# LOG#096. Group theory(XVI).

Given any physical system, we can perform certain “operations” or “transformations” with it. Some examples are well known: rotations, traslations, scale transformations, conformal transformations, Lorentz transformations,… The ultimate quest of physics is to find the most general “symmetry group” leaving invariant some system. In Classical Mechanics, we have particles, I mean point particles, and in Classical Field Theories we have “fields” or “functions” acting on (generally point) particles. Depending on the concrete physical system, some invariant properties are “interesting”.

Similarly, we can leave the system invariant and change the reference frame, and thus, we can change the “viewpoint” with respect to we realize our physical measurements. To every type of transformations in space-time (or “internal spaces” in the case of gauge/quantum systems) there is some mathematical transformation $F$ acting on states/observables. Generally speaking, we have:

1) At level of states: $\vert \Psi\rangle \longrightarrow F\vert \Psi\rangle =\vert \Psi'\rangle$

2) At level of observables: $O\longrightarrow F(O)=O'$

These general transformations should preserve some kind of relations in order to be called “symmetry transformations”. In particular, we have conditions on 3 different objects:

A) Spectrum of observables:

$O_n\vert \phi_n\rangle =O_n\vert \phi_n\rangle \leftrightarrow O'\vert \phi'_n\rangle=O_n\vert \phi'\rangle$

These operators $O, O'$ must represent observables that are “identical”. Generally, these operators must be “isospectral” and they will have the same “spectrum” or “set of eigenvalues”.

B) In Quantum Mechanics, the probabilities for equivalent events must be the same before and after the transformations and “measurements”. In fact, measurements can be understood as “operations” on observables/states of physical systems in this general framework. Therefore, if

$\displaystyle{\vert\Psi\rangle =\sum_n c_n\vert \phi_n\rangle}$

where $\vert\phi_n\rangle$ is a set of eigenvectors of O, and

$\displaystyle{\vert\Psi'\rangle =\sum_n c'_n\vert\phi'_n\rangle}$

where $\vert\phi'_n\rangle$ is a set of eigenvectors of O’, then we must verify

$\vert c_n\vert^2=\vert c'_n\vert^2\longleftrightarrow \vert \langle \phi_n\vert \Psi\rangle\vert^2=\vert\langle\phi'_n\vert \Psi'\rangle\vert^2$

C) Conservation of commutators. In Classical Mechanics, there are some “gadgets” called “canonical transformations” leaving invariant the so-called Poisson brackets. There are some analogue “brackets” in Quantum Mechanics: the commutators are preserved by symmetry transformations in the same way that canonical transformations leave invariant the classical Poisson brackets.

These 3 conditions constrain the type of symmetries in Classical Mechanics and Quantum Mechanics (based in Hilbert spaces). There is a celebrated theorem, due to Wigner, saying more or less the mathematical way in which those transformations are “symmetries”.

Let me define before two important concepts:

Antilinear operator.  Let A be a linear operator in certain Hilbert space H. Let us suppose that $\vert \Psi\rangle,\vert\varphi\rangle \in H$ and $\alpha,\beta\in\mathbb{C}$. An antilinear operator A satisfies the condition:

$A\left(\alpha\vert\Psi\rangle+\beta\vert\varphi\rangle\right)=\alpha'A\left(\vert\Psi\rangle\right)+\beta^*A\left(\vert\varphi\rangle\right)$

Antiunitary operator.  Let A be an antilinear opeartor in certain Hilbert space H. A is said to be antiunitary if it is antilinear and

$AA^+=A^+A=I\leftrightarrow A^{-1}=A^+$

Any continuous family of continuous transformations can be only described by LINEAR operators. These transformations are continuously connected to the unity matrix/identity transformation leaving invariant the system/object, and this identity matrix is in fact a linear transformation itself. The product of two unitary transformations is unitary. However, the product of two ANTIUNITARY transformations is not antiunitary BUT UNITARY.

Wigner’s theorem. Let A be an operator with eigenvectors $B=\vert\phi_n\rangle$ and $A'=F(A)$ another operator with eigenvectors $B'=\vert\phi'_n\rangle$. Moreover, let us define the state vectors:

$\displaystyle{\vert \Psi\rangle=\sum_n a_n\vert\phi_n\rangle}$ $\displaystyle{\vert\varphi\rangle=\sum_n b_n\vert\phi_n\rangle}$

$\displaystyle{\vert\Psi'\rangle=\sum_n a'_n\vert\phi'_n\rangle}$ $\displaystyle{\vert\varphi'\rangle=\sum_n b'_n\vert\phi'_n\rangle}$

Then, every bijective transformation leaving invariant

$\vert \langle \phi_n\vert \Psi\rangle\vert^2=\vert\langle\phi'_n\vert \Psi'\rangle\vert^2$

can be represented in the Hilbert space using some operator. And this operator can only be UNITARY (LINEAR) or ANTIUNITARY(ANTILINEAR).

This theorem is relative to “states” but it can also be applied to maps/operators over those states, since $F(A)=A'$ for the transformation of operators. We only have to impose

$A\vert\phi_n\rangle =a_n\phi_n\rangle$

$A'\vert\phi'_n\rangle=a_n\vert\phi'_n\rangle$

Due to the Wigner’s theorem, the transformation between operators must be represented by certain operator $U$, unitary or antiunitary accordingly to our deductions above, such that if $\vert\phi'_n\vert=U\vert\phi_n\vert$, then:

$A'\vert\phi'_n\rangle=A'U\vert\phi_n\rangle=a_n U\vert\phi_n\rangle$

$U^{-1}A'U\vert\phi_n\rangle=a_n\vert\phi_n\rangle$

This last relation is valid vor every element $\vert\phi_n\rangle$ in a set of complete observables like the basis, and then it is generally valid for an arbitrary vector. Furthermore,

$U^{-1}A'U=A$

$A\rightarrow A'=U^{-1}AU$

There are some general families of transformations:

i) Discrete transformations $A_i$, both finite and infinite in order/number of elements.

ii) Continuous transformation $A(a,b,\ldots)$. We can speak about uniparametric families of transformations $A(\alpha)$ or multiparametric families of transformations $A(\alpha_1,\alpha_2,\ldots,\alpha_n)$. Of course, we can also speak about families with an infinite number of parameters, or “infinite groups of transformations”.

Physical transformations form a group from the mathematical viewpoint. That is why all this thread is imporant! How can we parametrize groups? We have provided some elementary vision in previous posts. We will focus on continuous groups. There are two main ideas:

a) Parametrization. Let $U(s)\in F(\alpha)$ be a family of unitary operators depending continuously on the parameter $s$. Then, we have:

i) $U(0)=U(s=0)=I$.

ii) $U(s_1+s_2)=U(s_1)U(s_2)$.

b) Taylor expansion. We can expand the operator as follows:

$U(s)=U(0)+\dfrac{dU}{ds}\bigg|_{s=0}+\mathcal{O}(s^2)$

or

$U(s)=I+\dfrac{dU}{ds}\bigg|_{s=0}+\mathcal{O}(s^2)$

There is other important definitiion. We define the generator of the infinitesimal transformation $U(s)$, denoted by $K$, in such a way that

$\dfrac{dU}{ds}\bigg|_{s=0}\equiv iK$

Moreover, $K$ must be an hermitian operator (note that mathematicians prefer the “antihermitian” definition mostly), that is:

$I=U(s)U^+(s)=I+s\left(\dfrac{dU}{ds}\bigg|_{s=0}+\dfrac{U^+}{ds}\bigg|_{s=0}\right)+\mathcal{O}(s^2)$

$iK+(iK)^+=0$

$K=K^+$

Q.E.D.

There is a fundamental theorem about this class of operators, called Stone theorem by the mathematicians, that says that if $K$ is a generator of a symmetry at infinitesimal level, then $K$ determines in a unique way the unitary operator $U(s)$ for all value $s$. In fact, we have already seen that

$U(s)=e^{iKs}$

So, the Stone theorem is an equivalent way to say the exponential of the group generator provides the group element!

