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

Combine Hamiltonians of two different bases

  1. Oct 29, 2013 #1
    I'm trying to solve a dynamical quantum mechanics problem related to the Cs atom, but I'm having trouble in the following, and I'm afraid I'm doing it wrong.

    Say I have the matrix form of the Hamiltonian on a basis for a system [itex]| \psi \rangle[/itex] to be [itex]H_\psi[/itex], and another system with bases [itex]| \phi \rangle[/itex] with Hamiltonian [itex]H_\phi[/itex].

    Now I would like to introduce interactions between the first and seconed systems, which will become (I suppose) [itex]| \psi \phi\rangle[/itex].

    1) How do I combine those hamiltonians in matrix formalism before adding the new hamiltonian that introduces the interactions?

    2) Say the first and second system are complete angular momenta basis in Zeeman eigen-states (so [itex]|F_1 m_{F_1} \rangle[/itex] and [itex]|F_2 m_{F_2}\rangle[/itex]). If the Wigner-D rotation matrix for the first Hamiltonian/basis is [itex]\mathcal{D}^{F_1}_{m_{F_1} m_{F_1}^\prime}[/itex], and for the second system is [itex]\mathcal{D}^{F_2}_{m_{F_2} m_{F_2}^\prime}[/itex]. How can I rotate each system and then combine them to introduce interactions between them?

    Thank you for any efforts.
     
  2. jcsd
  3. Oct 29, 2013 #2
    You suppose correct. For this situation we create a Hilbert Space by the tensor product of basis of the two spaces containing Hamiltonians [itex]\hat{H}_\psi[/itex] and [itex]\hat{H}_\phi[/itex](which are spanned by [itex]| \psi \rangle[/itex] and [itex]| \phi \rangle[/itex] respectively).

    So the new Hilbert Space is now spanned by basis [itex]| \psi_i \phi_j\rangle[/itex] or [itex]| \psi_i\rangle \otimes |\phi_j\rangle[/itex] (sum over i and j). The Hamiltonian of the system now becomes [itex]\hat{H}_\psi \otimes I + I \otimes \hat{H}_\phi + \hat{H}_{int}[/itex]. If you are not aware of this algebra, I would recommend Quantum Computation and Quantum Information by Neilsen and Chaung, Chapter 2.
     
    Last edited: Oct 29, 2013
  4. Oct 29, 2013 #3
    Thank you so much for your answer. Let me though put an example to make this clear, because I got some result that I don't believe. Say I have the Hamiltonian matrices [itex]H_1[/itex] and [itex]H_2[/itex]

    [tex]H_1=\left(
    \begin{array}{cccc}
    \text{s1} & 0 & 0 & 0 \\
    0 & \text{s2} & 0 & 0 \\
    0 & 0 & \text{s3} & 0 \\
    0 & 0 & 0 & \text{s4} \\
    \end{array}
    \right)[/tex]

    [tex]H_2=\left(
    \begin{array}{ccc}
    \text{t1} & 0 & 0 \\
    0 & \text{t2} & 0 \\
    0 & 0 & \text{t3} \\
    \end{array}
    \right)[/tex]

    So now we have:

    [tex]H_1 \otimes I_3 = \left(
    \begin{array}{cccccccccccc}
    \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} \\
    \end{array}
    \right)[/tex]

    and

    [tex]H_2 \otimes I_3 = \left(
    \begin{array}{cccccccccccc}
    \text{t1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & \text{t2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & \text{t3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & \text{t1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & \text{t2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & \text{t3} & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & \text{t1} & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t2} & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t3} & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t1} & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t2} & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t3} \\
    \end{array}
    \right)[/tex]

    So now the new Hamiltonian is

    [tex]H_1 \otimes I_3 + I_4 \otimes H_2 = \left(
    \begin{array}{cccccccccccc}
    \text{s1}+\text{t1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & \text{s1}+\text{t2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & \text{s1}+\text{t3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & \text{s2}+\text{t1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & \text{s2}+\text{t2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & \text{s2}+\text{t3} & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & \text{s3}+\text{t1} & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s3}+\text{t2} & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s3}+\text{t3} & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4}+\text{t1} & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4}+\text{t2} & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4}+\text{t3} \\
    \end{array}
    \right)[/tex]

    Is that how it's done?
     
