Spin and polarization basis problem?

[Mentz114]

It is well established that Fermionic or quantum spin is described by the Pauli matrices and their algebra which give the basis vectors and operators. Consider a superposition of spin-up and spin-down states
$\psi_S = \cos(\alpha)|z_+\rangle + \sin(\alpha)|z_-\rangle$ and project into $|z_+\rangle$ with $\hat{Z_{+}}=|z_+\rangle \langle z_+|$. The probability
of projection is $\langle \psi_S|\hat{Z_{+}}|\psi_S\rangle = \cos(\alpha)^2$. Experiment shows this to be valid.

Repeat the exercise in the polarization context. Start with a superpositon $\psi_P = \cos(\alpha)|H\rangle + \sin(\alpha)|V\rangle$ and repeat the calculation to get
$\langle \psi_P|\hat{H}|\psi_P\rangle = \cos(\alpha)^2$. Wrong, this contradicts Malus law.

I would guess that this has happened because the spaces of the spin basis vectors and the polarization basis vectors are different. This has to be taken into account in some way ( halving or doubling angles ?) or errors will arise.

I fear that the correlation given by $-\cos(2(a-b))$ is one. In practice, using $\cos(a-b)$ does not make much difference.

( Richard Gill of Leiden University pointed out to me that there was a problem with the spin/polarization contexts which caused confusion, but the logic above is mine so I must take the blame.)

 

[Jilang]

The angle must be halved for polarisation as 90 degrees leads the to orthogonal states, but for spins it’s 180 degrees.

 

[Mentz114]

The angle must be halved for polarisation as 90 degrees leads the to orthogonal states, but for spins it’s 180 degrees.

That's my point. It looks as if I got a wrong result. The steps in the two calculations are identical but one of them is wrong.

Where is the mistake ?

The mistake must be in using spin-1/2 bases when photons are spin-1. How do you build this into the calculation ?

 

[PeterDonis]

The steps in the two calculations are identical

No, they're not. They look identical, formally, but the angle $\alpha$ has a different meaning in the two cases. To see this, consider: the way you have written the superpositions, $\alpha$ is defined such that $\cos \alpha = 1$ corresponds to the "up" basis state, $z^+$ or $H$, and $\sin \alpha = 1$ corresponds to the "down" basis state, $z^-$ or $V$. So those values of $\alpha$ represent the physical angles at which you have to set your measuring device to measure those two basis states. For fermion spins, those angles are 0 degrees for "up" and 180 degrees for "down"; but for photon polarizations, those angles are 0 degrees for "up" and 90 degrees for "down". So $\alpha$ is representing two different ranges of measurement angles in the two cases.

The angle must be halved for polarisation as 90 degrees leads the to orthogonal states, but for spins it’s 180 degrees.

Actually, if we just look at the cosine and sine values, we expect $\alpha$ to be 90 degrees for the "down" state, since that's where $\sin \alpha = 1$. So formally, it's the photon case that gives $\alpha$ the straightforward meaning expressed by the calculation interpreted at face value, and it's the fermion case that requires extra explanation, to see why $\sin \alpha = 1$ corresponds to a measurement angle of 180 degrees instead of 90 degrees.

 

[Mentz114]

No, they're not. They look identical, formally, but the angle $\alpha$ has a different meaning in the two cases. To see this, consider: the way you have written the superpositions, $\alpha$ is defined such that $\cos \alpha = 1$ corresponds to the "up" basis state, $z^+$ or $H$, and $\sin \alpha = 1$ corresponds to the "down" basis state, $z^-$ or $V$. So those values of $\alpha$ represent the physical angles at which you have to set your measuring device to measure those two basis states. For fermion spins, those angles are 0 degrees for "up" and 180 degrees for "down"; but for photon polarizations, those angles are 0 degrees for "up" and 90 degrees for "down". So $\alpha$ is representing two different ranges of measurement angles in the two cases.
Actually, if we just look at the cosine and sine values, we expect $\alpha$ to be 90 degrees for the "down" state, since that's where $\sin \alpha = 1$. So formally, it's the photon case that gives $\alpha$ the straightforward meaning expressed by the calculation interpreted at face value, and it's the fermion case that requires extra explanation, to see why $\sin \alpha = 1$ corresponds to a measurement angle of 180 degrees instead of 90 degrees.

Thanks, Peter, I think you make my case. The abstract notation hides this. It only relies on basis vectors being 'orthogoal'. But othogonal in the two cases means $\theta^{\perp}=\theta + \pi$ or $\theta^{\perp}=\theta + \pi/2$.

I'm doing some reading in groups and QFT and it looks like simple QM can't handle this. The QFT solution cannot assume $SO(3)$ is isomorphic to $SU(2)\otimes SU(2)$. This seems to be 'accidental' and does not reflect the difference in the physical rotations.

I may misinterpreting stuff so I have to leave this to experts.

 

[vanhees71]

SU(2) is the double cover of SO(3). Also you should be aware that massless quantum fields are quite different from massive ones. So a massive vector particle (like a $\rho$ meson) has three polarization states, as expected from the dimensionality of the representation of SU(2), $2J+1$. It's usual spin, defined in the rest frame of the particle.

For mass 0 there is no rest frame and for any spin $\geq 1/2$ you have only 2 polarization states (there's no polarization degree of freedom for scalar fields, of course). A convenient basis are the helicity eigenstates, i.e., for photons the left- and right-circular polarized modes.

Also the difference concerning the angles, nicely explained by @PeterDonis in #4 is due to the fact that SU(2) is the double cover of SO(3), i.e., you have to go twice around (angle $4 \pi$) to get back to the same vector. Of course, it doesn't matter in the sense of states, because pure states are represented by unit rays rather than unit vectors in Hilbert space. This makes half-integer spin possible since it allows for substituting the classical rotation group SO(3) by its covering group SU(2) in QT.

 

[PeterDonis]

I'm doing some reading in groups and QFT and it looks like simple QM can't handle this.

I think it can as long as you're ok with just declaring by fiat that the meaning of $\alpha$ is different for the two cases: basically, $\alpha = \theta s$, where $\theta$ is the actual measurement angle and $s$ is the spin magnitude (1/2 for fermions, 1 for photons). I'm not sure whether I agree that simple QM cannot explain why this is the case. See below.

The QFT solution cannot assume $SO(3)$ is isomorphic to $SU(2)\otimes SU(2)$.

That's because it isn't. $SO(3)$ is locally isomorphic to (i.e., has the same Lie algebra as) $SU(2)$; but globally, $SU(2)$ is the double cover of $SO(3)$, so a 360 degree rotation, which is equivalent to the identity in $SO(3)$, is not in $SU(2)$. That's the simplest group theoretic way of describing where spin-1/2 comes from. AFAIK, this works in non-relativistic QM.

In QFT, we use the fact that $SO(4)$, or $SO(3, 1)$ if we take the Lorentz signature of the metric into account, is isomorphic to $SU(2) \otimes SU(2)$. This is used to construct the representation theory of the Lorentz group from the representation theory of $SU(2)$. And you need to take SR into account to do this, because the Lorentz group doesn't arise in non-relativistic QM. Locally, you can treat $SU(2) \otimes SU(2)$ as equivalent to (i.e., having the same Lie algebra as) $SO(3) \otimes SO(3)$, so everything from the first item above gets subsumed into the QFT treatment.

 

[Mentz114]

I think it can as long as you're ok with just declaring by fiat that the meaning of $\alpha$ is different for the two cases: basically, $\alpha = \theta s$, where $\theta$ is the actual measurement angle and $s$ is the spin magnitude (1/2 for fermions, 1 for photons). I'm not sure whether I agree that simple QM cannot explain why this is the case. See below.

I agree, if we are in spin-1/2 context we write the full angle in the superposition. For photons we write the half-angle.

So someone who uses it incorrectly gets a wrong answer.

Now look at chapter 9 in Gerry and Knight "Introductory Quantum Optics" where they seem to make exactly this error and get the wrong answer for photons.

This is a fine book but I believe ( for other reasons) that the correlation got from their calculation is wrong. If I'm wrong I will grovel apprpriately.

 

[stevendaryl]

I agree, if we are in spin-1/2 context we write the full angle in the superposition. For photons we write the half-angle.

Don't you have that backwards? It's the spin-1/2 particles that use half-angles:

If you have a spin-1/2 particle that is spin-up in direction $\vec{s}$, then its spin state in terms of $|\mathcal{z}_+\rangle$, $|\mathcal{z}_-\rangle$ (spin-up and spin-down relative to the z-axis) is given by:

$|\psi\rangle = e^{-i \phi/2} cos(\theta/2) |\mathcal{z}_+ \rangle + e^{+i \phi/2} sin(\theta/2) |\mathcal{z}_-\rangle$

where $\theta$ and $\phi$ are defined by:

$s_x = sin(\theta) cos(\phi)$
$s_y = sin(\theta) sin(\phi)$
$s_z = cos(\theta)$

That is, $\theta$ is the angle that $\vec{s}$ makes with the z-axis, and $\phi$ is the angle that its projection onto the x-y plane makes with the x-axis.

To see that this works, you can check that $\vec{s} \cdot \vec{\sigma} |\psi\rangle = +1 |\psi\rangle$

where $\vec{\sigma}$ is the the three Pauli spin matrices.

 

[Mentz114]

Don't you have that backwards? It's the spin-1/2 particles that use half-angles:
[]
where $\vec{\sigma}$ is the the three Pauli spin matrices.

Good reply.

The problem is that with that $\langle \psi_S|\hat{Z_{+}}|\psi_S\rangle =\cos(\alpha/2)^2$ and $\langle \psi_P|\hat{H}|\psi_P\rangle = \cos(\alpha)^2$ which are both wrong.

 

[stevendaryl]

Good reply.

The problem is that with that $\langle \psi_S|\hat{Z_{+}}|\psi_S\rangle =\cos(\alpha/2)^2$ and $\langle \psi_P|\hat{H}|\psi_P\rangle = \cos(\alpha)^2$ which are both wrong.

Why do you say they're wrong? What answers are you expecting?

 

[Mentz114]

Why do you say they're wrong? What answers are you expecting?

For intrinsic spin - I understand that in the SG apparatus, the probability of being projected from a thermal state into $|Z_{+}\rangle$ is $\cos(\theta)^2$ where $\theta$ is the angle between the objects spin axis and the direction of the magnetic field $\vec{z}$.

For photons and polarizer projections I expect Malus law.

 

[stevendaryl]

For intrinsic spin - I understand that in the SG apparatus, the probability of being projected from a thermal state into $|Z_{+}\rangle$ is $\cos(\theta)^2$ where $\theta$ is the angle between the objects spin axis and the direction of the magnetic field $\vec{z}$.

For photons and polarizer projections I expect Malus law.

I don't know what you mean by "from a thermal state". But the Stern Gerlach experiment involves half-angles, as described in equation 6.28 of this online book:
http://physics.mq.edu.au/~jcresser/Phys301/Chapters/Chapter6.pdf

Let us suppose that the experiment is repeated many times over for each setting of the angle θ_i in order to obtain, experimentally, the fraction, or in other words, the probability, of atoms emerging from the final Stern-Gerlach device in either of the two beams. The experimental result obtained is that
Probability of atoms emerging in the S_z = $\frac{\hbar}{2}$ beam = cos²(θ_i/2)
Probability of atoms emerging in the S_z = $-\frac{\hbar}{2}$ beam = sin²(θ_i/2)​

 

[vanhees71]

Good reply.

The problem is that with that $\langle \psi_S|\hat{Z_{+}}|\psi_S\rangle =\cos(\alpha/2)^2$ and $\langle \psi_P|\hat{H}|\psi_P\rangle = \cos(\alpha)^2$ which are both wrong.

Why doe you they both are wrong? If that would be true, nobody would use quantum theory anymore! For the polarization states you can use any two basis states you like. Perferrably you use a orthonormal system of two vectors. From a physics point of view the most intuitive ones are the left and right circular polarization states, which are the helicity eigenstates, i.e., the projection of the total angular momentum (there's no way to split a photon's angular momentum in spin and orbital part that is well defined, i.e., gauge invariant, although even some textbooks claim so, but it's a mathematial fact!) to the direction of the photon's momentum. Another convenient basis are two linear polarization states at perpendicular directions.

