Pre-QFT 1: The quantum harmonic oscillator
[2017-09-24 Sun]
This is a standard piece of theory in quantum mechanics, and it is
crucial for understanding QFT. I am including it here so I can look it
up whenever I need it.
The classical system called the simple harmonic oscillator involves a particle in a 1-dimensional space (coordinate \(q\)) moving under the influence of a potential which is quadratic in \(q\). We'll write \(p\) for the momentum of the particle along the \(q\)-axis. In suitable units (to make the constants as simple as possible) the Hamiltonian for the system is \[H=\frac{1}{2}p^2+\omega^2q^2,\] where \(\omega\) is a constant. When we quantise this system, it makes life slightly easier if we rewrite this Hamiltonian as \[H=\omega\frac{1}{\sqrt{2\omega}}\left(\omega q-ip\right) \frac{1}{\sqrt{2\omega}}\left(\omega q+ip\right).\] We apply canonical quantisation to this system:
- taking as our Hilbert space of states the space of square-integrable functions in \(q\);
- replacing \(q\) with the operator \(\hat{q}\): \[\hat{q}(\psi(q))=q\psi(q);\]
- replacing \(p\) with the operator \(\hat{p}\): \[\hat{p}(\psi)=-i\partial_q\psi.\]
The Hamiltonian becomes \[\hat{H}=\omega a^\dagger a,\] where \begin{align*} a&=\frac{1}{\sqrt{2\omega}}(\omega\hat{q}+i\hat{p})\\ a^\dagger&=\frac{1}{\sqrt{2\omega}}(\omega\hat{q}-i\hat{p})\\ \end{align*} (here the dagger denotes Hermitian conjugation). Equivalently, \begin{align*} \hat{q}&=\frac{1}{\sqrt{2\omega}}\left(a+a^\dagger\right)\\ \hat{p}&=-i\sqrt{\frac{\omega}{2}}\left(a-a^\dagger\right). \end{align*}
Note that before we quantise, the ordering of \(q\) and \(p\)
is not important, so we have to make choice about how we order them
when we quantise. Different choices will give different Hamiltonians
(differing by a constant, which emerges from the canonical
commutation relation as the qs and ps move past one another). The
choice we made here is called normal ordering, where we have
written the Hamiltonian as an expression in \(a\) and \(a^\dagger\),
and taken all the \(a^\dagger\) terms to the left. Later, in QFT, we
will have an infinite number of simple harmonic oscillators to
handle at the same time, and the extra constants that would appear
from a different choice of ordering would give rise to an annoying
infinite quantity.
The operators \(a\) and \(a^\dagger\) satisfy the following
commutation relations:
\begin{align*}
\left[a,a^\dagger\right]
&=\frac{1}{2\omega}[\omega\hat{q}+i\hat{p},\omega\hat{q}-i\hat{p}]\\
&=-i[\hat{q},\hat{p}]\\
&=1.
\end{align*}
Moreover, they satisfy the following commutation relations with the
Hamiltonian \(\hat{H}\):
\begin{align*}
[\hat{H},a]
&=\omega a^\dagger a^2-\omega aa^\dagger a\\
&=\omega[a^\dagger,a]a\\
&=-\omega a,
\end{align*}
\begin{align*}
[\hat{H},a^\dagger]
&=\omega a^\dagger aa^\dagger-\omega \left(a^\dagger\right)^2a\\
&=\omega a^\dagger[a,a^\dagger]\\
&=\omega a^\dagger.
\end{align*}
There is a state \(|0\rangle\) called the vacuum state which
satisfies \[a|0\rangle=0,\quad\langle 0|0\rangle=1.\] This is an
eigenstate of the quantised Hamiltonian with eigenvalue zero:
\[\hat{H}|0\rangle=\omega a^\dagger a|0\rangle=0.\]
Explicitly, in our chosen Hilbert space \(L^2(\mathbf{R})\), this is
the unique (normalised) solution to the differential equation
\(a\psi=0\), that is \[\partial_q\psi=-\omega q\psi.\] This solution
is \(\psi=Ce^{-\omega q^2/2}\), where \(C\) is a suitably chosen
normalisation factor.
We can act on this state using \(\left(a^\dagger\right)^n\) to generate
further states
\[|n\rangle=\left(a^\dagger\right)^n|0\rangle,\]
These are also eigenstates of \(\hat{H}\); using the commutation
relation \([a,a^\dagger]=1\) and the fact that \(a|0\rangle=0\), we get
\begin{align*}
\hat{H}|n\rangle
&=\omega a^\dagger a \left(a^\dagger\right)^n|0\rangle\\
&=\omega \left(a^\dagger\right)^n|0\rangle
+\omega \left(a^\dagger\right)^2a\left(a^\dagger\right)^{n-1}|0\rangle\\
&=2\omega \left(a^\dagger\right)^n|0\rangle
+\omega \left(a^\dagger\right)^3a\left(a^\dagger\right)^{n-2}|0\rangle\\
&=\cdots\\
&=n\omega|n\rangle.
\end{align*}
The norms of these states can be computed recursively as follows:
\begin{align*}
\langle n|n\rangle
&=\langle 0|a^n\left(a^\dagger\right)^n|0\rangle\\
&=\langle n-1|n-1\rangle
+\langle 0|a^{n-1}a^\dagger a\left(a^{\dagger}\right)^{n-1}|0\rangle\\
&=\cdots\\
&=n\langle n-1|n-1\rangle\\
&=n!.
\end{align*}
so normalised eigenstates are
\(\frac{1}{\sqrt{n!}}\left(a^\dagger\right)^n|0\rangle\).
We can find explicit wavefunctions in our Hilbert space which
represent these eigenstates by applying \(\left(a^\dagger\right)^n\)
(considered as the differential operator
\(\frac{1}{2\omega}\left(\omega\hat{q}-i\hat{p}\right)\)) to the
Gaussian vacuum state \(\psi=Ce^{-\omega q^2/2}\). For example:
\begin{align*}
|1\rangle
&=a^\dagger|0\rangle\\
&=\frac{1}{2\omega}\left(\omega q+\partial_q\right)\psi\\
&=\frac{1}{2\omega}\left(\omega Cqe^{-\omega q^2/2}+C\omega q e^{-\omega q^2/2}\right)\\
&=Cqe^{-\omega q^2/2}.
\end{align*}
In general, these states are of the form \(P_n(q)e^{-\omega q^2/2}\)
where \(P_n(q)\) is the Hermite polynomial of degree \(n\). These
functions form an orthonormal basis for \(L^2(\mathbf{R})\), so we
have completely solved the quantum mechanical problem of finding the
spectrum of the quantum Hamiltonian.
Classically, this system arises from something like a ball on a spring
obeying Hooke's law and ignoring friction: the ball, released, will
oscillate along the \(q\) axis with frequency \(\omega\). There is a
continuum of states depending on how much we stretch the spring before
we release the ball (the more we stretch it, the larger the potential
energy stored in the system, so the higher the energy of the
state). Quantum mechanically, we see a discrete collection of states
with energies being nonnegative integer multiples of \(\omega\). In
quantum field theory, the thing which is oscillating will be (roughly
speaking) the value of the (Fourier transform of the) field at a
point. See my introductory post on QFT for more about this.