We can generalize the above considerations to finite multiparametric operators. The generator would be defined, for a multiparametric family of group elements $G(\alpha_1,\alpha_2,\ldots,\alpha_n)$. Then,

$iK_{\alpha_j}=\dfrac{\partial G}{\partial_{\alpha_j}}\bigg|_{\alpha_j=0}$

There are some fundamental properties of all this stuff:

1) Unitary transformations $G(\alpha_1,\alpha_2,\ldots,\alpha_n)$ form a Lie group, as we have mentioned before.

2) Generators $K_{\alpha_j}$ form a Lie algebra. The Lie algebra generators satisfy

$\displaystyle{\left[K_i,K_j\right]=\sum_k c_{ijk}K_k}$

3) Every element of the group or the multiparametric family $G(\alpha_1,\alpha_2,\ldots,\alpha_n)$ can be written (likely in a non unique way) such that:

$G\left(\alpha_1,\alpha_2,\ldots,\alpha_n\right)=\exp \left( iK_{\alpha_1}\alpha_1\right)\exp \left( iK_{\alpha_2}\alpha_2\right)\cdots \exp \left( iK_{\alpha_n}\alpha_n \right)$

4) Every element of the multiparametric group can be alternatively written in such a way that

$e^{iK_\alpha\alpha}e^{iK_\beta\beta}=e^{K_\alpha\omega_1(\alpha,\beta)}e^{iK_\beta\omega_2(\alpha,\beta)}$

where the parameters $\omega_1, \omega_2$ are functions to be determined for every case.

What about the connection between symmetries and conservation laws? Well, I have not discussed in this blog the Noether’s theorems and the action principle in Classical Mechanics (yet) but I have mentioned it already. However, in Quantum Mechanics, we have some extra results. Let us begin with a set of unitary and linear transformations $G=T_\alpha$. These set can be formed by either discrete or continuous transformations depending on one or more parameters. We define an invariant observable Q under G as the set that satisfies

$Q=T_\alpha Q T^+_\alpha,\forall T_\alpha\in G$

Moreover, invariance have two important consequences in the Quantum World (one “more” than that of Classical Mechanics, somehow).

1) Invariance implies conservation laws.

Given a unitary operator $T^+=T^{-1}$, as $Q=T_\alpha Q T^+_\alpha$, then

$QT_\alpha=T_\alpha Q$ and thus

$\left[Q,T_\alpha\right]=0$

If we have some set of group transformations $G=T_\alpha$, such the so-called hamiltonian operator $H$ is invariant, i.e., if

$\left[H,T_\alpha\right]=0,\forall T_\alpha\in G$

Then, as we have seen above, these operators for every value of their parameters are “constants” of the “motion” and their “eigenvalues” can be considered “conserved quantities” under the hamiltonian evolution. Then, from first principles, we could even have an infinite family of conserved quantities/constants of motion.

This definifion can be applied to discrete or continuous groups. However, if the family is continuous, we have additional conserved constants. In this case, for instance in the uniparametric group, we should see that

$T_\alpha=T(\alpha)=\exp (i\alpha K)$

and it implies that if an operator is invariant under that family of continuous transformation, it also commutes with the infinitesimal generator (or with any other generator in the multiparametric case):

$Q=T_\alpha QT^+_\alpha \leftrightarrow \left[Q,K\right]=0$

Every function of the operators in the set of transformations is also a motion constant/conserved constant, i.e., an observable such as the “expectation value” would remain constant in time!

2) Invariance implies (not always) degeneration in the spectrum.

Imagine a hamiltonian operator $H$ and an unitary transformation $T$ such as $\left[H,T\right]=0$. If

$H\vert \alpha\rangle=E_\alpha\vert \alpha\rangle$

then

1) $\vert\beta\rangle=T\vert\alpha\rangle$ is also an eigenvalue of H.

2) If $\vert\alpha\rangle$ and $\vert\beta\rangle$ are “linearly independent”, then $E_\alpha$ is (a) degenerated (spectrum).

Check:

1st step. We have

$\left[H,T\right]=0\longrightarrow HT=TH$

$H(T\vert\alpha\rangle)=T(H\vert\alpha\rangle)=E_\alpha T\vert\alpha\rangle$

2nd step. If $\vert\alpha\rangle$ and $\vert\beta\rangle$ are linarly independent, then $T\vert\alpha\rangle=c\vert\alpha\rangle$ and thus

$H(T\vert\alpha\rangle)=H(c\vert\alpha\rangle)=cE_\alpha\vert\alpha\rangle$

If the hamiltonian $H$ is invariant under a transformation group, then it implies the existence (in general) of a degeneration in the states (if these states are linearly independent). The characteristics features of this degeneration (e.g., the degeneration “grade”/degree in each one of these states) are specific of the invariance group. The converse is also true (in general). The existence of a degeneration in the spectrum implies the existence of certain symmetry in the system. Two specific examples of this fact are the kepler problem/”hydrogen atom” and the isotropic harmonic oscillator. But we will speak about it in other post, not today, not here, ;).

# LOG#036. Action and relativity.

The hamiltonian formalism and the hamiltonian H in special relativity has some issues with the definition. In the case of the free particle one possible definition, not completely covariant, is the relativistic energy

$\boxed{E=\sqrt{m^2c^4+c^2p^2}=H}$

There are two others interesting scalars in classical relativistic theories. They are the lagrangian L and the action functional S. The lagrangian is obtained through a Legendre transformation from the hamiltonian:

$\boxed{L=pv-H}$

From the hamiltonian, we get the velocity using the so-called hamiltonian equation:

$\dot{\mathbf{q}}=\mathbf{v}=\dfrac{\partial H}{\partial \mathbf{p}}=c^2\dfrac{\mathbf{p}}{E}$

Then,

$L=\dfrac{E}{c^2}\mathbf{v}^2-E=E\left(\dfrac{v^2}{c^2}-1\right)=-\dfrac{E}{\gamma^2}=-\dfrac{m\gamma c^2}{\gamma^2}=-\dfrac{mc^2}{\gamma}$

and finally

$\boxed{L=-mc^2\sqrt{1-\dfrac{v^2}{c^2}}=-\dfrac{mc^2}{\gamma}=-mc\sqrt{-\dot{X}^2}}$

The action functional is the time integral of the lagrangian:

$\boxed{S=\int Ldt}$

However, let me point out that the above hamiltonian in SR has some difficulties in gauge field theories. Indeed, it is quite easy to derive that a more careful and reasonable election for the hamiltonian in SR should be zero!

In the case of the free relativistic particle, we obtain

$S=-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt$

Using the relation between time and proper time (the time dilation formula):

$dt=\gamma d\tau\rightarrow \dfrac{dt}{\gamma}=d\tau$

direct substitution provides

$-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt=-mc^2\int d\tau$

And defining the infinitesimal proper length in spacetime as $ds=cd\tau$, we get the simple and wonderful result:

$\boxed{S=-mc\int ds}$

Sometimes, the covariant lagrangian for the free particle is also obtained from the following argument. The proper length is defined as

$ds^2=d\mathbf{x}^2-c^2dt^2$

The invariant in spacetime is related with the proper time in this way:

$ds^2=-c^2d\tau^2=d\mathbf{x}^2-c^2dt^2$

Thus, dividing by $dt^2$

$-c^2\dfrac{d\tau^2}{dt^2}=\mathbf{v}^2-c^2$

and

$d\tau^2=\gamma^{-2}dt^2=\dfrac{1}{\gamma^2}dt^2=\left(1-\dfrac{\mathbf{v}^2}{c^2}\right)dt^2$

so

$d\tau=\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}dt$

$cd\tau=ds=\sqrt{c^2-\mathbf{v}^2}dt=\sqrt{-\dot{X}^2}dt$

that is

$\boxed{ds=cd\tau=\sqrt{-\dot{X}^2}dt=\sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=\sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}$

and the free coordinate action for the free particle would be:

