Symmetries in Quantum Mechanics

In summary: However, because the rotation is a two-dimensionaltransform, the range of f changes from R^2 to R^3, but the range of gremains the same. So, the inner product between f and g is unchanged,despite the fact that f is now in a different range from g.In summary, the QM symmetries represented by unitary operators can be represented by simple scalar transformations. Similarity transformations are equivalent to another representation where the D on the RHS is not a unitary operator. QFT uses the Lorentz symmetry, which is the most common symmetry in the context of particles. The Lorentz symmetry between a particle and its environment is maintained even when
  • #1
LAHLH
409
1
Hi,

In QM symmetries can be represented by unitary operators. For example for rotations: [tex] \hat{U}_{R}\psi(\vec{x})=\psi(R^{-1}\vec{x}) [/tex], which is simple enough, as it just says that the vale of the rotated wavefunction at some point is the value of the old wavefunction at the pre-rotated point. Similary for a translation we could have [tex] \hat{T}(a)\psi(x)=\psi(x-a) [/tex] and so on. This much I understand.

Also we have similarity that is to say any representation D of a group is equivalent to another representation [tex] D'=SDS^{-1} [/tex] where S is just a non singular matrix that relates the basis vectors [tex] \vec{e}_i=S_{ji}\vec{f}_j [/tex] say. So this kind of equivalence is just looking at the operator in another basis effectively.

Now when we come to QFT, obviously the main symmetry of interest in Lorentz transformations and it is quite common to see written, [tex] U(\Lambda)^{-1}\Psi(x)U(\Lambda)=D(\Lambda)\Psi(\Lambda^{-1}x)[/tex]. I think what may be confusing me is that now [tex] \Psi[/tex] is not a state, it is an operator. If I'm not mistaken then, we are in the Heisenburg picture, so the states are staying fixed instead of transforming as in my first paragraph results (i.e. [tex] \hat{U}_{R}\psi(\vec{x})=\psi(R^{-1}\vec{x}) [/tex] etc), and instead these unitary operators are acting to evolve the operators themselves, as usual in Heisenburg. So this relation isn't a similarity transformation after all?

I don't understand however, why we need the D on the RHS, and why is this D not unitary?

just not quite 100% clear on what's going on here.
 
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  • #2
In your first equation, where you wrote using wavefunctions (so no operators, actually operator-valued distributions), the wavefunction is a scalar in respect to the orthogonal group. It should be generalized to account for the spin in a Galilei-invariant framework, i.e. use Galilei spinors.

The equation using operators is already contains the Lorentz spinors, as it was derived from the representation theory of the special-relativistic group, the restricted Poincare group (actually its universal covering).

If you generalize the wavefunction approach to account for the spin of the particle, then you should get a similar form as in the s-rel case.
 
  • #3
I'm afraid you lost me somewhat :s

I think my point about not actually a similarity transform wasn't true after thinking some more, because the change in the wavefunction or Hilbert space basis, is completely analogous to the usual change of basis one thinks of when looking at a similarity transform. So the Heisenberg pic with the operators transforming under the group is just like the usual sim transform on ops.

I still don't really understand the RHS, and perhaps I should have asked the questions in terms of just some operator being transformed under SO(3) or something easier to visualise instead of Lorentz. Just don't really understand what this D is really doing to the operator and why isn't it the same unitary rep as we have representing the group on the LHS etc
 
  • #4
Something I wrote about relativistic wave functions years ago might help, or it might further confuse the issue.

Code:
First, though, consider the inner product for standard
non-relativistic quantum mechanics. If f and g are wavefunctions, then
their inner product is

         /
<f|g> =  |f*(x) g(x) dx.
         /

Note that f and g are elements of an (subset of) infinite-dimensional
function space F given by

F = { f | f : R^3 -> C },

where R^3 represents all of space, and all the elements of F satisfy
the Schroedinger equation. So, wave functions are complex-valued
functions of space. Note that the inner product of wave functions is
defined on elements of F, not on elements of C, which is the range of
the functions in F.

Now let's move to the relativistic case. The wave functions are still
going to be functions that have common domains and ranges, and the
inner product is still going to be defined on wave function space,
and not on the common range of the functions in wave function space.
Wave function space is still going to be infinite-dimensional. It
would be strange if in going from the non-relativistic to the
relativistic we go from an infinite-dimensional wave function space
to a finite-dimensional wave-function space. What changes? The domain
and range of the wave functions. The domain for all the wavefunction
changes to all of spacetime (R^4 in a particular inertial coordinate
system), and the common range is (Dirac-)spinor space D, which is
4-dimensional. Thus, wave function space is now

G = { f | f : R^4 -> D },

where all the elements G satisfy the Dirac equation.