Let's take the photon going in $z$ direction. Then two possible polarization states are
$$|\hat{x} \rangle, \quad \hat{y} \rangle.$$
If a photon is linearly polarized in an angle of $\varphi$ relative to the $x$ axis its polarization state is given by
$$|\hat{\varphi} \rangle=\cos \varphi |\hat{x} \rangle + \sin \varphi |\hat{y} \rangle,$$
The probability that such a photon passes a polarization filter oriented in $x$ direction is (Born's rule!)
$$P_x(\varphi)=|\langle \hat{x}|\hat{\varphi} \rangle|^2=\cos^2 \varphi,$$
which is nothing else than Malus's Law in the language of QED aka quantum optics of single-photon states.

The most general polarization state is characterized by three real quantities,
$$|\psi \rangle=\lambda_1 |\hat{x} \rangle + \lambda_2 |\hat{y} \rangle,$$
where $\lambda_1,\lambda_2 \in \mathbb{C}$ (which would make 4 real parameters), restricted by the normalization constraint ($|\lambda_1|^2+|\lambda_2|^2=1$).

Another way to see that it's three real parameters, determining a general polarization state is that you can use one angle and two phases for that parametrization, i.e.,
$$\psi \rangle=\cos \vartheta \exp(-\mathrm{i} \alpha_x) |\hat{x} \rangle + \sin \vartheta \exp(-\mathrm{i} \alpha_y) |\hat{y} \rangle.$$

For more details, see the Wikipedia article:

https://en.wikipedia.org/wiki/Photon_polarization

 

[Mentz114]

The probability that such a photon passes a polarization filter oriented in xx direction is (Born's rule!)

Px(φ)=|⟨^x|^φ⟩|2=cos2φ,​

P_x(\varphi)=|\langle \hat{x}|\hat{\varphi} \rangle|^2=\cos^2 \varphi,
which is nothing else than Malus's Law in the language of QED aka quantum optics of single-photon states.

Well this for me is a shock. So we actually use a different form of Malus ?
Obviously this is the source of my confusion.

Thank you very much for the reply.

 

[Mentz114]

I don't know what you mean by "from a thermal state". But the Stern Gerlach experiment involves half-angles, as described in equation 6.28 of this online book:
http://physics.mq.edu.au/~jcresser/Phys301/Chapters/Chapter6.pdf

Let us suppose that the experiment is repeated many times over for each setting of the angle θ_i in order to obtain, experimentally, the fraction, or in other words, the probability, of atoms emerging from the final Stern-Gerlach device in either of the two beams. The experimental result obtained is that
Probability of atoms emerging in the S_z = $\frac{\hbar}{2}$ beam = cos²(θ_i/2)
Probability of atoms emerging in the S_z = $-\frac{\hbar}{2}$ beam = sin²(θ_i/2)​

Thanks. If we use half-angles for spin-1/2 and as G&K show polarization is su(2), why do their amplitudes use the whole angle ?
I'm sure I'll be able to work that out (!).

Thanks for the replies.

 

[Mentz114]

The replies above are thought provoking and more web searches found this post ( url below) where the answer by (the estimable) 'QMechanic' seems to address my doubts and confirm what @stevendaryl and @vanhees71 have said. The point at which angles get changed is made explicit.

https://physics.stackexchange.com/q...er-2-used-for-a-bloch-sphere-instead-of-theta

It will take me ages to absorb everything.

 

[vanhees71]

Well this for me is a shock. So we actually use a different form of Malus ?
Obviously this is the source of my confusion.

Thank you very much for the reply.

It always help to define, what you mean clearly in terms of formulae. So what's what you call Malus's Law? To be honest, I've never heard this name for the polarization properties of em. waves/photons before. According to the Wikipedia article

https://en.wikipedia.org/wiki/Polarizer

it's the cosine-square law, I've written down in my previous posting. I've never heard the name Malus's Law for it.

