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# Content
$\newcommand{\L}{\mathcal L}$
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## Repe Harmonic oscillator
Consider $H = \frac{1}{2}m \omega^{2}x^{2}+\frac{1}{2m}p^{2}$
How can we solve this in qm?
In principle the solution is super painful (Hermite polynomials)
However there is a nice trick. What we actually want to do is we want to diagonalize the hamiltonian. We thus try to decompose it:
We want to write it as the product of two operators (instead of as a sum)
Looking at the structure $H = A^{2} + B^{2}$
we can attempt to write
$H = (A +iB)(A -iB)$
Which would give us:
$$H = \underbrace{(\sqrt{\frac{m\omega^{2}}{2}} x + i \sqrt{\frac{1}{2m}}p)(\sqrt{\frac{m\omega^{2}}{2}} x - i \sqrt{\frac{1}{2m}}p)}_{\frac{1}{2}m\omega^{2} + \frac{1}{2m}p^{2} -i \sqrt{\frac{m\omega^{2}}{2}}\sqrt{\frac{1}{2m}}xp +i (\ldots) px } - \ldots$$
We note that due to the commutator we still need to subract a part
$$H = (\sqrt{\frac{m\omega^{2}}{2}} x + i \sqrt{\frac{1}{2m}}p)\left(\sqrt{\frac{m\omega^{2}}{2}} x - i \sqrt{\frac{1}{2m}}p\right)- i\sqrt{m\omega^{2}}\sqrt{\frac{1}{2m}}[x,p]$$
Using $[x,p] = i\hbar$ we get. We also introduce new operators $\hat a$ and $\hat a^{\dagger}$
$$H = \underbrace{\left(\sqrt{\frac{m\omega^{2}}{2}} x + i \sqrt{\frac{1}{2m}}p\right)}_{a^{\dagger}c}\left(\sqrt{\frac{m\omega^{2}}{2}} x - i \sqrt{\frac{1}{2m}}p\right)+ \hbar\sqrt{m\omega^{2}}\sqrt{\frac{1}{2m}}$$
We use the constant $c$ to make our commutators nice:
$[a,a^{\dagger}] = 1$
Thus:
$a = \sqrt{\frac{m\omega}{2\hbar}}(x + \frac{i}{m\omega} p)$
$a^{\dagger} = \sqrt{\frac{m\omega}{2\hbar}}(x - \frac{i}{m\omega} p)$
And:
$x = \sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+ a) = x_{zpf} (a^{\dagger}+ a)$
$p = i\sqrt{\frac{\hbar m\omega}{2}}(a^{\dagger}- a)$
We now get:
$H = \hbar \omega (a^{\dagger}a + \frac{1}{2})$
and $a^{\dagger}a = N$
with the nice property that:
$[N, a^{\dagger}] = a^{\dagger}$
$[N, a] = -a$
Which allows us to do:
$\hat N a^{\dagger}\ket n = \hat N \sqrt{n+1} \ket{n+1}$ (note that for now $n+1$ is simply a label, where we do not know $\hat N$ yet)
$(a^{\dagger}\hat N + a^{\dagger})\ket n = \hat N \sqrt{n+1} \ket{n+1}$
$(a^{\dagger}\ket{n} N + \sqrt{n+1}\ket {n +1}= \hat N \sqrt{n+1} \ket{n+1}$
$\sqrt{n+1}\ket{n+1} N + \sqrt{n+1}\ket {n +1}= \hat N \sqrt{n+1} \ket{n+1}$
$(N+1)\ket {n +1}= \hat N \ket{n+1}$
Which shows that the $a^{\dagger}$ operator indeed increases $N$ by one.
# Field Quantisation
In the coulomb gauge we have:
$$\nabla \cdot \vec A = 0$$
In free space we also have:
$$\square \cdot \vec A = \frac{1}{c^{2}}\dede{^{2}}{t^{2}} \vec A - \nabla^{2} \vec A = 0$$
We know by now that this means that $\vec A$ can be solved by plane waves.