So, wave functions are spinor-valued functions of spacetime. Note that
while D is finite-dimensional, G isn't!. Again, the inner product
needs to be defined on wave function space, not on the common range of
the wavefunctions.(Sorry for repeating myself.)

I think part of the confusion comes from the following. Consider the
action of a rotation on non-relativistic wave functions f and g. Under
a rotation R, f -> f' and g -> g' defined by

f'(x) = f( R^(-1) x), g'(x) = g( R^(-1) x),

where x is a vector in R^3. Clearly, <f'|g'> = <f|g> for every f and g,
so the rotation R has a unitary action on wave function space F.

Now consider the action of a boost on relativistic wave functions f and
g. Under a boost B, f -> f' and g -> g' defined by

f'(x) = L f( B^(-1) x), g'(x) = L g( B^(-1) x),

where L is the transformation on spinor-space D that corresponds to the
boost B. At the the risk of being pedantic, x and x' = B^(-1) x are
4-vectors, f(x') is a spinor, f'(x) = L f(x') is a
spinor, and f and f' are wavefuntions. Unfortunately, some books
abbreviate the above by leaving out the arguments, i.e., they write
f' = L f. This gives the impression that wave functions are spinors
(elements of D), but they're not, they're spinor-valued functions
(elements of G).

In the non-relativistic case, leaving out the arguments results in
f' = f, which is non-sensical, but actually just tries to say that a
rotation leaves elements of C (complex scalars) unchanged. This doesn't
happen in the relativistic case because elements of D (spinors) are not
left unchanged by the action of the Lorentz group.

For the relativistic wave functions above, an appropriately defined
inner product on the infinite-dimensional wave function space G
results in <f'|g'> = <f|g> for every f and g, so the boost B has a
unitary action on wave function space G.
 
  • #5
In QM, you can also view the action of symmetries on the operators instead of the states. For example,

[tex] U(R)^\dag \vec P U(R) = R \vec P[/tex]

Where [itex]\vec P[/itex] is the vector operator of momentum, [itex]R[/itex] is an ordinary 3-dimensional rotation matrix, and [itex]U(R)[/itex] is the unitary operator in hilbert space that corresponds to [itex]R[/itex].

In quantum field theory, you just have an operator or set of operators located at each point in space. The operators [itex]U[/itex] in hilbert space need to be unitary but there is no reason that the matrix [itex]D[/itex] has to be unitary. Of course, in the above case, [itex]D(R) = R[/itex] does in fact happen to be orthogonal, but there is no general reason why it has to be.
 
  • #6
matonski said:
In QM, you can also view the action of symmetries on the operators instead of the states. For example,

[tex] U(R)^\dag \vec P U(R) = R \vec P[/tex]

Where [itex]\vec P[/itex] is the vector operator of momentum, [itex]R[/itex] is an ordinary 3-dimensional rotation matrix, and [itex]U(R)[/itex] is the unitary operator in hilbert space that corresponds to [itex]R[/itex].

In quantum field theory, you just have an operator or set of operators located at each point in space. The operators [itex]U[/itex] in hilbert space need to be unitary but there is no reason that the matrix [itex]D[/itex] has to be unitary. Of course, in the above case, [itex]D(R) = R[/itex] does in fact happen to be orthogonal, but there is no general reason why it has to be.

hmm, I see this. I think what's worrying me is why we need to act on the LHS with U(R) and just R on the RHS. e.g. why not have [tex] U(R)^\dag \vec P U(R) = U(R) \vec P[/tex]. If U(R) is the operator corresponding to the group element, then what exactly is the operator R alone on the RHS? In classical mech we know what a vector is, and indeed part of the definition of a vector is to transform under the usual rotation matrix in this way, but why do we demand this 'vector of operators' to behave this way.

I'm struggling to get across exactly what I don't understand about this, as I think I understand most of the pieces but not the whole as well as I would like. It's definitely not about the particularly symmetry (wether it being Lorentz/SO(3)/translations etc, just this scheme in general, so you're example of a vector operator is probably a good one to avoid uneccessary complications)
 
  • #7
Perhaps its easier to go back to viewing the transformations on states and instead look at expectation values. Under a rotation,

[tex]
|\psi \rangle \to U(R) |\psi \rangle
[/tex]

Thus

[tex]
\langle \psi | P^i |\psi \rangle \to \langle \psi | U(R)^\dag P^i U(R) |\psi \rangle = R^{ij} \langle \psi| P^j | \psi \rangle
[/tex]

In other words, [itex]\langle \psi| P^i | \psi \rangle[/itex] is an ordinary, 3-dimensional vector. The equation states that the expectation value of [itex]P^i[/itex] in the rotated state is obtained by taking a linear combination of the expectation values of the [itex]P^i[/itex]s in the unrotated state.
 
