Reduced density matrices and Lorentz transformation

[emma83]

Hello,

If I understand well, since a Lorentz transformation applied on a particle induces a Wigner rotation which depends on the momentum, the spin reduced density matrix that is naively done by tracing out the momentum has no (Lorentz) transformation law. Only the overall system can be Lorentz-transformed correctly.

But what I don't understand is the difference between massive and massless particles, i.e. my former statement seems to be actually only a problem for photons and not for massive particles. Why ?

Thanks a lot for your help!

 

[samalkhaiat]

Hi Emma, thanks for the e-mail and sorry to keep you waiting! I am not an expert on quantum cryptography/optics. However, I might be able to help you on the difference between the massive and the massless irreducible representations of Poicare' group. Ok, let us do it anyway.

Let $q_{a}$ be a 4-vector living on the surface

$$p_{a}p^{a} = m^{2} \geq 0 , \ \ p^{0}> 0$$

in momentum space. Let $\mathcal{V}_{q} \subset \mathcal{H}$ be the space of one-particle states having the 4-momentum $q_{a}$;

$$P_{a}|q \rangle = q_{a} |q \rangle , \ \ \forall |q \rangle \in \mathcal{V}_{q}$$

Every momentum eigenstate realizes a one-dimensional representation of the translation

$$U(0,a) |p \rangle \equiv \exp (i a_{a}P^{a}) |p \rangle = \exp (i a_{a}p^{a}) |p \rangle$$

Therefore, to construct IRs of the whole Poicare' group, it is sufficient to find the related representations of the Lorentz transformations

$$U(\omega , 0) = \exp (- \frac{i}{2}\omega_{ab}M^{ab}) ,$$

as follows from the multiplication law

$$(\Lambda , a) = (1 , a) (\Lambda , 0)$$

According to Poicare' algebra, the operator $U(\Lambda , 0)$ transforms some state $|q \rangle$ into another $|\bar{q}\rangle$ where

$$\bar{q}^{a} = \Lambda^{a}{}_{b}q^{b} \ \ \ (1)$$

Next, we define the set $H(q)$ of group elements $(\Lambda , a)$ such that

$$U(\Lambda , a) \mathcal{V}_{q} \in \mathcal{V}_{q} \ \ \ (2)$$

Clearly H(q) forms a subgroup of the Poincare' group. It is called the little group (or the stability subgroup) for $\mathcal{V}_{q}$.
To find this subgroup, we simply set $\bar{q} = q$ in (1) as required by the condition (2). This leads to

$$q^{a} = ( \delta^{a}{}_{b} + \omega^{a}{}_{b})q^{b}$$

or

$$\omega^{a}{}_{b}q^{b} = 0$$

which has the general solution

$$\omega_{ab} = \epsilon_{abcd}q^{c}n^{d} \ \ \ (3)$$

where n is an arbitrary 4-vector.

Thus, a typical element of the little group H(q) looks like

$$u(n,a) = \exp (ia_{a}P^{a} - \frac{i}{2}\epsilon^{abcd}q_{c}n_{d}M_{ab}) \ \ (4)$$

We are now in good posision to introduce the Pauli-Lobanski vector into our story.It is defind by

$$W^{a} = \frac{1}{2}\epsilon^{abcd}M_{bc}P_{d} \ \ \ (5)$$

For later use, we list some of its properties, viz

$$W_{a}P^{a} = 0, \ \ [W_{a}, P_{b}] = 0 \ \ \ (6a)$$

$$[M^{ab} , W^{c}] = i \eta^{ac}W^{b} - i\eta^{bc}W^{a} \ \ (6b)$$

$$[W^{a} , W^{b}] = i \epsilon^{abcd}W_{c}P_{d} \ \ \ (6c)$$

On the subspace $\mathcal{V}_{q}$, elements of H(q) can now be written in terms of the P-L vector as

$$u_{q}(n) = e^{i\alpha} \exp (in_{a}W^{a}) \ \ (7)$$

where $\alpha = a_{a}q^{a}$. This equation, together with eq(6c), suggests that the components of the P-L vector form a Lie algebra restricted to $\mathcal{V}_{q}$. And, the little group H(q) acts on $\mathcal{V}_{q}$ as $U(1) \times G$, where $G$ is a Lie group generated (on $\mathcal{V}_{q}$) by the P-L vector. Below, we will see that $G = SO(3) \approx SU(2)$ in the massive case, and $G = SO(2) \approx U(1)$ in the massless case.