  5. Oct 29, 2013 #4
    You have done correctly with some typos.

    [tex]H_1=\left(
    \begin{array}{cccc}
    \text{s1} & 0 & 0 & 0 \\
    0 & \text{s2} & 0 & 0 \\
    0 & 0 & \text{s3} & 0 \\
    0 & 0 & 0 & \text{s4} \\
    \end{array}
    \right),[/tex]

    and define identity operator on first space

    [tex]I_1=\left(
    \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
    \end{array}
    \right).[/tex]
    Similarly
    [tex]H_2=\left(
    \begin{array}{ccc}
    \text{t1} & 0 & 0 \\
    0 & \text{t2} & 0 \\
    0 & 0 & \text{t3} \\
    \end{array}
    \right),[/tex]
    and identity operator on second space
    [tex]I_2=\left(
    \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{array}
    \right).[/tex]

    The total Hamiltonian now should be [itex]\hat{H}_1 \otimes I_2 + I_1 \otimes \hat{H}_2 + \hat{H}_{int}[/itex]. We can evaluate terms as follows

    [tex]H_1 \otimes I_2 = \left(
    \begin{array}{cccccccccccc}
    \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} \\
    \end{array}
    \right),[/tex]
    as you have done correctly.
    Second term is
    [tex]I_1 \otimes H_2 = \left(
    \begin{array}{cccccccccccc}
    \text{t1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & \text{t2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & \text{t3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & \text{t1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & \text{t2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & \text{t3} & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & \text{t1} & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t2} & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t3} & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t1} & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t2} & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t3} \\
    \end{array}
    \right).[/tex]
    (you did correctly but it seems lot of typesetting made you overlook the order [itex]I_1 \otimes H_2[/itex] :)).
    And you can add these matrices to get total Hamiltonian matrix (without interaction).
     
  6. Oct 30, 2013 #5
    Hmmmmmmm... Thank you so much. Though I thought this approach would solve my problem, but apparently it doesn't.

    I have Hyperfine states with [itex]F=3,4[/itex] (where [itex]F=I+J[/itex] is the sum of the total angular momentum and the nuclear spin), and I want to separate them, do a rotation with Wigner-D matrices, and combine them back. Do you know how this could be done?
     
  7. Oct 30, 2013 #6
    I am sorry, I don't know what Wigner-D matrices are.
     
  8. Oct 30, 2013 #7
    Thank you, I appreciate your help so far. You could though (if you're interested) look at this further. Wigner-D matrices are simply the matrices that provide rotations in the angular momentum basis with Euler angles.

    So you have a basis of [itex]F=3[/itex] and [itex]m_F=-3,...,3[/itex], the Wigner-D matrix will be [itex]D_{m_F m_F^\prime}^{F=3}[/itex], where [itex]m_F, m_F^\prime[/itex] will be the matrix elements, and we'll have [itex]7\times 7=49[/itex] matrix elements for this basis.

    That's the idea simply, so now, with that given, I know how to rotate the basis F=3 or F=4, but I don't know how to rotate them both when they're combined. You see where my problem is?
     
  9. Oct 30, 2013 #8
    Please give one demonstration. Say I need to rotate basis F=3. How will I proceed?
     
  10. Oct 30, 2013 #9
    Thank you for your time. Wigner D-Matrix for that case will be the following, starting from the eigen-vector [itex]|F=3;m_F=-3 \rangle[/itex] to [itex]|F=3;m_F=3 \rangle[/itex]