We now consider a box with length $L$.
The above equations are solved (classically) by:
$$\vec A(\vec x,t) = \frac{1}{\sqrt{L^{3}}}\sum\limits_{\vec k, \lambda} q_{\vec k,\lambda}(t)\vec e(\vec k, \lambda)e^{i\vec k \cdot \vec x} + c.c.$$
Here $q$ are the fourier expansion coefficients (which are non-zero for $\vec k$ which contain energy)
And $\lambda \in \{1,2\}$ is the polarization index.
$\vec e$ is the polarization vector, which is orthogonal to $\vec k$ to enforce $\nabla \cdot A = 0$
We now want to find the total energy in this system. By using:
$H_{tot} =\frac{1}{8\pi}\int (|E|^{2} + |B|^{2}) d^{3}x$
(and a bunch of math...)
we can find that the energy decomposes into individual modes.
(from now on I will not write $\vec k$ for brevity )
$H = \sum\limits_{k,\lambda} \frac{\omega_{k}^{2}}{2\pi c^{2}} |q_{k,\lambda}|^{2}$
#### Sidenote:
This is not surprising if you remember the "Parseval" (or Plancherel) theorem from MMP I, which states that the Fourier transform of the absolute square of a function is equal to the absolute square of the Fourier transform.
The shape of the argument is as follows $H = \int |E|^{2} dx = \int |\tilde E|^{2} d\omega = \int | \delta_{\omega}|^{2}d \omega$
## Quadratures
Seeing that we have a quadratic energy contribution for every $k$ and $\lambda$ lends itself to a description with harmonic oscillators.
For a single mode we thus have:
$$\frac{\omega_{k}^{2}}{2\pi c^{2}} |q_{k,\lambda}|^{2} = \frac{1}{2m} P^{2}_{k,\lambda} + \frac{1}{2}\omega_{k,\lambda}^{2}m Q^{2}_{k,\lambda}$$
To make our lives simple we set $m = 1$ (note the units)
$$ |q_{k,\lambda}|^{2} = \frac{2\pi c^{2}}{\omega_{k}^{2}} \left(\frac{1}{2} P^{2}_{k,\lambda} + \frac{1}{2}\omega_{k,\lambda}^{2}Q^2_{k,\lambda} \right)$$
Using $a^{2} + b^{2} = (a + ib)(a-ib)$ we get (we use this trick all the time):
$$q_{k,\lambda}(t) = \sqrt{\pi c^{2}} \left(Q_{k,\lambda}(t)+ \frac{i}{\omega_{k}}P_{k,\lambda}(t)\right)$$
We check that $P,Q$ are actually conjugate to eachother. We thus want that:
$\dot Q = -\{H, Q\}$ and $\dot P =-\{H,P\}$ (note that the brackets here are still classical liouville brackets)
$\{H,Q\} = \dede{H}{Q}\dede{Q}{P} - \dede{H}{P}\dede{Q}{Q} = Q \dede{Q}{P} - P = -P$
Which shows the above statement.
## Quantisation
We now do canonical quantisation:
$$[Q,P] = i\hbar \{Q,P\} = i\hbar$$
Because all modes are orthogonal (they decompose into a sum in the hamiltonian) we also get:
$[Q_{k,\lambda},P_{k,\lambda}] = i\hbar \delta_{k,k'} \delta_{\lambda,\lambda'}$
So we now have one harmonic oscillator per point in _k-space_ (not real space), which corresponds to photon in this fourier mode.
We solve the harmonic oscillator (the standard way we always do it) and get:
$H_{k,\lambda} = \hbar\omega_{k}(\hat a_{k,\lambda}^{\dagger}\hat a_{k,\lambda} + \frac{1}{2})$
With:
$$\hat a = \sqrt{\frac{\omega}{2\hbar}}\hat Q_{k,\lambda} + \frac{i}{\sqrt{2\hbar\omega}}\hat P_{k,\lambda} = \sqrt{\frac{\omega_{k}}{2\pi\hbar c^{2}}}\hat q_{k,\lambda}$$
$$\hat a^{\dagger} = \sqrt{\frac{\omega}{2\hbar}}\hat Q_{k,\lambda} - \frac{i}{\sqrt{2\hbar\omega}}\hat P_{k,\lambda}$$
The creation and annihilation operator respectively
Note that they are not hermitian.
We can now rebuild our vector potential (but now as an operator):
$$\vec{\hat A}(\vec x) = \frac{1}{\sqrt{L^{3}}} \sum\limits_{k,\lambda} \sqrt{\frac{2\pi\hbar c^{2}}{\omega_{k}}}\left(\hat a_{k,\lambda} \vec e_{k,\lambda} e^{ikx} + h.c. \right)$$
We have now done a bit of a circle:
$A \to q \to (Q,P) \to (\hat Q,\hat P)\to \hat a \to \hat q \to \hat A$
But now all our observables are operators!
## Results
We find that between different frequencies (or k's) there is no uncertainty (because $\delta_{k,k'}$ in the canonical quantisation). This means that we can measure the electric field everywhere at once!
However $[A,\dede{}{t} A] \neq 0$ (which we also see from $[Q_{k,\lambda},P_{k,\lambda}] = i\hbar$)
This means we cannot measure both the $B$ and $E$ field of a mode simultaneously. The same is also true for the position.
## States
Note that due to the mode independence we can write our states as tensorproducts over these modes:
In the single oscillator case we have:
$\ket n = \frac{a^{\dagger n}}{\sqrt{n!}}\ket 0$
For more modes we just write:
$$\ket{n_{k_{1},\lambda_{1}}, \ldots n_{k_{m},\lambda_{m}}} = \frac{a_{k_{1},\lambda_{1}}^{\dagger n_{k_{1},\lambda_{1}}}}{\sqrt{n_{k_{1},\lambda_{1}}!}}\ket 0 \otimes \ldots \otimes \frac{a_{k_{m},\lambda_{m}}^{\dagger n_{k_{1},\lambda_{1}}}}{\sqrt{n_{k_{1},\lambda_{1}}!}}\ket 0$$
Due to the multi-linearity we can collect all prefactors at the front:
$$\ket{n_{k_{1},\lambda_{1}}, \ldots n_{k_{m},\lambda_{m}}} = \frac{a_{k_{1},\lambda_{1}}^{\dagger n}}{\sqrt{n!}} \cdot \ldots \cdot \frac{a_{k_{m},\lambda_{m}}^{\dagger n}}{\sqrt{n!}}\ket 0 \otimes \ldots \otimes \ket 0$$
Or for brevity
$$\ket{n_{k_{1},\lambda_{1}}, \ldots n_{k_{m},\lambda_{m}}} = \frac{a_{k_{1},\lambda_{1}}^{\dagger n}}{\sqrt{n!}} \cdot \ldots \cdot \frac{a_{k_{m},\lambda_{m}}^{\dagger n}}{\sqrt{n!}}\ket {0,\ldots, 0}$$
We often choose the order of the modes of interest beforehand, and then later just use the same order, without specifying it again.
Consider the energy in the field:
$\bra{N_{k_{1},\lambda_{1}} \ldots} H \ket{N_{k_{1},\lambda_{1}} \ldots} = \sum\limits \hbar \omega_{k} \left(N_{k, \lambda} + \frac{1}{2}\right) = \sum\limits \hbar \omega N + \bra 0 H \ket 0$
We can thus think about the field as a collection of particles with energy $\hbar \omega$, this is the photon!
Anihilation operator thus destroys a photon in a specific mode, while creation operator creates one.
