# Symmetry in quantum mechanics

1. Mar 25, 2014

### AndrewShen

I have several questions about symmetry in quantum mechanics.

1. It is often said that the degeneracy is the dimension of irreducible representation. I can understand that if the Hamiltonian has a symmetric group G, then the state space with the same energy eigenvalue will carry a representation of G. However, why this representation is usually irreducible?
2. Wigner proved that any symmetric group representation in quantum mechanics must be either unitary or anti-unitary. Is it true that the representation of continuous symmetric group must be unitary and cannot be anti-unitary?
3. What is the difference between geometric symmetry and dynamical symmetry? By dynamical symmetry I mean for example the SO(4) symmetry of hydrogen. Some text refers dynamical symmetry to "internal" symmetry. How to state the definition of dynamical symmetry strictly? If it can't be obtained by rotation, translation and inversion, then how do to find it generally?

2. Mar 25, 2014

### ChrisVer

1. we prefer the irreducible representation because it does avoid mixing of states...in other words you can distinguish states among themselves since they will exist in an irrep. Eg for something quite familiar, think of the states coming by spin (l=1)x(l=1). The result is a 3*3=9dim representation, which be decomposed in irreps, giving you that (l=2)+(l=1)+(l=0) or 5+3+1. If you don't reduce it, probably your states of l=2,1,0 will be mixed under the action of the repr.
Another way to state it is by Peter-Weyl theorem, stating that the state of squared integrable functions on G is the Hilbert space direct sum over finite dimensional irreducible representations.

2.I was 100% wrong....

3. An intrinsic symmetry is a symmetry existing in internal degrees of freedom. For example the fact that the wavefunction is symmetric under a total phase shift. I guess that the geometrical symmetries are those associated with Poincare group and its transformations, or more specially the Lorentz Group SO(3,1) whereas the dynamical symmetries are those we call gauge symmetries.

Last edited: Mar 25, 2014
3. Mar 25, 2014

### vanhees71

ad 1) A unitary representation is called irreducible if you can reach any normalized state ket of the corresponding space of states by acting with symmetry transformations on one arbitrary such normalized state. Since by definition the Hamiltonian is a Casimir operator of the symmetry group, acting with an arbitrary symmetry transformation on an eigenstate of the Hamiltonian leads to another eigenstate of the same eigenvalue. Thus the representation of the symmetry group decomposes into "block matrices" in the energy-eigenbasis where each eigenspace to a given eigenvalue builds one "block". If the representation acting on one such block is not irreducible within this subspace, there must be additional symmetry operations that can be added to the so far considered symmetry group, leading to a larger group.

E.g. for the hydrogen atom in addition to the rotations as a symmetry group you have more symmetry transformations, generated by the Runge-Lenz vector augmenting the symmetry group from SO(3) to SO(4) (for the bound states). A complete set of energy eigenvectors for the bound states is given by the vectors $|n,l,m \rangle$, which is a common eigenvector of the possible complete set of compatible observables $H$ (eigenvalues $E_n=-1 \text{Ry}/n$, $n \in \{1,2,\ldots \}$), $\vec{L}^2$ (eigenvalues $l(l+1)$, $l \in \{0,1, \ldots, n-1\}$), and $L_z$ (eigenvalues $m \in \{0,1,\ldots,l \}$).

This explains why the energy eigenvalues only depend on the principal quantum number $n$ and not also on the orbital-angular momentum quantum number $l$. The eigenspace to eigenvalue $E_n$ is $n^2$ dimensional. For each $n \in \{1,2,\ldots \}$ all angular-momentum eigenvectors for $l \in \{0,1,\ldots,(n-1) \}$ contribute to the basis of this eigenspace, and that are
$$\sum_{l=0}^{n-1} (2l+1)=n^2$$
orthonormalized basis vectors for each eigenspace.

ad 2) For a continuous symmetry group you can connect the group identity with a continuous path $G(g,t)$, $t \in [0,1]$ with any given group element $g$. For the representation we have
$$U[G(g,t)] c |\psi>=C(t) U[G(g,t)] |\psi \rangle,$$
where $C(t)$ for each $t$ is either $c$ or $c^*$. If the group representation is a continuous function of $t$, $C(t)$ must be the same $c$ or $c^*$ for all $t$ since for a non-real $c$ otherwise there'd be a jump. But now we have $U[G(g,t=0)]=U(\text{id})=\hat{1}$, and this means that $C(t)=c=\text{const}$. Thus the representation must be necessarily unitary for all group elements.

ad 3) An "intrinsic symmetry" is just defined as any symmetry operation that is not an operation referring on a spac-time symmetry. It's just a definition. There is nothing special about them otherwise.

4. Mar 26, 2014

### AndrewShen

Thanks! And also thanks to ChrisVer.

However, I still don't understand why the representation is essentially irreducible.

