Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Physical interpretation of unitary transformation

  1. Dec 29, 2009 #1
    what actually happens physically ...when we make transpose of a matrix...and in unitary transformation we transpose the matrix and take the conjugate.......physically what type of change happens in it.
  2. jcsd
  3. Dec 29, 2009 #2


    User Avatar

    I guess the answer depends what trait you take as "physical" traits, here's some spontaneous thoughts.

    One important thing is that in terms of the normal QM entropy of von-neumann, unitary transformations are for example "information preserving". When we consider unitary time evolution, the initial and final states are somehow equally a priori probable if that's measured in terms of von-neumann entropy.

    Also, generally physical expectation values are invariant during unitary transformations.

    This is interesting in the context of information processes where in general one can loose, gain or maintain information. So unitary processes in QM, IMHO describe a special case of information processes, where the state of information changes, but in a way that maintain total entropy. One could even think of all states generated by unitary transformations as defining a sort of equiprobable class of states.

    So non-information preserving transformations are IMO more "interesting".

  4. Dec 30, 2009 #3
    the action S or hamiltonian invariant under transformations corresponds to the symmetry of the system. these transformations may be continuous( U(1), SU(2)) or discrete.

    while the state may not retain these symmetry--symmetry breaking.
  5. Jan 3, 2010 #4
    Well, consider what happens under a unitary transformation when dealing with finite matrices. Say you have unitary matrix U, and you are transforming matrix H. U^-1 HU is the unitary transformation. Recall a unitary matrix is composed of orthonormal columns. This means its adjoint is its inverse (assuming a square matrix). So, HU gives the matrix H acting on each column of U. So if we call the columns |a_n>, then since they are also the eigenvectors of U, U=|u_n><u_n| as n runs through all the columns (this summation will be implied from now on). Notice U is just a projection operator. And U^-1=|u_n*><u_n*| where * denotes the complex conjugate. Now, lets represent H with its eigenvectors |H_k>, and eigenvalues E_k: H=|H_k>E_k<H_k|. So, you can see that U^-1HU is simply equal to |P_k*>E_k<P_k| where |P_k> denotes the projection of |H_k> onto U space. U^-1HU is clearly (and by definition) isomorphic to H. So in conclusion, what you are doing is representing H in terms of the space represented by U.

    Recall that if H commutes with U U^-1 HU=H (proved by multiplying both sides by U), H is invariant. That's where the symmetry stuff comes from. Go ahead an play around with the this (like sharing eigenstuff when H and U commute-stuff like that). You should right away see the mathematical origins of much of the quantum physics theorems.
    Good question! It has lots of interesting answers!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook