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I am studying Poincare group and encountered the term massless representations of the Poincare group. I know Poincare group is studied by the studying the little group of various momenta, massless and massive. Does massless representations of Poincare group mean representations of the little group for $p^{\mu} = (1,1,0,0) $.(first coordinate is the energy).

This is the second question: If we are given a Lagrangian for a field(say Electromagnetic field), then we can study the representations of the relevant field. To study these representations of these, we can study the representations of the little group for various cases and using the method of induced representations, we can get the representations. Did I get the picture right ?

My question is

1.We know oen of the Casimir elements of the poicare group is $p^{\mu}p_{\mu} = m^2$ . How does one identify the m with mass of the field we are studying ?

2.If the Lagrangian involved mass-term,(like in Proca equation). How do the representations change? I don't see any reason why the representations should change because of the mass. (relevant explanation for the scalar field is enough for me)

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    $\begingroup$ Related post by OP: physics.stackexchange.com/q/454621/2451 $\endgroup$
    – Qmechanic
    Jan 22, 2019 at 5:50
  • $\begingroup$ Sorry but nothing much explained there. And I asked that question too. $\endgroup$
    – user183683
    Jan 22, 2019 at 5:57

2 Answers 2

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You are asking how to connect the "mass term" for the field in the Lagrangian with the "mass value" for the particle state given by the irreducible unitary Poincaré representation it transforms in.

The connection is through the four-momentum operator. Doing canonical quantization starting from the Lagrangian and without ever thinking about any representations on particles, you can show that the four-momentum operator for a scalar field can be expressed in terms of creation/annihilation operators as $$ P^\mu = \int p^\mu a^\dagger(\vec p) a(\vec p) \frac{\mathrm{d}^3 p}{(2\pi)^3}$$ so that $$ P^2 = P_\mu P^\mu = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{\mathrm{d}^3 q}{(2\pi)^3}\left(q_\mu p^\mu a^\dagger(\vec q)a(\vec q)a^\dagger(\vec p)a(\vec p)\right).$$ Using $a(\vec q) a^\dagger(\vec p) = (2\pi)^3 \delta(\vec p - \vec q) + a^\dagger(\vec p)a(\vec q)$, we have that $$ P^2 = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\left(m^2 a^\dagger(\vec p)a(\vec p) + \text{term involving two annihilators on the right}\right)$$ and applying this operator to a one-particle state corresponding to the field (i.e. created by $a^\dagger$) therefore yields $m^2$ since $\int a^\dagger a$ is just a number operator.

Therefore, the mass value from the Lagrangian is indeed the Casimir value that appears in the Poincaré representation for a one-particle state.

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How do we know, what's what?

First: remember what the components of $p_μ$ stand for. Explicitly, with $c$ intact $$p_0 = -E, \quad p^0 = -\frac{E}{c^2}, \quad \left(p_1, p_2, p_3\right) = 𝐩 = -\left(p^1, p^2, p^3\right),$$ if adopting the metric given by the line element $$g_{μν} dx^μ dx^ν = c^2 dt^2 - \left(dx^2 + dy^2 + dz^2\right),$$ and coordinates $$x^0 = t, \quad x^1 = x, \quad x^2 = y, \quad x^3 = z.$$ Then $p^μ p_μ = E^2/c^2 - |𝐩|^2$.

Next, note the transforms under boost, given infinitesimally by: $$Δ𝐩 = -𝞄\frac{E}{c^2}, \quad ΔE = -𝞄·𝐩,$$ for an infinitesimal boost $𝞄$. This comes directly out of the Lie algebra for the Poincaré group. In finite form, this transform becomes: $$𝐩 → γ\left(𝐩 - 𝐯\frac{E}{c^2}\right), \quad E → γ\left(E - 𝐯·𝐩\right) \quad \left(γ = \frac{1}{\sqrt{1 - (|𝐯|/c)^2}}\right),$$ if the boost velocity $𝐯$ is chosen to be collinear with $𝐩$.

Now, let's look at the cases.

If we choose the boost $𝐯 = c^2𝐩/E$, and if it is admissible, i.e. $|𝐯| < c$, then the transform yields: $$𝐩 → 𝟬, \quad E → E_0 = E\sqrt{1 - \frac{|𝐩|^2c^2}{E^2}},$$ which transforms you to the rest frame.

Now, use's Einstein's Equation: $E = mc^2$, on the rest energy $E_0$, to get: $$E\sqrt{1 - \frac{|𝐩|^2c^2}{E^2}} = mc^2.$$ Thus: $$E^2 - |𝐩|^2c^2 = \left(mc^2\right)^2,$$ and, from that, you get $p^μ p_μ = E^2/c^2 - |𝐩|^2 = (mc)^2$.

If, on the other hand, $𝐩 ≠ 𝟬$, and we choose the boost $𝐯 = E𝐩/|𝐩|^2$, and if it is admissible, i.e. $|𝐯| < c$, then the transform yields: $$𝐩 → 𝐩_∞ = 𝐩\sqrt{1 - \frac{E^2}{|𝐩|^2c^2}}, \quad E → 0,$$ which is the infinite speed frame.

Then, $𝐩_∞$ is an instantaneous impulse across space, and $|𝐩|^2 - E^2/c^2 = |𝐩_∞|^2$. Thus $p^μ p_μ = -|𝐩_∞|^2$: minus impulse-squared. That's what this class - called the "tachyons" - actually is.

Finally, if neither case applies, then there is neither a rest frame nor an infinite speed frame and one has - instead - the limiting case of $m = 0$ and $𝐩_∞ = 𝟬$, both at once. In all frames, $E^2 = |𝐩|^2c^2$ and $p^μ p_μ = 0$. But technically, there really is no concept of any "rest mass" $m$ nor impulse $𝐩_∞$, since there's no rest frame or infinite speed frame at all. Likewise, for the tachyons, there is no concept of any "rest mass", since there is no rest frame.

Wigner, in his 1939 paper on the classification, also mentioned the homogeneous case, where $p_μ = 0$ in all frames. They are invariant under spatial translation and time translation. But, he made no further mention of them, except the special case where it is also isotropic and boost-invariant, i.e. the "vacuum".

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