In this post we will be exploring two mathematical operators known as the creation and annihilation operators. We’ll begin with one of the simplest dynamical systems possible: the simple harmonic oscillator (SHO), and show how this system can induce very simple quantum effects.

On its own, the SHO is not very interesting to study.

Unless you disturb it.

Spring
Figure 1: The Spring System as a physical analog of a quantum harmonic oscillator.

Once it’s disturbed a number of things begin to happen. First, it attempts to restore itself back to how it was before you disturbed it, and how quickly it does this is dictated by the spring constant k.

Second, the way in which it restores itself depends to two properties: the mass m (to which is applied the disturbance force F), and the spring constant k. At this point I will assume that the system has no damping, in other words, once the initial disturbance is set in motion the system will oscillate forever.

When you have a system like this where you have two factors (disturbative and restorative) that are constantly pulling and pushing against each other, the position of the mass at any given time t can be found, even if it has been oscillating for years. This position, denoted x, will depend on time t, and its value is given by

x(t) = A\cos\left(\sqrt{\displaystyle\frac{k}{m}} t + \phi\right)

where \phi is just a number that lets you say that the initial disturbance was a certain distance from a zero position (you can ignore it for now). Also note that the factor A out the front is just the disturbance magnitude, also known as the amplitude.

So where did this equation come from? Long ago Newton did some work that allowed us to equate a force F with the resultant acceleration a that force would give to a motion-resistance mass m. This being Newton’s second law

F = ma

This is actually a second-order ordinary differential equation and this property becomes more apparent when acceleration is written as the second derivative of position with respect to time:

F = m\displaystyle\frac{\mbox{d}^2x}{\mbox{d}t}

and one can re-arrange the two above equations so they look like a proper differential equation:

\displaystyle\frac{d^2 x}{dt} = -\displaystyle\frac{k}{m}x

which has the solution x(t) = A\cos\left(\sqrt{\frac{k}{m}} t + \phi\right).

It is interesting that mathematics says that the position of the mass under these conditions is dictated by a periodic, sinusoidal pattern. In other words, the mass will continue to oscillate up and down in a very periodic way and this position now can be described by 1) the initial displacement of the disturbance (the amplitude A), 2) the initial phase of the position \phi, 3) and the frequency \sqrt{\frac{k}{m}}, which is itself dependent on the restoration force and the mass. So far all very easy concepts.

Energy

Now let’s talk about the mechanical energy E of this system. Why? Because in many cases in physics it is often more useful to talk about the energy of a system rather than the values of all its individual components (like position, velocity, acceleration, angular frequency, amplitude, phase, etc…) – plus you get little cheats like conservation of energy and symmetries that help you perform calculations.

The energy can always be expressed as the sum of two parts: the first being the kinetic energy and the second begin the potential energy. As we all know, a mass undergoing oscillatory motion will move between having only potential energy (at the peaks and troughs) and only kinetic energy (when it is moving the fastest, midway between the peaks and troughs) – all the while the sum of the two (the total energy of the system) remains perfectly constant.

These two components of total mechanical energy is given by

\displaystyle E_{\mbox{tot}} = E_{\mbox{kin}} + E_{\mbox{pot}}

Further, it is well known that

\displaystyle E_{\mbox{kin}} = \frac{1}{2}Mp^2

E_{\mbox{pot}} = \displaystyle\frac{1}{2}Kx^2

where one would normally write M = \frac{1}{m} but I’ve decided to write it this way to show how similar the two equations are. Note that kinetic energy is dependent on the momentum of the mass whilst the potential energy is dependent on the position of the mass. This makes sense, since potential energy is maximised when the mass is not moving (i.e. it has reached maximum amplitude, which is a specific position of the system). The kinetic energy is maximised when the mass is moving the fastest, and so this equation needs to know about the momentum of the mass.

Some Quantum Mechanics

Let’s leave the SHO behind for a moment and recall the Schrödinger equation:

\displaystyle H\psi = E_{\mbox{tot}}\psi

A couple of things here: The H on the left hand side is called the Hamiltonian of the system which is supposed to encapsulate all the information about the energy of the system (which is why E appears in the same spot on the right-hand side). Schrödinger says that if it is a quantum system then we can expand the above equation as

\left( -\displaystyle\frac{1}{2}M\hbar^2 \displaystyle\frac{\partial^2 }{\partial x^2} + \frac{1}{2}Kx^2 \right)\psi = E_{\mbox{tot}}\psi

But this is ugly, and will probably scare a lot of people away.

