ;\quad \alpha \in \mathbb{C} \]
E.g., The ground state \(\hat{a}|0\rangle =0|0\rangle =0\). To solve for the eigenstate, write
\[ |\alpha \rangle =\sum_{n=0}{C_n\left( \alpha \right) |n\rangle}\]
Then,
\[\begin{split}
\begin{align*}
\hat{a}|\alpha \rangle &=\sum_{n=0}{C_n\left( \alpha \right) \hat{a}|n\rangle}\\
&=\sum_{n=0}{C_n\left( \alpha \right) \sqrt{n}|n-1\rangle}\\
&=\sum_{n=0}{C_{n+1}\left( \alpha \right) \sqrt{n+1}|n\rangle}\\
\end{align*}
\end{split}\]
versus
\[ \alpha |\alpha \rangle =\sum_{n=0}{C_n\left( \alpha \right) \alpha |n\rangle}\]
we conclude
\[ C_{n+1}\left( \alpha \right) =\frac{\alpha C_n\left( \alpha \right)}{\sqrt{n+1}}=\frac{\alpha ^2C_{n-1}\left( \alpha \right)}{\sqrt{\left( n+1 \right) n}}=\cdots =\frac{\alpha ^{n+1}C_0\left( \alpha \right)}{\sqrt{\left( n+1 \right) !}} \]
Note: \(|\alpha\rangle\) is a superposition of \(\{|n\rangle\}\). It’s not an energy eigenstate (unless \(\alpha=0\)).
We have
\[ |\alpha \rangle =C_0\left( \alpha \right) \sum_{n=0}{\frac{\alpha ^n}{\sqrt{n!}}|n\rangle}\]
Normalization:
\[\begin{split}
\begin{align*}
1&=\langle \alpha |\alpha \rangle =\left| C_0\left( \alpha \right) \right|^2\sum_{m,n=0}{\frac{\left( \alpha ^* \right) ^m\alpha ^n}{\sqrt{m!}\sqrt{n!}}\langle m|n\rangle}\\
&=\left| C_0\left( \alpha \right) \right|^2\sum_{m=0}{\frac{\left| \alpha \right|^{2m}}{m!}}\\
&=\left| C_0\left( \alpha \right) \right|^2e^{\left| \alpha \right|^2}
\end{align*}
\end{split}\]
\[ C_0\left( \alpha \right) =e^{-\left| \alpha \right|^2/2}\]
\[ |\alpha \rangle =e^{-\left| \alpha \right|^2/2}\sum_{n=0}{\frac{\alpha ^n}{\sqrt{n!}}|n\rangle}\]
This looks almost like the exponential! Recall
\[ |n\rangle =\frac{\hat{a}^{\dagger}}{\sqrt{n}}|n-1\rangle =\frac{\hat{a}^{\dagger 2}}{\sqrt{n\left( n-1 \right)}}|n-2\rangle =\cdots =\frac{\hat{a}^{\dagger n}}{\sqrt{n!}}|0\rangle \]
So,
\[\begin{split}
\begin{align*}
|\alpha \rangle &=e^{-\left| \alpha \right|^2/2}\sum_{n=0}{\frac{\alpha ^n\hat{a}^{\dagger n}}{\sqrt{n!}\sqrt{n!}}|0\rangle}\\
&=e^{-\left| \alpha \right|^2/2}e^{\alpha \hat{a}^{\dagger}}|0\rangle
\end{align*}
\end{split}\]