  • #8
LAHLH,

The reason for the transformation law

[tex] U(\Lambda)^{-1}\Psi(x)U(\Lambda)=D(\Lambda)\Psi(\Lambda^{-1}x)[/tex]

is very simple. We *intentionally* defined quantum field [tex]\Psi(x)[/tex] in order to have exactly this transformation law. The only place where this logic is explained with full clarity is vol. 1 of Weinberg's QFT textbook.

The main problem in QFT is to build a relativistically invariant interaction operator. The idea is to build this interaction as a product of x-dependent "quantum field" operators. It can be proven that such a product satifies relativistic conditions if the factors [tex]\Psi(x)[/tex] have the above transformation rule plus some additional properties (e.g., commutativity at space-like intervals). Weinberg shows how the usual expressions for quantum field operators follow from these requirements. In this logic, quantum fields (and their transformation laws) do not have any physical meaning. They are just convenient mathematical objects: factors used to construct the interaction operator.

Eugene.
 
  • #9
Thank you all for the help, this seems a lot clearer to me now.
 
  • #10
LAHLH said:
Hi,

In QM symmetries can be represented by unitary operators. For example for rotations: [tex] \hat{U}_{R}\psi(\vec{x})=\psi(R^{-1}\vec{x}) [/tex], which is simple enough, as it just says that the vale of the rotated wavefunction at some point is the value of the old wavefunction at the pre-rotated point. Similary for a translation we could have [tex] \hat{T}(a)\psi(x)=\psi(x-a) [/tex] and so on. This much I understand.

Also we have similarity that is to say any representation D of a group is equivalent to another representation [tex] D'=SDS^{-1} [/tex] where S is just a non singular matrix that relates the basis vectors [tex] \vec{e}_i=S_{ji}\vec{f}_j [/tex] say. So this kind of equivalence is just looking at the operator in another basis effectively.

Now when we come to QFT, obviously the main symmetry of interest in Lorentz transformations and it is quite common to see written, [tex] U(\Lambda)^{-1}\Psi(x)U(\Lambda)=D(\Lambda)\Psi(\Lambda^{-1}x)[/tex]. I think what may be confusing me is that now [tex] \Psi[/tex] is not a state, it is an operator. If I'm not mistaken then, we are in the Heisenburg picture, so the states are staying fixed instead of transforming as in my first paragraph results (i.e. [tex] \hat{U}_{R}\psi(\vec{x})=\psi(R^{-1}\vec{x}) [/tex] etc), and instead these unitary operators are acting to evolve the operators themselves, as usual in Heisenburg. So this relation isn't a similarity transformation after all?

I don't understand however, why we need the D on the RHS, and why is this D not unitary?

just not quite 100% clear on what's going on here.


Remember that free field [itex]\phi(x)[/itex] connects the vacuum to the free one-particle state and no others. So, the C-number

[tex]\Phi(x) = \langle \Phi | \phi (x) | 0 \rangle [/tex]

represents the wave function describing the one-particle state [itex]|\Phi \rangle[/itex]. Therefore, the transformed state;

[tex]|\bar{\Phi}\rangle = U |\Phi \rangle[/tex]

can be represented by the wave function

[tex]\bar{\Phi}(\bar{x}) = \langle \bar{\Phi}|\phi (\bar{x}) | 0 \rangle = \langle \Phi | U^{\dagger} \phi (\bar{x}) U | 0 \rangle \ \ (1)[/tex]

(use has been made of the invariance of the vacuum [itex]U|0 \rangle = | 0 \rangle[/itex])

Now, if the C-number wave function transforms according to some (finite-dimensional matrix) representation, D, of the symmetry group in question, i.e., if

[tex]\bar{\Phi}(\bar{x}) = D \Phi (x) = D \langle \Phi | \phi (x) | 0 \rangle[/tex]

then it follows from eq(1), that the field operator transforms as

[tex]\bar{\phi}(\bar{x}) \equiv U^{\dagger} \phi (\bar{x}) U = D \phi (x)[/tex]

See also post #9 in

www.physicsforums.com/showthread.php?t=172461

regards

sam
 
  • #11
samalkhaiat said:
Now, if the C-number wave function transforms according to some (finite-dimensional matrix) representation, D, of the symmetry group in question, i.e., if

[tex]\bar{\Phi}(\bar{x}) = D \Phi (x) = D \langle \Phi | \phi (x) | 0 \rangle[/tex]

then it follows from eq(1), that the field operator transforms as

[tex]\bar{\phi}(\bar{x}) \equiv U^{\dagger} \phi (\bar{x}) U = D \phi (x)[/tex]

The problem with this argument is that for wavefunctions the transformation matrix D is always unitary, while for quantum fields it is not unitary in most cases. Also, for wave functions (in the momentum representation) there is such a thing as "Wigner rotation". It is absent in the field transformation law.