From (7), we see that (up to phase factors) the little group H(q) is generated by the P-L vector. Since all states in $\mathcal{V}_{q}$ describe particles with the same momentum. Therefore, on physical ground, any two linearly independent states $|q,1\rangle , \ |q,2 \rangle \in \mathcal{V}_{q}$ sould correspond to different spin projections, and transform into each other under the action of H(q) {or G}.This indicates that the spin properties of the particles are determined by the P-L vector.
We know that the spectrum of spin projections contains only finite number of values; particles of infinite spin (= infinite-dimensional IRs) do not seem to exist in nature. Therefore, on physical ground, all subspaces $\mathcal{V}_{q}$ must be finite dimensional and H(q) (or G) acts irreducibly on $\mathcal{V}{_q}$.
Now one can use Wigner's methods of induced representations to show that the unitarity and irreducibility of the representations of Poincare' group are directly related to the same properties of the little group.[see Wienberg, Vol 1, sec.(2.5), P.(62-66)].
Thus, in order to find physically admissible IRs of the Poincare' group, it is sufficient to construct all unitary finite-dimensional representations of the little group H(q) corresponding to an arbitrary momentum on the surface

$$p^{2} = m^{2}\geq 0, \ \ p_{0} > 0$$

1) Massive IRs

we chooce the 4-momentum $q_{a}=(m , 0,0,0)$ of a particle at rest. When acting on $\mathcal{V}_{q}$, the components of the P-L vector are (using eq(5) and eq(6a))

$$W_{0} = 0, \ \ W_{i} = m J_{i}$$

where

$$J_{i} = \frac{1}{2}\epsilon_{ijk}M^{jk}$$

Hence

$$W^{2} = -m^{2}J^{2}$$

From eq(6b) or (6c), we find the algebra

$$[J_{i},J_{j}] = i\epsilon_{ijk}J_{k}$$

This is nothing but the angular momentum algebra $su(2) \approx so(3)$. Thus

$$H(q) = U(1) \times SU(2)$$

As you know, the finite-dimensional IRs of SU(2) are characterized by the eigenvalues of its Casimir operator;

$$J^{2} = j(j+1)I, \ \ j = 0,1/2,1,...$$

and for a given j, the dimension of the representation is (2j+1):

$$W_{3}|\vec{q}=0, s \rangle = |m|s |\vec{q} = 0, s \rangle$$
$$s = -j, -j+1, ... ,j$$

It is clear that the projection of the spin is not Poincare' invariant, while the spin value j does not depend on the choice of the frame of reference. Hence, in the massive case, IRs of the Poincare group are classified by mass and spin:

$$W^{2}= - m^{2}j(j+1) I$$

2) Massless IRs : $P^{a}P_{a} = 0$

On the upper light-cone surface $p^{2} = 0, \ \ p_{0}> 0$

we take the standard momentum

$$q_{a} = \omega (1,0,0,1), \ \ \omega \neq 0 \ \ \ (8)$$

For simplicity, we set $\omega = 1$.

As before, we use eq(5) & (6a) to find the components of the P-L vector on $\mathcal{V}_{q}$:

$$W_{0} = W_{3} = M_{12} \ \ \ (9a)$$

$$W_{1} = M_{32} + M_{02} \equiv E_{1} \ \ \ (9b)$$

$$W_{2} = M_{31} + M_{01} \equiv E_{2} \ \ \ (9c)$$

Thus, on $\mathcal{V}_{q}$, we have

$$-(W_{a}W^{a}) = (E_{1})^{2} + (E_{2})^{2}$$

So, mathematically, $P_{a}P^{a}= 0$ does not imply $W^{a}W_{a}= 0$.

From eq(6b) or (6c), it follows that the (little group) algebra generated by $E_{1}, E_{2}$ and $M_{12}$ is isomorphic to the Lie algebra of the group E(2) of translations and rotations on a 2-dimensional plane;

$$[E_{1} , E_{2}] = 0$$

$$[M_{12}, E_{1}] = -iE_{2}, \ \ [M_{12}, E_{2}] = iE_{1}$$

The Casimir operator of E(2) is $(E_{1})^{2} + (E_{2})^{2}$, and in any unitary IR, it takes the form

$$(E_{1})^{2} + (E_{2})^{2} = \mu^{2} I, \ \ \mu^{2} \geq 0$$

If $\mu^{2} > 0$, IRs of E(2) are labelled by the points $\vec{e}= (e_{1},e_{2})$ of the one-sphere

$$e_{1}^{2} + e_{2}^{2} = \mu^{2}$$

where the states

$$|\vec{e}\rangle : E_{1}|\vec{e}\rangle = e_{1}|\vec{e}\rangle , \ \ E_{2}|\vec{e}\rangle = e_{2}|\vec{e}\rangle$$

represent a base in the space of representation. Thus, if $\mu^{2}> 0$, the generators $E_{1}$ and $E_{2}$ have continuous spectrum, and the representations are infinite-dimentional. They should be interpreted as particles of infinite spin! Since no such particles are found in nature, these representations are of no PHYSICAL interest. As we mentioned before, the subspace $\mathcal{V}_{q}$ should have a finite dimension. Therefore, we are left with no other possibility but $\mu^{2} = 0$ and $E_{1},E_{2}$ are trivial on $\mathcal{V}_{q}$. Thus

$$W_{1} = W_{2} = 0 \ \ \ (10)$$

and on physical ground, we conclude that $W_{a}W^{a}= 0$.

Therefore, in the massless case, eq(7) becomes

$$u_{q} = e^{i\alpha} \exp (i \theta M_{12})$$

This means that (up to phase factors) the little group H(q) acts on $\mathcal{V}_{q}$ as a group of rotations in the Euclidean plane, i.e., as the $SO(2) \approx U(1)$ group generated by the only non-trivial operator $M_{12}$.
Since the action oh H(q) on $\mathcal{V}_{q}$ should be irreducible, this space consists of only one non-trivial state, i.e., one-dimensional;

$$M_{12}|q, h \rangle = h |q , h \rangle \ \ \ (11)$$

Indeed, since E(2) is non-compact Lie group, its only finite-dimensional unitary representation is one-dimensional.
Notice that $\exp (i \theta M_{12})$ describes space rotation along the direction of motion. Putting $\theta = 2 \pi$, we get

$$e^{2i \pi M_{12}}|q , h \rangle = e^{2i \pi h}|q , h \rangle$$

The phase factor must equal to ($\pm 1$) depending on whether the representation is single-valued or double-valued. Thus

$$h = 0 , \pm 1/2 , \pm 1 , ...$$

The quantity |h| is called the spin of massless particle, and h is the helicity quantum number.
From eq(9), (10) and (11), we see that the relation

$$W^{a} = h P^{a} \ \ (12)$$

holds in the reference system q = (1,0,0,1). Since W and P are Lorentz vectors, eq(12) holds in any coordinates system. Thus, the helicity is a Lorentz invariant and, a fortiori, Poincare; invariant characteristic of massless particles; the massless finite-dimensional IRs of the Poincare' group are classified by helicity.

Please note that the change of sign of h defines, in general, a different particle state which may not exist physically. The 2 helicity states of the photon mean that the photon is described by a reducible representation of the Poincare' group P(1,3); under space inversion, the states with h = $\pm 1$ transform into one another. These 2 states form an IR of P(1,3) + space inversion.
Now we have all that is needed to prove that the two polarization states of photon can not be described by Lorentz vector, and that the only admissible vector field of mass zero is a gradient of a scalar field of helicity zero.

regards

sam

 

[Entropia]

Hello,

You are correct in your understanding that Lorentz transformations on particles induce Wigner rotations that depend on the momentum. This means that the reduced density matrix, which is obtained by tracing out the momentum, does not have a simple Lorentz transformation law. This is because the reduced density matrix only describes the state of the particle, and not its momentum, which is affected by the Lorentz transformation.

The difference between massive and massless particles in this context is that massless particles, such as photons, have a zero rest mass and therefore always travel at the speed of light. This means that their momentum is always in the same direction as their velocity, and therefore Lorentz transformations do not affect their momentum. This is why the reduced density matrix for massless particles can be Lorentz-transformed correctly.

On the other hand, massive particles have a non-zero rest mass and their momentum is not necessarily in the same direction as their velocity. This means that their momentum is affected by Lorentz transformations, and therefore the reduced density matrix cannot be Lorentz-transformed in a simple way.

I hope this helps clarify the difference between massive and massless particles in terms of Lorentz transformations and reduced density matrices. Let me know if you have any further questions.