Essentially, Schrödinger is telling us that the total energy E_{\textup{tot}} of the (quantum) version of the SHO is equal to the ugly bit:

E_{\mbox{tot}} = -\displaystyle\frac{1}{2}M\hbar^2 \displaystyle\frac{\partial^2 }{\partial x^2} + \frac{1}{2}Kx^2

and this is not at all like the total energy with potential and kinetic energy as written a couple of equations above – it’s got all these extra bits!

Where did they come from?

They come from re-writing the momentum p of a particle by

p = -i\hbar\displaystyle\frac{\partial }{\partial x}

On its own this doesn’t do much because it is differential operator – so it must operate on something. I haven’t told you yet what it operates (spoiler: wavefunctions), so regard this as a just an instruction for now.

Schrödinger’s equation just equates two different forms of total energy of a system. On the left is H, a quantum version which encapsulates some sort of ‘quantumness’ due to the -i\hbar part, and the right E is the good ol’-fashioned total mechanical energy.

The Schrödinger equation has a solution as well, but you’re not going to like how we solve it but I promise that you will like the look of the final solution:

E = \left(n+\displaystyle \frac{1}{2}\right)\hbar\omega

Isn’t that nice?

Looks more simple than you might expect.

This is saying that given some funny quantum stuff going on with the momentum part of the SHO (which we haven’t explained yet) the total energy of the (quantum) system is not continuous.

Harmonic_Oscillator
Figure 1: The first 8 solutions of the quantum harmonic oscillator. Note the bottom density. Observe how the density is concentrated at the origin. This means that a particle that obeys the harmonic oscillator (in its ground state) spends the majority of its time at the bottom of the potential well, as one would expect for a state with very little energy. As the energy of the oscillator increases the probability density  peaks at the edges (note that all energy levels except the ground show this feature). This is where the “pendulum” is turning and has low velocity and therefore spends more time there. This is a perfect example of a quantum system reproducing classical physics in the limit of large quantum numbers, i.e. the correspondence principle.

Well that is weird. Why?

Because, as it is written, the energy of a SHO only has values for each integer n \in \mathbb{N}.

The total energy is this integer number \left(n + \frac{1}{2}\right)\hbar mutiplied by the angular frequency, which we already know is dependent on the mass of the particle and the restorative force that is driving it back toward equilibrium after that initial push.

So we didn’t do anything wrong here. All we did was say that the momentum (related to the kinetic energy part of the SHO) was a bit funny (quantised). Once we did that, however, the total energy of the system suddenly stops being continuous and has all these integer values, like a ladder, and in between: simply does not exist!

So what does the n represent in the equation for total energy of a quantum SHO? This number starts at n=0 (it can’t go negative because it is a natural number) and goes to infinity via all the natural numbers. At n=0 the total energy of the quantum SHO is simply

E = \displaystyle\frac{1}{2}\hbar\omega

For a mass of 1\mbox{kg} and a restorative force of 1 \mbox{kg/s}^2, the energy associated with the value n=0 is

\displaystyle E_0 = \frac{1}{2}\times\hbar\omega = \frac{1}{2}\frac{h}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2}6.626\times 10^{-34}\mbox{J.s}\times 1 = 3.313\times 10^{-34}

This value of this energy is called the zero point energy and, even though it clearly is not zero, it is still a very, very small amount of energy. This is not always the case though, there are systems where the zero point energy is actually zero, but not for a simple harmonic oscillator!

Our problem is that the SHO is a very simple system. It is so simple it shows up an awful lot in nature (nature likes simple things). For example, in quantum field theory, empty space is supposedly made up of an infinity of these tiny quantum simple harmonic oscillators. There are other overlapping fields too, like the fermionic and bosonic fields. Each of these are modelled as having a tiny SHO at each and every point in space. Given a forceful disturbance these fields can oscillate too (although damped) and when not oscillating they too have a zero point energy, and, you guessed it, since they are quantum they will have a non-zero energy (even though it’s called zero point energy!). At each point in space, the combination of all these fields contributes \frac{1}{2}\hbar\omega. Since in any given volume of space there are an infinite number of these tiny quantum SHOs, the zero point energy is infinite (when summed up) resulting in the need to renormalise. This non-zero zero point energy is also an explanation for the cosmological constant.

Er, isn’t this article about creation and annihilation?

