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

Simultaneously diagonalize two operators

  1. Mar 2, 2014 #1
    Most part of the fundamental quantum mechanics rely upon finding some operators [itex]\hat{X}[/itex] that commutes with hamiltonian and is able to simultaneously diagonalize [itex]\hat{X}[/itex] and hamiltonian.

    Actually what do you mean by diagonalize simultaneously??

    Is there any relation with diagonalize the corresponding matrix..??

    Can anybody explain me the concept using some example???

    Thank you.
     
  2. jcsd
  3. Mar 2, 2014 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    For matrices, it just means to find a basis of the vector space which are eigenbases for both matrices.

    For example, take the matrices

    [tex]A=\left(\begin{array}{cccc}
    1 & 0 & 1\\
    0 & 0 & 0\\
    1 & 0 & 1
    \end{array}\right)
    ~\text{and}~
    A=\left(\begin{array}{cccc}
    2 & 1 & 1\\
    1 & 0 & -1\\
    1 & -1 & 2
    \end{array}\right)[/tex]

    Then if we choose the basis
    [tex](1,0,1),~(-1,-1,1),~(1,-2,-1)[/tex]
    then we see that these vectors are eigenvectors for both matrices. So we have found a common eigenbasis for both matrices.

    If we then change our basis to this eigenbasis, then both matrices will appear to be in diagonal form.
     
  4. Mar 2, 2014 #3

    WannabeNewton

    User Avatar
    Science Advisor

    It just means we can find a complete set of simultaneous eigenvectors of both operators. For example if we look at the total angular momentum operator ##J^2## and relative to some coordinate system the ##z## component of angular momentum ##J_z## then we know ##[J^2,J_z] = 0##. We go through the usual raising and lowering operator jazz to get the complete set of simultaneous eigenvectors of these operators that we usually label ##\{|j,m \rangle \}## and the matrices ##\langle j',m' |J^2|j,m \rangle ## and ##\langle j',m' |J_z|j,m \rangle ## will both be diagonal matrices.
     
  5. Mar 2, 2014 #4
    I understood what you've told upto this line. But can you elaborate this line.
    How can I represent an operator in this eigenbasis..? And can you tell me the diagonalized form of any one of it?

    Thank you
     
  6. Mar 2, 2014 #5

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

  7. Mar 8, 2014 #6
  8. Mar 27, 2014 #7
    you are right, you could read abt similarity transformation!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook