Photon Chirality: Left, Right & Beyond

[Jeff Root]

TL;DR Summary

Are photons left- and right-handed?

Are photons left- and right-handed?

If so, does the handedness change when a photon is reflected?

Is there any other way the handedness can change?

-- Jeff, in Minneapolis

 

[vanhees71]

For a massless particle handedness (or chirality) has a very intuitive meaning, because it's the same as helicity, i.e., the projection of the total angular momentum (not spin!) to the direction of the particle's momentum. Photons can have helicity +1 ("right handed") or -1 ("left handed").

Under spatial reflections, i.e., $t \rightarrow t$, $\vec{x} \rightarrow -\vec{x}$ momentum transforms as a polar vector, i.e., $\vec{p} \rightarrow -\vec{p}$ but angular momentum as an axial vector, i.e., $\vec{J} \rightarrow \vec{J}$. Consequently the parity flips, i.e., $\lambda \rightarrow -\lambda$. Thus a "right handed" photon is mapped to a left handed one and vice versa.

 

[Jeff Root]

Thank you.

You kept it pretty brief and simple, but there is a great deal
about it that I don't understand. Let me start with the very
end, which seems like it might be easiest to explain.

What do you mean by "mapped to"? Something more than what it
means for one thing to be the mirror image of another, like the
letter b can be mapped to the letter d and vice versa?

You say that lambda goes to negative lambda. Since lambda is
wavelength and lengths are absolute values, I take it that you
are implying forward and backward directions. Can you clarify?
This seems so obvious, does it have some significance that is
not obvious? What do you mean by negative wavelength?

You say that t goes to t, which I take to mean that time is not
reversed when a photon is reflected. Is that that all the
notation means?

You say the vector x goes to negative x. I take that to mean
the spatial position of the photon is reversed. Intuitively,
that makes sense, but is it what your notation actually means?
I'm not really clear on what -x means.

I have never heard of "polar" or "axial" vectors before, so I
don't know what the difference implies. Actually, just from
the names, it sounds to me like you might have got them mixed
around. I would think that a "polar" vector might have to do
with rotation and angular momentum, while an "axial" vector
would have to do with linear motion and momentum. No?

You say the polar vector p goes to negative p. I take that to
mean the direction of the momentum is reversed. That seems
straightforward, although I wonder how the momentum of the
reflecting surface figures in.

You say the axial vector J is unchanged. I take that to mean
the direction of the angular momentum is unchanged, but I am
confused by it since the direction of travel has reversed and
presumeably the direction that the photon is "facing" has also
reversed. Maybe that is irrelevant because all of the quantities
you specify are relative to an (imaginary) observer, not relative
to the photon?

Is handedness/chirality/helicity the same thing as polarization?
Are they connected? Related? Totally separate?

Is spin a completely unrelated property? It of course sounds
like the same thing. How can photons have two (or three?)
different kinds of rotational properties?

-- Jeff, in Minneapolis

 

[PeroK]

@vanhees71 gives some great answers, but not always at the B level. $\lambda$ is parity, not wavelength.

 

[Jeff Root]

Oh! I don't recall ever seeing lamba represent parity. I presume
that is somebody's standard. I'll have to reconsider everything.
Thank you.

-- Jeff, in Minneapolis

 

[hutchphd]

Summary:: Are photons left- and right-handed?

Are photons left- and right-handed?

If so, does the handedness change when a photon is reflected?

Is there any other way the handedness can change?

-- Jeff, in Minneapolis

I propose a stupid experiment (which course amuses me greatly). It requires a set of Real 3D glasses...not Imax 3D glasses. The Real 3D filters Right and Left Circularly Polarized Light for each eye. If you get some glasses look in a mirror and close one eye.
No, you have to do the experiment...

 

[Jeff Root]

I doubt it will answer my questions, but I'll definitely try to find some
polarized filters and do this. I can imagine what I will (and won't) see.

Thank you, too.

-- Jeff, in Minneapolis

 

[vanhees71]

Thank you.

You kept it pretty brief and simple, but there is a great deal
about it that I don't understand. Let me start with the very
end, which seems like it might be easiest to explain.

What do you mean by "mapped to"? Something more than what it
means for one thing to be the mirror image of another, like the
letter b can be mapped to the letter d and vice versa?

"Handedness" or "chirality" is a property of physical quantities defined as how they change under space reflections. Take the description of a single particle in Newtonian mechanics. Then this space-reflection transformation is defined as mapping the position vector $\vec{x} \mapsto \vec{x}'=-\vec{x}$ and leaving time unchanged, $t \mapsto t'=t$.

From this you get $\vec{v}=\dot{\vec{x}} \mapsto \vec{v}'=-\vec{v}$. The angular momentum, $\vec{J}=m \vec{x} \times \dot{\vec{x}}$ of the particle thus behaves like $\vec{J} \mapsto \vec{J}'=\vec{J}$.

This shows that we have two kinds of vectors, namely such like $\vec{x}$ and $\vec{v}$ that change their sign under space reflections, called polar vectors and such like angular momentum, $\vec{J}$, that don't change their sign under space reflections, called axial vectors. That's because polar vectors usually refer to translations, i.e., the motion of the particle along the direction indicated by the direction of the vector but axial vectors refer to rotations around an axis. That's defined by the right-hand rule: pointing the thumb of your right hand in direction of the vector your fingers indicate the direction of rotation.

You say that lambda goes to negative lambda. Since lambda is
wavelength and lengths are absolute values, I take it that you
are implying forward and backward directions. Can you clarify?
This seems so obvious, does it have some significance that is
not obvious? What do you mean by negative wavelength?