How do we get the creation and annihilation operators from here? The trick is to consider Heisenberg’s uncertainty principle. His principle states that position x and momentum p do not commute (in other words, the difference between measuring position then momentum, and measuring momentum then position, is not zero). It should be classically, as the order of measurement should make no difference to the overall result, but in quantum mechanics, it does. So we say

px - xp = i\hbar > 0

In other words, if you measure momentum p then position x and subtract from it the same measurement of position x then momentum p, the difference is not zero, but this tiny little number i/\hbar. In essence, it doesn’t quite commute.

The non-commutativity of the factorisation of the Hamiltonian of the quantum SHO is off by a factor of the zero point energy.

If we now go back to the quantum Hamiltonian H (recall it is the quantum version of total energy of the quantum SHO):

\displaystyle H = \frac{1}{2}Mp^2 + \frac{1}{2}Kx^2

and put in the values of M and K:

H = \displaystyle\frac{1}{2m}p^2 + \displaystyle\frac{1}{2}m\omega^2x^2

then factorise

\displaystyle H = \frac{1}{2}m\omega^2\left(x - \displaystyle\frac{i}{m\omega}p\right)\left(x + \displaystyle\frac{i}{m\omega}p \right)

(note: the i is there just so that when we expand we get i^2 = -1 and the sign changes from a negative back to a positive)

When we expand we will inevitably get a xp factor and a px factor which we will want to cancel with the identity xp - px = 0, but we can’t because they don’t commute. So the factorisation above won’t quite work because when we expand we instead get:

\displaystyle H = \frac{1}{2}m\omega^2x^2 + \frac{1}{2m}p^2 + \frac{1}{2}i\omega\left(xp - px\right)

and the bit on the right (\frac{1}{2}i\omega\left(xp - px\right)) doesn’t go away and spoils all the fun, and so we end up have to tack on an additional bit to our Hamiltonian, so it looks like this:

\displaystyle H + \frac{1}{2}i\omega\left(xp - px\right) = H+\frac{1}{2}i\omega[x,p] = H+\frac{1}{2}i\omega(i\hbar) = H-\frac{1}{2}\hbar\omega = H - E_0 \neq H

OK, wow. This means a lot. The non-commutativity of the factorisation of the Hamiltonian of the quantum SHO is off by a factor of the zero point energy. The quantum-ness introduced in to the momentum p, which resulted in a non-zero zero point energy E_0, also causes the factorisation of the Hamiltonian to fail by the same amount.

Now look at the factors in the parentheses (from a couple of equations back):

\textup{factor 1:} \left(x - \displaystyle\frac{i}{m\omega}p\right)
\textup{factor 2:} \left(x + \displaystyle\frac{i}{m\omega}p\right)

we call these the creation and annihilation operators a^{\dagger} and a respectively. Using them, we can re-write the Hamiltonian, firstly with the factor out the front:

\displaystyle H = \frac{1}{2}m\omega^2\left[x^2 + \displaystyle\frac{1}{m^2\omega^2}p^2 \right]

then factorising the part in the parentheses (not forgetting about the extra zero point energy bit):

\displaystyle H = \frac{1}{2}m\omega^2\left[\left(x - \displaystyle \frac{i}{m\omega}p \right)\left( x + \displaystyle \frac{i}{m\omega}p \right) + \displaystyle \frac{\hbar}{m\omega^2}\right]

Substituting the symbols for the creation and annihilation operators, we simply equate together to two representations of the part inside the brackets to get:

\displaystyle a^{\dagger}a + \frac{\hbar}{m\omega^2} = x^2 + \displaystyle\frac{1}{m^2\omega^2}p^2

A little re-arranging and we get:

\displaystyle a^{\dagger}a = \left( x^2 + \frac{1}{m^2 \omega^2}p^2 \right) - \frac{\hbar}{m\omega^2}

We would prefer not to have the -\frac{\hbar}{m\omega^2} on the end. To remedy this, it is preferrable to re-define the creation and annihilation operators like this:

\displaystyle a^{\dagger} := \sqrt{\frac{m\omega}{2\hbar}} \left(x - \frac{i}{m\omega}p\right)
\displaystyle a := \sqrt{\frac{m\omega}{2\hbar}} \left(x + \frac{i}{m\omega}p\right)

Where the bits out the front, \sqrt{\frac{m\omega}{2\hbar}}, are called normalisation factors and are pretty much chosen to make the following calculation simpler, and also comes from the fact that when we expand the commutator [x,p] we get a new factor of \frac{\hbar}{m\omega} that we will want to cancel out.

