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:

Note that these operators satisfy \begin{align*} [\hat{q},\hat{p}]\psi&=-iq\partial_q\psi+i\partial_q(q\psi)\\ &=-iq\partial_q\psi+iq\partial_q\psi+i\psi, \end{align*} that is, \([\hat{q},\hat{p}]=i\). This is called the canonical commutation relation.

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.

Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at if you have something to share.

CC-BY-SA 4.0 Jonny Evans.