$\boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}dt=-mc^2\int \sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}$

Note, that since the election of time “t” is “free”, we can choose $t=\tau$ to obtain the generally covariant free action:

$\boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}d\tau=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}d\tau}$

Remark: the (rest) mass is the “coupling” constant for the free particle proper lenght to guess the free lagrangian

$\boxed{L=-mc^2\sqrt{-\dot{X}^2}}$

Now, we can see from this covariant action that the relativistic hamiltonian should be a feynmanity! From the equations of motion,

$P_\mu=\dfrac{\partial L}{\partial \dot{X}^\mu}=mc\dfrac{\dot{X}_\mu}{\sqrt{-\dot{X}^2}}$

The covariant hamiltonian $\mathcal{H}$, different from H, can be build in the following way:

$\mathcal{H}=P_\mu \dot{X}^\mu-L=mc\dfrac{\dot{X}_\mu \dot{X}^\mu}{\sqrt{-\dot{X}^2}}-mc\sqrt{-\dot{X}^2}=0$

The meaning of this result is hidden in the the next identity ( Noether identity or “hamiltonian constraint” in some contexts):

$\boxed{\mathcal{H}=P_\mu P^\mu+m^2c^2=0}$

since

$P_\mu P^\mu=m U_\mu mU^\mu=-m^2c^2$

This strange fact that $\mathcal{H}=0$ in SR, a feynmanity as the hamiltonian, is related to the Noether identity $E^\mu \dot{X}_\mu$ for the free relativistic lagrangian, indeed, a consequence of the hamiltonian constraint and the so-called reparametrization invariance $\tau'=f (\tau)$. Note, in addition, that the free relativistic particle would also be invariant under diffeomorphisms $x^{\mu'}= f^\mu (x)=f^\mu (x^\nu)$ if we were to make the metric space-time dependent, i.e., if we make the substitution $\eta_{\mu\nu}\rightarrow g_{\mu\nu} (x)$. This last result is useful and important in general relativity, but we will not discuss it further in this moment. In summary, from the two possible hamiltonian in special relativity

$H=E=\sqrt{\mathbf{p}^2c^2+(mc^2)^2}$

$\mathcal{H}=P_\mu P^\mu+m^2c^2=0$

the natural and more elegant (due to covariance/invariance) is the second one. Moreover, the free particle lagrangian and action are:

$\boxed{L=-mc^2\sqrt{-\dot{X}^2}}$

$\boxed{S=-mc^2\int d\tau=-mc\int ds=\int L dt}$

Remark: The true covariant lagrangian dynamics in SR is a “constrained” dynamics, i.e., dynamics where we are undetermined. There are more variables that equations as a result of a large set of symmetries ( reparametrization invariance and, in the case of local metrics, we also find diffeomorphism invarince).

The dynamical equations of motion, for a first order lagrangian (e.g., the free particle we have studied here), read for the lagrangian formalism:

$\boxed{\delta S=\delta \int L (q,\dot{q};t)dt =0\leftrightarrow\begin{cases}E(L)=0,\mbox{with E(L)=Euler operator}\\ E(L)=\dfrac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{q}}\right)=0\end{cases}}$

By the other hand, for the hamiltonian formalism, dynamical equations are:

$\boxed{\delta S=\delta \int H (q, p; t)dt =0\leftrightarrow\begin{cases}p=\dfrac{\partial L}{\partial \dot{q}},\;\mbox{with}\; \det\left(\dfrac{\partial ^2L}{\partial \dot{q}^i\partial \dot{q}^j}\right)\neq 0\\ \;\\ \dfrac{dq}{dt}=\dot{q}=\dfrac{\partial H}{\partial p}\\ \;\\\dfrac{dp}{dt}=\dot{p}=-\dfrac{\partial H}{\partial q} \\ \;\\ \dfrac{dH}{dt}=\dot{H}=\dfrac{\partial H}{\partial t}=-\dfrac{\partial L}{\partial t}\end{cases}}$