"A unitary representation is called irreducible if you can reach any normalized state ket of the corresponding space of states by acting with symmetry transformations on one arbitrary such normalized state. " Why this can always been done? Why can we reach any state from any state (of the same energy eigenvalue)? Is there any physical explanation?

"If the representation acting on one such block is not irreducible within this subspace, there must be additional symmetry operations that can be added to the so far considered symmetry group, leading to a larger group." Why is this statement is true? I always see this in the book. However, I haven't see any rigorous proof.

For the 3rd question. If it cannot be obtained from space-time opearation, is there any general approach to find them? For example, how do I construct R-L vector? By plain guess?

5. Mar 26, 2014

### The_Duck

If state A cannot be reached from state B by a symmetry operation, then state A and state B are not related by symmetry and so there is no reason for them to have the same energy. It would be quite a coincidence if two states had the same energy but were completely unrelated by symmetry. So we expect that all the states with the same energy are probably going to be related to each other by some symmetry, i.e. that they are all going to be in a single irreducible representation of some symmetry group.

If you find that there is indeed an "accidental" degeneracy of two energy eigenstates that don't seem to be related by symmetry, you should get suspicious and wonder if there is some subtle symmetry that you haven't noticed that actually relates them and explains the "accident."

6. Mar 26, 2014

### ChrisVer

A representation is called irreducible when it cannot be reduced to acting on smaller invariant subspaces... Or in other words, lets think of a vector space V, which is acted upon by a representation $\rho$, that is:
$\rho: V \rightarrow V$
$v^{i} \rightarrow (\rho)^{i}_{j} v^{j} , v^{i}\in V$
Now suppose that the representation acts on some of these v vectors in such a way that the result will also belong in some of these vectors. Then we are talking about an invariant subspace in the vector space V. Then indeed in matrix form, the representation can be reduced in block form, the block acting on the invariant subspace in order to keep it unchanged. Suppose for example you have an upper block form 2x2, that means that the first 2 components of your vector will be mixed just between each other.

2nd- I'm not sure.

As for the 3rd..R-L vector is not a gauge symmetry if I recall well.. It has to do with the spacetime symmetry of the problem...
A symmetry which doesn't come from the symmetries of spacetime for example, is those of standard model SU(3)SU(2)U(1) (each having its own geometry but not in spacetime space, but in the intrinsic space).
At least if someone finds any misconception, please correct me too...

7. Mar 26, 2014

### strangerep

There is a (reasonably general) theory of symmetries of differential equations -- relevant to physics as symmetries of the equations of motion. A "general symmetry" maps solutions of the equations among themselves. There are several, successively more complicated, types of such symmetries.

One could transform the independent variables to a new set of independent variables.
One could transform the dependent variables to a new set of dependent variables.
One could transform the set of all variables (i.e., dependent and independent) among themselves.
One could transform the set of all variables and derivatives of the dependent variables among themselves (so-called generalized symmetries).

(Btw, one sometimes encounters these concepts under the heading of "variational symmetries" wherein one looks for symmetries of the Action, rather than the equations of motion that result from extremizing the latter.)

The general approach to finding such (continuous) symmetries systematically is by Lie transformation methods. One can re-express the quantities in the differential equation in terms of infinitesimally-transformed variables, and impose a certain condition that ensures the transformed differential equation remains satisfied identically. This leads to (in general) a set of coupled partial differential equations for the parameters of the transformation. In various physically-interesting cases, this set can be solved without too much trouble to find the generators of all possible such transformations. Quantities like momentum, Hamiltonian, laplace-runge vector, etc, can be identified among the generators.

I found the following textbook to be a useful physicist-friendly introduction to the subject:

H. Stephani, "Differential Equations -- Their solution using symmetries",
Cambridge University Press, 1989C, ISBN 0-521-36689-5.

He also explains the meanings of "point symmetry", "geometric symmetry", "dynamical symmetry", "contact symmetry", "variational symmetry", "generalized symmetry". But note that the terminology seems not yet universal among other authors. E.g., P.J. Olver seems to use some of the terms differently in his (more advanced) textbook on the subject.

Stephani's emphasis is on finding solutions -- since if one can find a symmetry there's usually a transformation which can simplify the original D.E. significantly, and reduce its order. But the general technique of finding such symmetries is interesting in itself.

Unfortunately, it seems the subject of arbitrarily generalized symmetries (wherein one mixes derivatives of arbitrary order) still has a long way to go before it could be considered mature. But the basic theory is nevertheless enlightening in many cases of physical interest, including the area of generalized coherent states.

HTH.

