Creation/annihilation operators and trigonometric functions

In summary, there is a striking similarity between the expressions for creation and annihilation operators in terms of position and momentum operators and trigonometric expressions in terms of exponentials. This is due to the fact that the simple harmonic oscillator has a dynamical ##\mathrm{U}(1)## (or equivalently ##\mathrm{SO}(2)##) symmetry, which can be described using complex numbers in the ##(XP)##-plane. The operators ##\hat{a}## and ##\hat{a}^\dagger## can be interpreted as phase factors for this symmetry, with the Hamiltonian being invariant under rotations in this plane. This analogy can be further extended by setting ##\hbar=1## and
  • #1
jshtok
18
3
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?

Thank you in advance!
 
Physics news on Phys.org
  • #2
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,
$$\hat{a}=\hat{X}+\mathrm{i} \hat{P}.$$
Then the Hamiltonian reads
$$\hat{H}=\omega (\hat{a}^{\dagger} \hat{a} + \frac{1}{2} \hat{1}).$$
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}##.
 
  • Like
Likes dextercioby and jshtok
  • #3
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...
 
  • Informative
Likes jshtok
  • #4
hilbert2 said:
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.
 
  • #5
vanhees71 said:
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,
$$\hat{a}=\hat{X}+\mathrm{i} \hat{P}.$$
Then the Hamiltonian reads
$$\hat{H}=\omega (\hat{a}^{\dagger} \hat{a} + \frac{1}{2} \hat{1}).$$
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 Euler`s 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 can`t have ##e^{iz}e^{-iz}=1##.
 
  • #6
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.
 
  • Like
Likes jshtok
  • #7
vanhees71 said:
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.
 
  • #8
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.
 
  • Like
Likes vanhees71
  • #9
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}##.
 
  • #10
jshtok said:
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?
Something similar to that can be found in the complex-number representation of coherent states. See e.g. the book L.E. Ballentine, Quantum Mechanics A Modern Development, pages 542-545.
 
  • Like
Likes vanhees71

Related to Creation/annihilation operators and trigonometric functions

1. What are creation and annihilation operators?

Creation and annihilation operators are mathematical operators used in quantum mechanics to describe the creation and annihilation of particles. They are represented by the symbols a† and a, respectively, and are used to manipulate quantum states.

2. How do creation and annihilation operators relate to trigonometric functions?

Creation and annihilation operators can be expressed in terms of trigonometric functions, specifically sine and cosine. This is because these functions are related to the exponential function, which is often used to represent quantum states.

3. What is the significance of using trigonometric functions in quantum mechanics?

Trigonometric functions are used in quantum mechanics because they provide a mathematical framework for describing the behavior of quantum systems. They allow for the representation of complex quantum states and the calculation of probabilities for different outcomes.

4. How are creation and annihilation operators used in quantum field theory?

In quantum field theory, creation and annihilation operators are used to describe the creation and annihilation of particles in a quantum field. They are used to create and destroy particles, which then interact with each other to produce different physical phenomena.

5. Can creation and annihilation operators be applied to classical systems?

No, creation and annihilation operators are specific to quantum mechanics and cannot be applied to classical systems. They are used to describe the probabilistic behavior of quantum particles and do not have a classical counterpart.

Similar threads

  • Quantum Physics
Replies
3
Views
895
Replies
1
Views
758
Replies
24
Views
1K
Replies
4
Views
2K
  • Quantum Physics
Replies
9
Views
1K
  • Quantum Physics
2
Replies
56
Views
3K
  • Quantum Physics
Replies
1
Views
633
Replies
16
Views
1K
Replies
3
Views
868
  • Quantum Physics
Replies
9
Views
2K
Back
Top