\displaystyle\begin{array}{rcl} a^{\dagger} a &=& \sqrt{\displaystyle\frac{m\omega}{2\hbar}} \cdot \sqrt{\displaystyle\frac{m\omega}{2\hbar}} \left( x - \displaystyle\frac{i}{m\omega}p \right)\left( x + \displaystyle\frac{i}{m\omega}p \right)  \\ &=& \displaystyle\frac{m\omega}{2\hbar}\left(x^2 + \displaystyle\frac{i}{m\omega}xp - \displaystyle\frac{i}{m\omega}px - \displaystyle\frac{i^2}{m^2\omega^2}p^2 \right) \\ &=& \displaystyle\frac{m\omega}{2\hbar}\left(x^2 + \displaystyle\frac{i}{m\omega}(xp - px) + \displaystyle\frac{1}{m^2\omega^2}p^2 \right) \end{array}

And now, you must press the “I believe in Quantum Mechanics“-button (i.e. the Born postule) in which case fundamental relations between canonical conjugate quantities (like position and momentum) satisfy

\displaystyle [x,p] = i\hbar

noting that (xp - px) = [x,p] we have

\displaystyle\begin{array}{rcl} a^{\dagger} a &=& \displaystyle\frac{m\omega}{2\hbar}\left(x^2 + \displaystyle\frac{i}{m\omega}(i\hbar) + \displaystyle\frac{1}{m^2\omega^2}p^2 \right)  \\ &=& \displaystyle\frac{m\omega}{2\hbar}\left(x^2 - \displaystyle\frac{\hbar}{m\omega} + \displaystyle\frac{1}{m^2\omega^2}p^2 \right)  \\ &=& \displaystyle\frac{1}{\hbar\omega}\left(\frac{1}{2}m\omega^2\right)\left(x^2 - \displaystyle\frac{\hbar}{m\omega} + \displaystyle\frac{1}{m^2\omega^2}p^2 \right)  \\ &=&\displaystyle\frac{1}{\hbar\omega}\left(\displaystyle\frac{1}{2}m\omega^2x^2 - \displaystyle\frac{1}{2}\hbar\omega + \displaystyle\frac{1}{2m}p^2 \right) \end{array}

Now we just need to substitute in the expression for the Hamiltonian \displaystyle H:= \frac{1}{2}m\omega^2 x^2 + \frac{1}{2m}p^2 and we get

\displaystyle a^{\dagger} a = \frac{1}{\hbar\omega}\left( H - \frac{1}{2}\hbar \omega \right)

A little rearranging and we get the main result of this article:

\displaystyle H = \hbar\omega\left( a^{\dagger}a + \frac{1}{2} \right)

Whoa, whoa, hold on. This looks almost exactly the same as our solution to Schrödinger’s equation from the beginning of the article:

\displaystyle E =\hbar\omega\left(n+\frac{1}{2}\right)

except that we have replaced n with a^{\dagger}a. In fact, it is very important to say that

a^{\dagger}a = n

This allows us to write:

\displaystyle [a,a^{\dagger}] := aa^{\dagger} - a^{\dagger}a

we have

\displaystyle aa^{\dagger} = [a,a^{\dagger}] + a^{\dagger}a = 1 + a^{\dagger}a = 1 + N

Summary

We started with a very simple physical system: the simple harmonic oscillator (SHO) and used Newton’s second law to turn it in to a linear ordinary differential equation (ODE). Then we used kinetic and potential energy to write down the total energy of the system at any point.

Then we equated the total energy of the system to Schrödinger’s equation (essentially they should be the same, as they both express the energy of the system and they’re both differential equations) – and then made one important assumption: that momentum was quantised.

Once we did this we found that not only is the energy also quantised but also has a minimum called the zero-point energy.

Then we mixed in a little of Heisenberg’s Uncertainty principle and out popped two non-Hermitian operators a^{\dagger} and a called the creation and annihilation operators which was shown to have an equivalent explanation in terms of the quantised energy levels of the SHO.

See Also

Chapter 9, Harmonic Oscillator, Nuclear Engineering at MIT.

References

[1] Lancaster, T & Blundell, S. J., Quantum Field Theory for the Gifted Amateur, Oxford University Press (2014)