### Concept question:
What is the interpretation of the following operator acting on a state:
$a^{\dagger}_{k}a_{k'}$
for $k \neq k'$
What about
$a^{\dagger 2}_{k}a_{k'}^{2}$
# Quantum Matter in Quantum Fields
We remember the transition rate from quantum matter in classical fields
$$\Gamma_{i\to f}^{k\lambda} = \frac{2\pi}{\hbar} |\bra f \hat V \ket i|^{2} \delta(E_{f}-E_{i}-\hbar \omega ) + \frac{2\pi}{\hbar} |\bra f \hat V \ket i|^{2} \delta(E_{f}-E_{i}+\hbar \omega ) $$
We see here that the transition is proportional to the magnitude squared of $\hat V$. This means that both emission and absorbtion need photons in the field.
We _only_ have stimulated emission, no spontaneous emission
## SQ case:
We consider $H = H_{em} + H_{0}+H_{I}$
Only the interaction term actually matters so we can apply fgr to it:
$$\Gamma_{i\to f} \frac{2\pi}{\hbar}| \bra f H_{I}\ket i \delta_{E_{f}-E_{i}} $$
Consider now a combined photon-atom state:
$\ket {i, N^{i} \ldots}$
The interaction hamiltonian arises again as the product between the particle current density and the vector potential:
$$H_{I}= - \frac{q}{\sqrt{L^{3}}} \sum\limits \sqrt{\frac{2\pi \hbar}{\omega_{k}}}\left( j(-k)\vec e\hat a + h.c. \right) $$
The flip for the $k$ inside of $j$ comes from the fact that $A(x) \propto e^{ikx} + e^{-ikx}$
We thus get
$\int A(x)j(x) dx \propto \int e^{ikx}j(x) + e^{-ikx}j(x) dx$
Relabling the x in the later part we get
$\int A(x)j(x) dx \propto \int e^{ikx}j(x) + e^{ikx}j(-x) dx$
We see that due to this trick $\hat a$ always is combined with $j(-k)$ and $\hat a ^{\dagger}$ is always combined with $j(k)$.
We can thus only take the transition elements of:
$j(-k) \hat a$
and
$j(k) \hat a^{\dagger}$
respectively
$$\bra {f, N^{f} \ldots} j(-k) \hat a \ket {i, N^{i} \ldots}$$
The hilbert space for atom and field separate
$$\bra {f} j(-k) \ket{i}\bra{N^{f} \ldots} \hat a \ket {N^{i} \ldots}$$
Because $\hat a$ destroys a photon, this can only be non-zero for $N_{i} - 1 = N_{f}$
Similarly for the $h.c.$ part we get
$N_{i} + 1 = N_{f}$
#### Absorbtion
$\ket {i, N^{i} \ldots}$
$\ket {f, N^{i}-1 \ldots}$
The matrix element is thus:
$$\bra {f} j(-k) \ket{i}\bra{N^{i}-1 \ldots} \hat a \ket {N^{i} \ldots}$$
_Reminder_: $\hat a \ket n = \sqrt{n}\ket {n-1}$
_Reminder_: $\hat a^{\dagger} \ket n = \sqrt{n+1}\ket {n+1}$
$$\bra {f} j(-k)\ket i \sqrt{N_{k}}$$
#### Emission
$\ket {i, N^{i} \ldots}$
$\ket {f, N^{i}+1 \ldots}$
The matrix element is thus:
$$\bra {f} j(k) \ket{i}\bra{N^{i}+1 \ldots} \hat a^{\dagger} \ket {N^{i} \ldots}$$
_Reminder_: $\hat a \ket n = \sqrt{n}\ket {n-1}$
_Reminder_: $\hat a^{\dagger} \ket n = \sqrt{n+1}\ket {n+1}$
$$\bra {f} j(k)\ket i \sqrt{N_{k} + 1}$$
#### FGR
Applying fermis golden rule we get:
$$\Gamma_{emiss} \propto |\bra f j(k) \ket i|^{2} (N_{k} + 1)$$
$$\Gamma_{abs} \propto |\bra f j(-k) \ket i|^{2} (N_{k})$$
We see that for emission even for $N_{k} = 0$ we can still emit into that mode, due to the $+1$.
_note: If you are wondering about where the energy comes from, remember, that we are de-exciting an atom to create the photon_
# Application:
### Lasers:
### Fiber optics: