# Creation/annihilation operators and trigonometric functions

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jshtok
TL;DR Summary
I am wondering at the striking similarity between the expressions for creation/annihilation operators in terms of position and momentum operators and the expressions for sine and cosine in terms of the exponential.
Hello everyone,
I have noticed a striking similarity between expressions for creation/annihilation operators in terms position and momentum operators and trigonometric expressions in terms of exponentials. In the treatment by T. Lancaster and S. Blundell, "Quantum Field Theory for the Gifted Amateur", Chapter 2, eqns. 2.9-2.13, the creation/annihilation operators for energy levels of the simple harmonic oscillators are given as

## \hat{a} = \sqrt{\dfrac{m\omega}{2\hbar}} \left(\hat{x} + \dfrac{i}{m\omega}\hat{p}\right) ##

## \hat{a}^\dagger = \sqrt{\dfrac{m\omega}{2\hbar}} \left(\hat{x} - \dfrac{i}{m\omega}\hat{p}\right) ##

and the inverse formulae are

## \hat{x} =\sqrt{\dfrac{\hbar}{2m\omega}} (\hat{a} +\hat{a}^\dagger ) =\dfrac{1}{2}\sqrt{\dfrac{2\hbar}{m\omega}} (\hat{a} +\hat{a}^\dagger ) ##

## \hat{p} =-i\sqrt{\dfrac{\hbar}{2m\omega}} (\hat{a} +\hat{a}^\dagger ) =\dfrac{-i}{2}\sqrt{\dfrac{2\hbar}{m\omega}} (\hat{a} -\hat{a}^\dagger ) ##

Now, my observation is that the first pair of expressions have the same structure as the Euler's formula

## e^{iz} = \cos(z)+i\sin(z), \;e^{-iz} = \cos(z)-i\sin(z)##,

upon substitution

##e^{iz} \rightarrow \sqrt{\dfrac{2\hbar}{m\omega}}\hat{a}, \;e^{-iz} \rightarrow \sqrt{\dfrac{2\hbar}{m\omega}}\hat{a}^{\dagger},##
## \cos(z) \rightarrow \hat{x},\; \sin(z) \rightarrow \dfrac{1}{m\omega}\hat{p}##,

and the second pair of equations is recovered with the same substitution from the inverse formulae

## \cos(z) = \dfrac{1}{2}(e^{iz}+e^{-iz}), ##
## \sin(z)=\dfrac{-i}{2}(e^{iz}-e^{-iz}).##

Now, I realize that the structural similarity stems from the definition ## \hat{a}= \hat{x}+i\hat{p}##, but there seems to be a geometrical meaning to this. Can we indeed interpret the interplay between the position and momentum as the connection between trigonometric functions? What is the meaning of the commutation relations then? Are you familiar with any textbook treating this aspect of quantization?

Gold Member
2022 Award
The reason for this is that in fact we deal with a dynamical ##\mathrm{U}(1)## (or equivalently ##\mathrm{SO}(2)##) symmetry of the simple harmonic oscillator. This is immediately clear when you use new variables,
$$\hat{X}=\sqrt{m \omega/2} \hat{x}, \quad \hat{P}=\frac{1}{\sqrt{2m \omega}} \hat{p}.$$
Then the Hamiltonian gets
$$\hat{H}=\omega (\hat{X}^2+\hat{P}^2).$$
It's obvious that this Hamiltonian is invariant under rotations in the ##(XP)##-plane.

Now these ##\mathrm{SO}(2)## transformations are easier to describe by taking the ##(XP)##-plane as the Gaussian plane of complex numbers,
$$\hat{a}=\hat{X}+\mathrm{i} \hat{P}.$$
$$\hat{H}=\omega (\hat{a}^{\dagger} \hat{a} + \frac{1}{2} \hat{1}).$$
The last piece comes from the non-commutativity of ##\hat{X}## and ##\hat{P}##, which fulfill ##[\hat{X},\hat{P}]=\mathrm{i} \hat{1}/2##:
$$\hat{a}^{\dagger} \hat{a}=(\hat{X}-\mathrm{i} \hat{P})(\hat{X}+\mathrm{i} \hat{P})=\hat{X}^2+\hat{P}^2+\mathrm{i}[\hat{X},\hat{P}]=\hat{X}^2+\hat{P}^2-\frac{1}{2} \hat{1}.$$
Now the rotations are described by simple phase factors for ##\hat{a}## and ##\hat{a}^{\dagger}##.

dextercioby and jshtok
Gold Member
Then what would be the equivalent of ##z## in the case of creation and annihilation operators? The ##\sin z## and ##\cos z## commute if ##z## is a complex number, but the operators ##\hat{x}## and ##\hat{p}## don't.

In the Heisenberg picture the position and momentum operators have the form of a sine or cosine function of time multiplying a time-independent operator.

Edit: looks like vanhees71 got here first...

jshtok
jshtok
Then what would be the equivalent of ##z## in the case of creation and annihilation operators? The ##\sin z## and ##\cos z## commute if ##z## is a complex number, but the operators ##\hat{x}## and ##\hat{p}## don't.

In the Heisenberg picture the position and momentum operators have the form of a sine or cosine function of time multiplying a time-independent operator.

Edit: looks like vanhees71 got here first...
Thank you for your input. Indeed, looking at the Heisenberg picture is a good hint, but the position and momentum operator are both combinations of sine and cosine functions of \omega*t there, unlike the analogy here. You are right, of course, that there are no non-zero commutators in cos(z), sin(z), so I am missing something here. Still, the analogy looks to me quite potent. I will study the expressions in Heisenberg picture some more.

jshtok
The reason for this is that in fact we deal with a dynamical ##\mathrm{U}(1)## (or equivalently ##\mathrm{SO}(2)##) symmetry of the simple harmonic oscillator. This is immediately clear when you use new variables,
$$\hat{X}=\sqrt{m \omega/2} \hat{x}, \quad \hat{P}=\frac{1}{\sqrt{2m \omega}} \hat{p}.$$
Then the Hamiltonian gets
$$\hat{H}=\omega (\hat{X}^2+\hat{P}^2).$$
It's obvious that this Hamiltonian is invariant under rotations in the ##(XP)##-plane.

Now these ##\mathrm{SO}(2)## transformations are easier to describe by taking the ##(XP)##-plane as the Gaussian plane of complex numbers,
$$\hat{a}=\hat{X}+\mathrm{i} \hat{P}.$$
$$\hat{H}=\omega (\hat{a}^{\dagger} \hat{a} + \frac{1}{2} \hat{1}).$$
The last piece comes from the non-commutativity of ##\hat{X}## and ##\hat{P}##, which fulfill ##[\hat{X},\hat{P}]=\mathrm{i} \hat{1}/2##:
$$\hat{a}^{\dagger} \hat{a}=(\hat{X}-\mathrm{i} \hat{P})(\hat{X}+\mathrm{i} \hat{P})=\hat{X}^2+\hat{P}^2+\mathrm{i}[\hat{X},\hat{P}]=\hat{X}^2+\hat{P}^2-\frac{1}{2} \hat{1}.$$
Now the rotations are described by simple phase factors for ##\hat{a}## and ##\hat{a}^{\dagger}##.
Thank you for the detailed explanation. Indeed, if I adjust my earlier analogy, from
## \cos(z) \rightarrow \hat{x},\; \sin(z) \rightarrow \dfrac{1}{m\omega}\hat{p}##,
to
## \cos(z) \rightarrow \sqrt{\dfrac{m\omega}{2}}\hat{x},\; \sin(z) \rightarrow \dfrac{1}{\sqrt{2m\omega}}\hat{p}##, and set ##\hbar=1##, I get the Eulers formula with ##e^{iz} = \hat{a}## . So using your notation, if I write ## \hat{X}=\cos(z), \hat{P} = \sin(z) ## as components of the annihilation operator ## \hat{a}## in the complex XP plane, then ## \hat{a}## mimics the complex number ##e^{iz}## . However, the analogy breaks down here, since this complex plane is non-commutative and I cant have ##e^{iz}e^{-iz}=1##.

Gold Member
2022 Award
Why should you want to write ##\hat{a}=\exp(\mathrm{i} \hat{z})##?

Concerning the Heisenberg picture, it's of course clear that that's the most intuitive picture to use in view of its close similarity with classical mechanics. It's also very natural with regard to the probabilistic meaning of the quantum state: Quantum states refer to preparation procedures (aka "initial conditions") and thus are constant in time, and the operators representing observable move according to the equations of motion known from classical mechanics in terms of Poisson brackets.

In the case of the harmonic oscillator it's an even closer analogy since the equations of motion are linear, where the non-commutativity of the operators do not lead to much differences. As in the classical case, the solution of the Heisenberg equations of motion read
$$\hat{a}(t)=\hat{a}_0 \exp(-\mathrm{i} \omega t), \quad \hat{a}^{\dagger}(t)=\hat{a}_0^{\dagger} \exp(+\mathrm{i} \omega t).$$
This immediately shows that ##\hat{a}## annihilates and ##\hat{a}^{\dagger}## creates a normal-mode excitation of the oscillator.

jshtok
jshtok
Why should you want to write ##\hat{a}=\exp(\mathrm{i} \hat{z})##?

Concerning the Heisenberg picture, it's of course clear that that's the most intuitive picture to use in view of its close similarity with classical mechanics. It's also very natural with regard to the probabilistic meaning of the quantum state: Quantum states refer to preparation procedures (aka "initial conditions") and thus are constant in time, and the operators representing observable move according to the equations of motion known from classical mechanics in terms of Poisson brackets.

In the case of the harmonic oscillator it's an even closer analogy since the equations of motion are linear, where the non-commutativity of the operators do not lead to much differences. As in the classical case, the solution of the Heisenberg equations of motion read
$$\hat{a}(t)=\hat{a}_0 \exp(-\mathrm{i} \omega t), \quad \hat{a}^{\dagger}(t)=\hat{a}_0^{\dagger} \exp(+\mathrm{i} \omega t).$$
This immediately shows that ##\hat{a}## annihilates and ##\hat{a}^{\dagger}## creates a normal-mode excitation of the oscillator.
Thank you very much for the explanations. I think I have a clearer picture now.

Gold Member
It's also good to remember that when an oscillator is nonlinear, the Heisenberg picture operators ##\hat{x},\hat{p}## and the expectation values ##\langle\hat{x}\rangle,\langle\hat{p}\rangle## don't have the same time dependence as the equivalent classical variables. It's just a property of the harmonic oscillator that both evolve like sine/cosine functions.

vanhees71
Gold Member
2022 Award
That's an important point. What holds is Ehrenfest's theorem. Take the most simple example of paticle motion in one spatial dimension and a Hamiltonian
$$\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x}).$$
Then Ehrenfest's theorem tells us that for any state
$$\mathrm{d}_t \langle \hat{x} \rangle=\frac{1}{m} \langle \hat{p} \rangle, \quad \mathrm{d}_t \langle \hat{p} \rangle =-\langle V'(\hat{x}) \rangle.$$
Now obviously if and only if
$$\langle V'(\hat{x}) \rangle=V'(\langle \hat{x} \rangle)$$
this is the same as the classical equation of motion for the expectation value.

However, this is generally the case only if ##V'(\hat{x})## is at most linear in ##\hat{x}##, which is the case for the free particle, where ##V=0## to begin with, particle under constant acceleration, where ##V(\hat{x})=-m a \hat{x}##, or for the harmonic oscillator, where ##V'(\hat{x})=-m \omega^2 \hat{x}##.