Last edited: Mar 27, 2014
8. Mar 26, 2014

### samalkhaiat

1. True.
OK.
That need not be the case. Consider the tensor product state
$$| \vec{ J }_{ 1 } \ ; n_{ 1 } \rangle \otimes | \vec{ J }_{ 2 } \ ; n_{ 2 } \rangle \equiv | \vec{ J }_{ 1 } + \vec{ J }_{ 2 } \ ; n_{ 1 } n_{ 2 } \rangle , \ \ \ (1)$$
where $n_{ i } = 2 J_{ i } + 1$ is the number of components of the state, the number of members in a multiplet, the dimension of the representation space spanned by $| \vec{ J }_{ i } \ ; n_{ i } \rangle$ or simply the measure of the degeneracy. That tensor state, as well as the Hamiltonian which acts on it is reducible: Indeed, the total Hamiltonian is of the form
$$H = \left( H_{ 1 } \otimes I_{ n_{ 2 } \times n_{ 2 } } \right) \oplus \left( I_{ n_{ 1 } \times n_{ 1 } } \otimes H_{ 2 } \right) . \ \ \ (2)$$
Let me give you some examples from the group $SU(2)$ and $SO(3)$.
$$| \vec{ 1/2 } \ ; 2 \rangle \otimes | \vec{ 1/2 } \ ; 2 \rangle \equiv | \vec{ 1/2 } + \vec{ 1/2 } \ ; 4 \rangle . \ \ \ (3)$$
This state belongs to a 4-dimensional rep space, $\mathcal{ V }_{ 4 }$. If you want to work with it, you must use operators and generators matrices of dimension $4 \times 4$. The next thing you do is to test the reducibility of this representation space $\mathcal{ V }_{ 4 }$; can you write it as a direct sum of non-intersecting and invariant subspaces? The above case, like all tensor product spaces, is reducible. Indeed you should know from the theory of angular momentum that
$$| \vec{ 1/2 } ; \ 2 \rangle \otimes | \vec{ 1/2 } \ ; 2 \rangle \equiv | \vec{ 1/2 } + \vec{ 1/2 } \ ; 4 \rangle = a | 1 \ ; 3 \rangle + b | 0 \ ; 1 \rangle . \ \ \ (4)$$
Mathematically speaking, this means that the 4-dimensional representation (vector) space $\mathcal{ V }_{ 4 }$ is the direct sum of two non-intersecting invariant subspaces: A 3-dimentional vector space, $\mathcal{ V }_{ 3 }$ spanned by $| 1 ; 3 \rangle$ and a 1-dimensional vector space, $\mathcal{ V }_{ 1 }$ spanned by $| 0 ; 1 \rangle$,
$$\mathcal{ V }_{ 2 } \otimes \mathcal{ V }_{ 2 } = \mathcal{ V }_{ 4 } = \mathcal{ V }_{ 3 } \oplus \mathcal{ V }_{ 1 } .$$
Physics textbooks write this as (the sum of triplet and singlet)
$$[2] \otimes [2] = [4] = [3] \oplus [1] ,$$
$$( \frac{ 1 }{ 2 } ) + ( \frac{ 1 }{ 2 } ) = ( 1 ) + ( 0 )$$
I collected both these notations in Eq(4). The two states $| 1 \ ; 3 \rangle$ and $| 0 \ ; 1 \rangle$ form two distinct multiplets; they do not mix with each other and no group element can take one to the other.
Some times we write Eq(4) using $SU(2)$ 2-vectors and tensors. In fact, Eq(4) is nothing but the following tensor identity $( a , b = 1 , 2 )$:
$$V_{ a } V_{ b } \equiv T_{ a b } = \frac{ 1 }{ 2 } ( T_{ a b } + T_{ b a } ) + \frac{ 1 }{ 2 } ( T_{ a b } – T_{ b a } ) . \ \ (5)$$
If you count the number of independent components in each tensor, you find
$$[2] \otimes [ 2 ] = [ 4 ] = [ 3 ] \oplus [ 1 ] .$$
In principle, the tensor method is capable of generating all irreducible representations of a given group:
1) construct all tensors with a given rank from the defining (fundamental) representations.
2) divide them into (totally)symmetric and (totally)anti-symmetric parts.
3) use the invariant tensors of the group to subtract all possible traces from the (totally)symmetric tensor and/or reduce the rank of the (totally)anti-symmetric tensor, and
4) identify the resulting parts with new irreducible representation.
Let us apply this program on the group of rotation in 3 dimension, $SO(3)$. The fundamental representation is given by the 3-vector $V_{ a } \sim [3]$. Now, we take the tensor product of $V_{ a }$ with $V_{ b }$ and call the resulting tensor $T_{ a b }$. Using the tensor identity Eq(5), we divide this tensor into symmetric and anti-symmetric tensors:
$$V_{ a }V_{ b } \equiv T_{ a b } = S_{ ( a b ) } + A_{ [ a b ] } .$$
By counting the number of components, we notice that this is just
$$[ 3 ] \otimes [ 3 ] = [ 9 ] = [ 6 ] \oplus [ 3 ] .$$
In 3 dimension, the rotation group has two invariant tensors: the symmetric $\delta_{ a b }$ and the totally anti-symmetric symbol $\epsilon_{ a b c }$. So, the invariant trace $S = \delta_{ a b } S_{ ( a b ) } = S_{ c c }$ can be subtracted from the symmetric tensor $S_{ ( a b ) } \sim [ 6 ]$ as follow
$$S_{ ( a b ) } = \left( S_{ ( a b ) } - \frac{ 1 }{ 3 } \delta_{ a b } S \right) + \frac{ 1 }{ 3 } \delta_{ a b } S .$$
So, we have managed to decompose the $[ 6 ]$ into the direct sum $[ 5 ] \oplus [ 1 ]$. Also, we can identify the anti-symmetric tensor $A_{ [ a b ] }$ with $\epsilon_{ a b c } V_{ c } \sim [ 3 ]$. For second rank tensors, no further reduction is possible. So, we are left with 3 irreducible tensors: traceless symmetric, which belongs to $[ 5 ]$, a vector in the $[ 3 ]$ and a scalar $S \in [ 1 ]$:
$$V_{ a }V_{ b } \equiv T_{ a b } = \left( S_{ ( a b ) } - \frac{ 1 }{ 3 } \delta_{ a b } S \right) + \epsilon_{ a b c } V_{ c } + \frac{ 1 }{ 3 } \delta_{ a b } S .$$
In terms of states in the Hilbert space, the above equation corresponds to
$$| \vec{ 1 } \ ; 3 \rangle \otimes | \vec{ 1 } \ ; 3 \rangle = | \vec{1} + \vec{1} \ ; 9 \rangle = a | \vec{2} \ ; 5 \rangle + b | \vec{1} \ ; 3 \rangle + c | \vec{0} \ ; 1 \rangle .$$
This means that any $9 \times 9$ matrix acting on the direct product state can be put in block diagonal form with $5 \times 5$, $3 \times 3$ and $1 \times 1$ matrices on the diagonal.
One last important remark: states belonging to irreducible subspaces of equal dimension needs not be the same states. For example, consider combining the spins of three spin-1 particles:
$$| \vec{1} \ ; 3 \rangle \otimes | \vec{1} \ ; 3 \rangle \otimes | \vec{1} \ ; 3 \rangle = | \vec{3} \ ; 7 \rangle \oplus (2) | \vec{ 2 } \ ; 5 \rangle \oplus (3) | \vec{1} \ ; 3 \rangle \oplus | \vec{0} \ ; 1 \rangle .$$
Here, we have one spin-3 state (spanning the 7-dimensional irreducible subspace $[7]$), two different spin-2 states (each spanning irreducible, inequivalent subspace of dimension $[5]$), and three spin-1 states and one spin-0 state in the trivial representation space $[1]$.