Maybe it's a bad habit to label helicity with $\lambda$, which is also used for wave-lengths. So rather let me call it $h$. What is meant by helicity is $h=\vec{p} \cdot \vec{J}/(|\vec{p}| |\vec{J}|)$. For photons it turns out that in a state of definite helicity $h$ can take the values $h \in \{1,-1\}$. $h=+1$ means it's a photon that is left-circular polarized, for $h=-1$ it's right-circular polarized.

Now as for the case of point particles, also for photons momentum is a polar vector and angular momentum an axial one, which implies that under space reflections $h \mapsto h'=-h$. One calls such a quantity a "pseudo-scalar quantity", which means that it doesn't change under rotations (that makes it a "scalar") but it flips sign under space reflections (making it more specifically a "pseudo-scalar"). So under space reflections a left-circular (or right-handed) photon transforms to a right-circular (or left-handed) one and vice versa. Since photons are massless, chirality and helicity are the same.

You say that t goes to t, which I take to mean that time is not
reversed when a photon is reflected. Is that that all the
notation means?

Yes, it describes how the quantities under consideration behave when doing a space reflection (as defined above for the point-particle observables).

You say the vector x goes to negative x. I take that to mean
the spatial position of the photon is reversed. Intuitively,
that makes sense, but is it what your notation actually means?
I'm not really clear on what -x means.

For photons position is not a good quantity. It's rather for momentum, which is a well-defined observable also for massless particles with spin 1, and momentum is a polar vector, i.e., $\vec{p} \mapsto \vec{p}'=-\vec{p}$ under space reflections, i.e., you map a photon with momentum $\vec{p}$ to one which has momentum $\vec{p}'=-\vec{p}$.

I have never heard of "polar" or "axial" vectors before, so I
don't know what the difference implies. Actually, just from
the names, it sounds to me like you might have got them mixed
around. I would think that a "polar" vector might have to do
with rotation and angular momentum, while an "axial" vector
would have to do with linear motion and momentum. No?

As explained above, it's the other way around: polar vectors flip sign under space reflections (and usualy describe in some way a displacement; in this case of the photon's momentum it's the direction of propgation of the phase of the corresponding electromagnetic wave) and axial vectors like angular momentum just stay the same under space reflections.

You say the polar vector p goes to negative p. I take that to
mean the direction of the momentum is reversed. That seems
straightforward, although I wonder how the momentum of the
reflecting surface figures in.

There is no reflecting surface. It's all about different situations: One is the original situation, i.e., a photon with a momentum $\vec{p}$ and some helicity $h$. Now you ask, how the situation looks like when you prepare a photon in the space-reflected state. As it turns out in this situation you get a photon with the opposite momentum and also with opposite helicity.

You say the axial vector J is unchanged. I take that to mean
the direction of the angular momentum is unchanged, but I am
confused by it since the direction of travel has reversed and
presumeably the direction that the photon is "facing" has also
reversed. Maybe that is irrelevant because all of the quantities
you specify are relative to an (imaginary) observer, not relative
to the photon?

Momentum and helicity refer to a photon state. As any relativistic momentum also the photon momentum forms a four-vector together with the photon's energy, i.e., $(p^{\mu})=(E/c,\vec{p})$ transforms with a Lorentz transformation as a four-vector when transforming from one observer's inertial reference frame to another observer's inertial reference frame. For a photon you have $E=c |\vec{p}|$, i.e., $p_{\mu} p^{\mu} = (E/c)^2-\vec{p}^2=0$, i.e., it's a light-like vector and thus the photon is massless.

Is handedness/chirality/helicity the same thing as polarization?
Are they connected? Related? Totally separate?

That's a pretty tricky question. For a massless quantum like the photon chirality and helicity are the same. That is, because you cannot flip the helicity of a massless particle by just changing the inertial frame of reference. That's because the photon always goes with the speed of light in any inertial frame of reference, i.e., you cannot overtake it as an observer, and thus if a photon has helicity $h$ in one inertial frame it must necessarily have the same helicity in any other inertial frame. In the particle-physicist's slang one says "helicity is a good (intrinsic) quantum number" for a photon, i.e., it's independent of the observer and thus can serve as an "intrinsic property of the photon".

For a massive particle you can of course also define helicity, but it's not a "good quantum number", because you can flip the sign by a fast enough Lorentz boost. So indeed for a massive particle helicity depends on the observer.

Sometimes you can define chirality separately. E.g., for a Dirac particle like an electron a state with certain chirality is defined as the eigenstates of the matrix $\gamma_5=\mathrm{i} \gamma^0 \gamma^1 \gamma^2 \gamma^3$. The eigenvalues are $\pm 1$ referring to (left-handed and right-handed states). The associated current $\bar{\psi} \gamma_5 \psi$ flips sign under space reflections and thus under space-reflection a left-handed Dirac particle becomes a right-handed one and vice versa. Chirality is a "good quantum number" also for massive particles. For massless particles chirality and helicity eigenstates are the same.

Is spin a completely unrelated property? It of course sounds
like the same thing. How can photons have two (or three?)
different kinds of rotational properties?

-- Jeff, in Minneapolis

Spin is related to chirality, but it's a pretty tricky business in the relativistic context. Massive particles with spin $s$ always have $2s+1$ polarization states referring to the possible eigenvalues of the spin component $s_z$, which can take the values $m_s \hbar$ with $m_s \in \{-s,-s+1,ldots,s-1,s \}$.

Massless particles are even more special. Here you always have only two polarization degrees of freedom, i.e., for a massless particle with spin $s$ you have only the helicities $\pm s$ as a basis for the polarization states.