Very stupid question about Wigner's Theorem

  • Context: Graduate 
  • Thread starter Thread starter Petr Mugver
  • Start date Start date
  • Tags Tags
    Stupid Theorem
Click For Summary
SUMMARY

Wigner's representation theorem asserts that any invertible transformation between rays of a Hilbert space that preserves transition probabilities can be represented by either a unitary or antiunitary transformation. In the context of a one-dimensional Hilbert space, the functions f(z)=z (unitary) and g(z)=z* (antiunitary) induce the same identity transformation on rays, leading to confusion regarding their equivalence. The discussion clarifies that while these transformations may coincide in a projective space of rays, they remain distinct in a vector space context. The theorem does not require unitary and antiunitary transformations to be distinct across all Hilbert spaces.

PREREQUISITES
  • Understanding of Wigner's representation theorem
  • Familiarity with unitary and antiunitary transformations
  • Knowledge of Hilbert spaces and rays in quantum mechanics
  • Basic concepts of complex functions and transformations
NEXT STEPS
  • Study the implications of Wigner's representation theorem in higher-dimensional Hilbert spaces
  • Explore the differences between unitary and antiunitary transformations in quantum mechanics
  • Investigate the role of projective spaces in quantum state representation
  • Examine examples of invertible transformations in multi-dimensional Hilbert spaces
USEFUL FOR

Quantum physicists, mathematicians specializing in functional analysis, and students studying quantum mechanics who seek a deeper understanding of transformation properties in Hilbert spaces.

Petr Mugver
Messages
279
Reaction score
0
Wigner's representation theorem says that any invertible transformation between rays of a Hilbert space that preserves transition probabilities can be implemented by a transformation on the Hilbert space itself, which is either unitary or antiunitary, depending on the particular transformation considered, right?

For example, you can find in any QM book that almost all symmetries are represented by linear operators, the only significant exception being time inversion, right?

I present here the most trivial example I can imagine: a one-dimensional Hilbert space. The two complex function of complex variable

f(z)=z\qquad\textrm{and}\qquad g(z)=z^*

are, respectively, unitary and antiunitary, and they both induce the same transformation between rays of the Hilbert space H=\mathbb{C}, the identity transformation.

I know the solution of this apparent paradox should be easy, but I really can't see it! I know the proof of the theorem, and it doesn't help!

Any hint woukd be appreciated.
 
Last edited:
Physics news on Phys.org
Petr Mugver said:
I present here the most trivial example I can imagine: a one-dimensional Hilbert space. The two complex function of complex variable

f(z)=z\qquad\textrm{and}\qquad g(z)=z^*

are, respectively, unitary and antiunitary, and they both induce the same transformation between rays of the Hilbert space H=\mathbb{C}, the identity transformation.

I don't understand your example. How is the operation of complex conjugation
equivalent to the identity transformation? If you're saying the z (a complex number) is
a complex multiple of z*, then effectively there's only one state (ray), and all (invertible)
transformations are the identity transformation. The transformations are different if
the Hilbert space is considered as a vector space, but coincide if we revert to a projective
space of rays.

Wigner's theorem doesn't say (iirc) that the unitary and antiunitary transformations
must be distinct for every possible type of Hilbert space.

Maybe try a slightly less trivial example of a 2D Hilbert space?
 
strangerep said:
I don't understand your example. How is the operation of complex conjugation
equivalent to the identity transformation? If you're saying the z (a complex number) is
a complex multiple of z*, then effectively there's only one state (ray), and all (invertible)
transformations are the identity transformation. The transformations are different if
the Hilbert space is considered as a vector space, but coincide if we revert to a projective
space of rays.

Yes, using the definition of rays, the two transformations coincide (you have to consider the Hilbert space as a space on complex numbers, but that's what you usually do in QM). In a one-dimensional Hilbert space there are actually two rays, because (but this probably depends on the definition you adopt) z=0 is a ray all alone. Not only the invertible transformations induce the identity on rays, it is sufficient, for example, that f(0)=0 and that f(z) is different from zero for z different from zero (you can find horrible functions that belong to this class!) and the induced map will be the identity anyway!

strangerep said:
Wigner's theorem doesn't say (iirc) that the unitary and antiunitary transformations
must be distinct for every possible type of Hilbert space.

Well, if a transformation is unitary, then it can't be antiunitary and vice versa...so I suppose they must be different!
 
Petr Mugver said:
Wigner's representation theorem says that any invertible transformation between rays of a Hilbert space that preserves transition probabilities can be implemented by a transformation on the Hilbert space itself, which is either unitary or antiunitary, depending on the particular transformation considered, right?d.

No. Should be: "of a Hilbert space of dimension at least two."
 
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
965
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K