Eugene.
 
  • #12
OK this is what I think I have gleaned so far.

(Firstly from Ballentine) So for every symmetry we have a transformation of the QM operators and the states, [tex] A\to A'[/tex] and [tex] |\Psi\rangle \to |\Psi'\rangle [/tex]

a)If [tex] A|\phi_n \rangle =a_n |\phi_n\rangle [/tex] then after the transformation [tex] A'|\phi^{'}_n \rangle =a_n |\phi^{'}_n\rangle [/tex]. i.e. they have same eigenvalues because they represent equivalent observables just in rotated frame.
b) If we have a state vector given by [tex] |\Psi\rangle=\sum c_n|\phi_n\rangle[/tex], then the rotated state vector will be of form [tex] |\Psi'\rangle=\sum c^{'}_n|\phi^{'}_n\rangle[/tex]. The two state vectors must obey [tex] |c^2|=|c'^{2}|[/tex], or equivalentley [tex]|\langle \phi_n|\Psi\rangle|^2=|\langle \phi^{'}_n|\Psi^{'}\rangle|^2[/tex]. This must be so, because they express the probability of equivalent events just in difference reference frames.

Next there is a theorem due to Wigner that says any mapping of the vector space onto itself preseveing [tex]|\langle\phi|\Psi\rangle|[/tex] may be implemented by operator U such that [tex] |\Psi\rangle\to |\Psi'\rangle= U|\Psi\rangle [/tex] and [tex] |\phi\rangle\to |\phi'\rangle= U|\phi\rangle [/tex], with U being either unitary or antiunitary. It turns out due to a few technical details that we must choose U unitary for continuous symmetries.

Now returning to [tex] A|\phi_n \rangle =a_n |\phi_n\rangle [/tex] and [tex] A'|\phi^{'}_n \rangle =a_n |\phi^{'}_n\rangle [/tex] and using [tex] |\phi'\rangle= U|\phi\rangle [/tex], we get:
[tex] A'|\phi^{'}_n \rangle =a_n |\phi^{'}_n\rangle [/tex]
[tex] A'U|\phi_n \rangle =a_n U|\phi_n\rangle [/tex]
[tex] U^{-1}A'U|\phi_n \rangle =a_n |\phi_n\rangle [/tex]

Hence [tex] A=U^{-1}A'U [/tex] this leads to [tex] A\to A^{'}=UAU^{-1} [/tex]
 
  • #13
Next we know that the expectation value of say a vector operator is an ordinary vector, hence say:

[tex] \langle \Psi |A^{'}_i(\vec{x})|\Psi\rangle =R_{ij} \langle \Psi|A_j(R^{-1}\vec{x})|\Psi \rangle [/tex]

But now we just use [tex] A\to A^{'}=UAU^{-1} [/tex]

[tex] \langle \Psi |UA(\vec{x})U^{-1}|\Psi\rangle =R_{ij} \langle \Psi|A_j(R^{-1}\vec{x})|\Psi \rangle [/tex]

which is valid for any state, so

[tex] UA(\vec{x})U^{-1} =R_{ij} A_j(R^{-1}\vec{x}) [/tex]

hmm but the U inverse is usually on the left...perhaps this is related to active and passive trans somehow?
 

1. What are symmetries in quantum mechanics?

Symmetries in quantum mechanics refer to the properties or transformations of a physical system that remains unchanged under certain operations. These operations can include rotations, translations, and reflections.

2. How are symmetries related to conservation laws in quantum mechanics?

Symmetries in quantum mechanics are closely related to conservation laws. In fact, Noether's theorem states that for every continuous symmetry in a physical system, there exists a corresponding conservation law.

3. Can symmetries be broken in quantum mechanics?

Yes, symmetries can be broken in quantum mechanics. This can occur when the physical system is in a state that does not exhibit the symmetry, or when the symmetry is broken due to external factors such as interactions with other particles.

4. How do symmetries impact the behavior of particles in quantum mechanics?

Symmetries play a crucial role in determining the behavior of particles in quantum mechanics. They can dictate the allowed energy levels, the selection rules for transitions between energy levels, and the properties of particles such as spin and charge.

5. Are there different types of symmetries in quantum mechanics?

Yes, there are different types of symmetries in quantum mechanics. These can include spatial symmetries, which involve transformations in physical space, and internal symmetries, which involve transformations in abstract spaces such as isospin or color space.

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