 

[Mentz114]

It always help to define, what you mean clearly in terms of formulae. So what's what you call Malus's Law? To be honest, I've never heard this name for the polarization properties of em. waves/photons before. According to the Wikipedia article

https://en.wikipedia.org/wiki/Polarizer

it's the cosine-square law, I've written down in my previous posting. I've never heard the name Malus's Law for it.

I have a paper that refers to it about six times.

The thing is I seem to have got it wrong from the start. What I'm calling Malus law does not use the half-angle as I've been assuming.

When I correct this everything falls into place.

I apologise for this incredibly stupid mistake - but I rejoice that my classical model now agrees in every way with the (excellent) G&K treatment.

Thanks for the help.

 

[vanhees71]

Sigh! Why is it so difficult to convince students to cite properly. It's, so to say, physics 100 to learn to cite! The sentence "I have a paper that refers to it about six times". Is useless. It's wasted bandwidth for my internet connection. Why don't you write: In the paper "Author, Journal, volume, page (year)". If you are very kind you also give the DOI, so that I simply can click to get the paper. In any case, I could check, what's written in the paper and maybe help you with your problem.

Of course, there are no half-angles in electrodynamics, because the electromagnetic field is a massless spin-1 field. Half-angles come into the game for spin 1/2 particles/quanta, not spin-1 particles/quanta.

 

[Mentz114]

Sigh! Why is it so difficult to convince students to cite properly. It's, so to say, physics 100 to learn to cite! The sentence "I have a paper that refers to it about six times". Is useless. It's wasted bandwidth for my internet connection. Why don't you write: In the paper "Author, Journal, volume, page (year)". If you are very kind you also give the DOI, so that I simply can click to get the paper. In any case, I could check, what's written in the paper and maybe help you with your problem.

Of course, there are no half-angles in electrodynamics, because the electromagnetic field is a massless spin-1 field. Half-angles come into the game for spin 1/2 particles/quanta, not spin-1 particles/quanta.

I did not think you would be interested !

Shuang Zhao, Hans De Raedt and Kristel Michielsen
Event-by-event simulation of Einstein-Podolsky-Rosen-Bohm experiments
PACS numbers: 03.65.-w , 02.70.-c , 03.65.Ta
(in press Foundations of physics)
http://arxiv.org/abs/0712.3693.pdf

'Malus' is mentioned on pages 3, 15, 16 (twice) , 20, 21, 24 and 31. That is 8 mentions.

My own effort is concise, correct and much more interesting ! But I could be biased.

You have helped me by pointing out that Malus law does not use half-angles. I still cannot work out where I got that idea.

 

[vanhees71]

Obviously their Eq. (11) is wrong (I guess they simply forgot a ^ in their LaTeX, and it's still wrong in the journal version of the paper)
$$\vec{S} \cdot \vec{a}=\cos \alpha \cos \xi + \sin \alpha \sin \xi=\cos(\alpha-\xi),$$
and the probability is the square of this, i.e., $\cos^2(\alpha-\xi)$. Obviously nobody corrected the proofs carefully enough :-(.

 

[Mentz114]

Obviously their Eq. (11) is wrong (I guess they simply forgot a ^ in their LaTeX, and it's still wrong in the journal version of the paper)
$$\vec{S} \cdot \vec{a}=\cos \alpha \cos \xi + \sin \alpha \sin \xi=\cos(\alpha-\xi),$$
and the probability is the square of this, i.e., $\cos^2(\alpha-\xi)$. Obviously nobody corrected the proofs carefully enough :-(.

Yes, bad proofing. But the correct value is used and the extract below is true.

It is easy to see that for fixed $\xi_n=\xi$ and $\alpha$, this algorithm generates events
such that $2<x_{n,i}>-1=\cos(\xi-\alpha)^2$ where $<X>$ denotes the average over many realizations of the variables $r_n,\xi_n$.
In this case, the input-output relation of the simulation model agrees with Malus law Eq. (11).

But they don't say that this also agrees with the singlet state, which I've shown explicitly ( I think).