What is the Fourier conjugate of spin?

[redtree]

TL;DR Summary

Momentum space is the Fourier conjugate space of position. Intrinsic angular momentum, i.e., spin, space is the Fourier conjugate space of what?

Momentum $\vec{p}$ and position $\vec{x}$ are Fourier conjugates, as are energy $E$ and time $t$.

What is the Fourier conjugate of spin, i.e., intrinsic angular momentum? Angular position?

 

[atyy]

Position and momentum operators don't commute.

Similarly, the different "directions" of spin don't commute.

See Eq 702, 703, 70:
http://farside.ph.utexas.edu/teaching/qmech/Quantum/node87.html

 

[ftr]

Position and momentum operators don't commute.

Similarly, the different "directions" of spin don't commute.

See Eq 702, 703, 70:
http://farside.ph.utexas.edu/teaching/qmech/Quantum/node87.html

I am not sure how that answers the OP. Is it possible to elaborate. Thanks

 

[Swamp Thing]

Why can't I think of questions like that?

 

[kith]

The Fourier transform is continuous and spin operators are discrete and finite. What makes observables "conjugate" in this case? Well, an analogue which has all the properties of the continuous case simply doesn't exist. For example, the trace of $[S_z, A]$ is zero for all observables $A$ so the commutator of $S_z$ and its "conjugate" can't be identical to $i \hbar 1$ (i.e. the Heisenberg uncertainty inequality for all possible pairs of observables depends on the state).

For the position operator and the momentum operator, we have the canonical commutation relations. For spin (component) operators, the fundamental relations are what @atyy linked to in post #2. So is there something like a Fourier transform involved? Indeed, this reddit post argues that $S_z$ and $S_x$ can be viewed as connected by a discrete Fourier transform.

 

[DrClaude]

For the position operator and the momentum operator, we have the canonical commutation relations. For spin (component) operators, the fundamental relations are what @atyy linked to in post #2. So is there something like a Fourier transform involved? Indeed, this reddit post argues that $S_z$ and $S_x$ can be viewed as connected by a discrete Fourier transform.

Interesting, but what about $S_y$?

 

[redtree]

Spin is a measure of energy: intrinsic angular momentum. Other forms of energy are the Fourier conjugates of variables in position space. It seems theoretically problematic that no position space representation exists for spin. Nor do I understand why the observation that spin is measured as discrete and finite, i.e., that it is quantized, would differentiate spin energy from other forms of quantized energy.

I don't see how this doesn't point to a fundamental theoretical problem with our understanding of spin. I don't mind being wrong; I would just like to know why.

 

[Nugatory]

Spin is a measure of energy: intrinsic angular momentum.

Angular momentum, whether intrinsic or not, is not energy. They're different things, separately conserved.

It seems theoretically problematic that no position space representation exists for spin

It's not if we consider the Hilbert space that we're working with. The position and momentum eigenstates are two different ways of spanning the same subspace so we can use either one as a basis to describe any state within that subspace; naturally there is a transformation between them. The subspace spanned by the spin eigenstates is a different subspace so it's not meaningful to consider a transformation between spin and position/momentum.

(An analogy: I can describe points in New York City in terms of uptown/crosstown or north-south/east-west and convert between the two descriptions; but there's nothing problematic about not being able to use either to describe altitude above sea level).

 

[strangerep]

What is the Fourier conjugate of spin, i.e., intrinsic angular momentum? Angular position?

Orientation (relative to some coordinate system) -- which I guess is what you meant by "angular position".

I've also seen it referred to as "pose", but Wikipedia defines that term to encompass both position and orientation.

As for commutation relations, try working in spherical coordinates and compute, say, $[\theta, \partial_\theta]$ .

HTH.

 

[vanhees71]

Note that "angle observables" are a very complicated subject. They cannot be treated in the usual sense as self-adjoint operators so easily!

The reason is quite simple to understand. Take a free particle constrained on a circle. In classical mechanics you describe it with an angle $\varphi$ in polar coordinates. The Lagrangian reads
$$L=\frac{m}{2} a^2 \dot{\varphi}^2,$$
where $a$ is the radius of the circle. The canonical momentum is
$$p_{\varphi}=\partial_{\dot{\varphi}} L=m a^2 \dot{\varphi}=l.$$
Now you can try to quantize this in the usual way introducing $\hat{\varphi}$ and $\hat{l}$ as a pair of canonically conjugate variables, subject to the Heisenberg commutation relations (working in natural units with $\hbar=1$),
$$[\hat{\varphi},\hat{l}]=\mathrm{i}.$$
Then in the usual way you can derive the wave-mechanics in $\varphi$ representation to get
$$\hat{\varphi} \psi(\varphi)=\varphi \psi(\varphi), \quad \hat{l} \psi(\varphi)=-\mathrm{i} \partial_{\varphi} \psi(\varphi).$$
Now in contradistinction to the case of a particle running along a line, leading to the usual position and momentum operators $\hat{x}$ and $\hat{p}$, here the wave functions are restricted to periodic functions (up to a phase, but let's forget this complication for now), i.e., you are lead to the Hilbert space $\mathrm{L}^2([0,2 \pi])$ of functions with periodic boundary conditions,
$$\psi(\varphi+2 \pi)=\psi(\varphi).$$
Now the dilemma is obvious, because
$$\hat{\varphi} \psi(\varphi)=\varphi \psi(\varphi)$$
won't be a periodic function anymore, and thus $\hat{\varphi}$ is not a self-adjoint operator on this Hilbert space. That's why there is not a naive "angle observable" in quantum mechanics in that case.