Yes, you can say that continuous symmetry groups have unitary representations on the Hilbert space. But, most discrete symmetry groups, such as charge conjugation and time reversal, act on the Hilbert space by anti-unitary operators. This is necessary because we need positive-definite norms in Hilbert space.

Every continuous group defines a differentiable manifold. So, all continuous symmetry groups are geometrical.
Usually people, who use the term “dynamical” symmetry, they also define what they mean by the term.

There is NONE.

STUDY GROUP THEORY.

Sam

Last edited: Mar 26, 2014
9. Mar 28, 2014

### andrien

I guess I am late,but anyway I will say something about the dynamical symmetry.If we are given eigenfunction ψ1 of hamiltonian with a eigenvalue E,we can generate a set of n independent eigenfunctions by operating with all the elements of group G (n≤ order of group).These functions will form a basis for a representation of G which may be reducible or irreducible.If we have however all symmetry transformation which will leave the hamiltonian invariant included in G,then the representation generated by these degenerate eigenfunctions must be an irreducible one.
Eigenfunctions transforming according to different representation of G, in general, have different eigenvalue.So if $ψ$ and $\phi$ belong to different irreducible representation,no operation of G will mix them.But it may happen that we have failed to include all symmetry transformation of hamiltonian in the group G,in which case the representation generated by degenerate eigenfunctions may be a reducible one.The basis functions belonging to two IR may be degenerate.However if we find this happening consistently,it will point to an overlooked symmetry.If we consider all possible symmetry transformation of hamiltonian,the basis functions belonging to different IR of G,in general, have different eigenvalues.
This is actually the case with hydrogen atom where the symmetry group is enlarged from SO(3) to SO(4),which was illustrated by Vanhees earlier.One can calculate the energy eigenvalues by constructing two casimir operators of SO(4).It is however still possible to have 'accidental' degeneracy which arises for some specific value of the parameters like intensity of magnetic field applied,nuclear charge etc. which cannot be attributed to any internal symmetry.