    [tex]
    D^{F=3}=\left(
    \begin{array}{ccccccc}
    e^{-3 i \alpha -3 i \gamma } \cos ^6\left(\frac{\beta }{2}\right) & -\sqrt{6} e^{-3 i \alpha -2 i \gamma } \cos ^5\left(\frac{\beta }{2}\right) \sin \left(\frac{\beta }{2}\right) & \sqrt{15} e^{-3 i \alpha -i \gamma } \cos ^4\left(\frac{\beta }{2}\right) \sin ^2\left(\frac{\beta }{2}\right) & -2 \sqrt{5} e^{-3 i \alpha } \cos ^3\left(\frac{\beta }{2}\right) \sin ^3\left(\frac{\beta }{2}\right) & \sqrt{15} e^{i \gamma -3 i \alpha } \cos ^2\left(\frac{\beta }{2}\right) \sin ^4\left(\frac{\beta }{2}\right) & -\sqrt{6} e^{2 i \gamma -3 i \alpha } \cos \left(\frac{\beta }{2}\right) \sin ^5\left(\frac{\beta }{2}\right) & e^{3 i \gamma -3 i \alpha } \sin ^6\left(\frac{\beta }{2}\right) \\
    \sqrt{6} e^{-2 i \alpha -3 i \gamma } \cos ^5\left(\frac{\beta }{2}\right) \sin \left(\frac{\beta }{2}\right) & \frac{1}{2} e^{-2 i \alpha -2 i \gamma } \cos ^4\left(\frac{\beta }{2}\right) (6 \cos (\beta )-4) & -\frac{1}{2} \sqrt{\frac{5}{2}} e^{-2 i \alpha -i \gamma } \cos ^3\left(\frac{\beta }{2}\right) (6 \cos (\beta )-2) \sin \left(\frac{\beta }{2}\right) & \sqrt{30} e^{-2 i \alpha } \cos ^2\left(\frac{\beta }{2}\right) \cos (\beta ) \sin ^2\left(\frac{\beta }{2}\right) & -\frac{1}{2} \sqrt{\frac{5}{2}} e^{i \gamma -2 i \alpha } \cos \left(\frac{\beta }{2}\right) (6 \cos (\beta )+2) \sin ^3\left(\frac{\beta }{2}\right) & \frac{1}{2} e^{2 i \gamma -2 i \alpha } (6 \cos (\beta )+4) \sin ^4\left(\frac{\beta }{2}\right) & -\sqrt{6} e^{3 i \gamma -2 i \alpha } \cos \left(\frac{\beta }{2}\right) \sin ^5\left(\frac{\beta }{2}\right) \\
    \sqrt{15} e^{-i \alpha -3 i \gamma } \cos ^4\left(\frac{\beta }{2}\right) \sin ^2\left(\frac{\beta }{2}\right) & \frac{1}{2} \sqrt{\frac{5}{2}} e^{-i \alpha -2 i \gamma } \cos ^3\left(\frac{\beta }{2}\right) (6 \cos (\beta )-2) \sin \left(\frac{\beta }{2}\right) & e^{-i \alpha -i \gamma } \cos ^2\left(\frac{\beta }{2}\right) \left(\frac{15}{4} (\cos (\beta )-1)^2+5 (\cos (\beta )-1)+1\right) & -\frac{2 e^{-i \alpha } \cos \left(\frac{\beta }{2}\right) \left(\frac{15}{4} (\cos (\beta )-1)^2+\frac{15}{2} (\cos (\beta )-1)+3\right) \sin \left(\frac{\beta }{2}\right)}{\sqrt{3}} & e^{i \gamma -i \alpha } \left(\frac{15}{4} (\cos (\beta )-1)^2+10 (\cos (\beta )-1)+6\right) \sin ^2\left(\frac{\beta }{2}\right) & -\frac{1}{2} \sqrt{\frac{5}{2}} e^{2 i \gamma -i \alpha } \cos \left(\frac{\beta }{2}\right) (6 \cos (\beta )+2) \sin ^3\left(\frac{\beta }{2}\right) & \sqrt{15} e^{3 i \gamma -i \alpha } \cos ^2\left(\frac{\beta }{2}\right) \sin ^4\left(\frac{\beta }{2}\right) \\
    2 \sqrt{5} e^{-3 i \gamma } \cos ^3\left(\frac{\beta }{2}\right) \sin ^3\left(\frac{\beta }{2}\right) & \sqrt{30} e^{-2 i \gamma } \cos ^2\left(\frac{\beta }{2}\right) \cos (\beta ) \sin ^2\left(\frac{\beta }{2}\right) & \frac{2 e^{-i \gamma } \cos \left(\frac{\beta }{2}\right) \left(\frac{15}{4} (\cos (\beta )-1)^2+\frac{15}{2} (\cos (\beta )-1)+3\right) \sin \left(\frac{\beta }{2}\right)}{\sqrt{3}} & \frac{1}{2} \left(5 \cos ^3(\beta )-3 \cos (\beta )\right) & -\frac{2 e^{i \gamma } \cos \left(\frac{\beta }{2}\right) \left(\frac{15}{4} (\cos (\beta )-1)^2+\frac{15}{2} (\cos (\beta )-1)+3\right) \sin \left(\frac{\beta }{2}\right)}{\sqrt{3}} & \sqrt{30} e^{2 i \gamma } \cos ^2\left(\frac{\beta }{2}\right) \cos (\beta ) \sin ^2\left(\frac{\beta }{2}\right) & -2 \sqrt{5} e^{3 i \gamma } \cos ^3\left(\frac{\beta }{2}\right) \sin ^3\left(\frac{\beta }{2}\right) \\
    \sqrt{15} e^{i \alpha -3 i \gamma } \cos ^2\left(\frac{\beta }{2}\right) \sin ^4\left(\frac{\beta }{2}\right) & \frac{1}{2} \sqrt{\frac{5}{2}} e^{i \alpha -2 i \gamma } \cos \left(\frac{\beta }{2}\right) (6 \cos (\beta )+2) \sin ^3\left(\frac{\beta }{2}\right) & e^{i \alpha -i \gamma } \left(\frac{15}{4} (\cos (\beta )-1)^2+10 (\cos (\beta )-1)+6\right) \sin ^2\left(\frac{\beta }{2}\right) & \frac{2 e^{i \alpha } \cos \left(\frac{\beta }{2}\right) \left(\frac{15}{4} (\cos (\beta )-1)^2+\frac{15}{2} (\cos (\beta )-1)+3\right) \sin \left(\frac{\beta }{2}\right)}{\sqrt{3}} & e^{i \alpha +i \gamma } \cos ^2\left(\frac{\beta }{2}\right) \left(\frac{15}{4} (\cos (\beta )-1)^2+5 (\cos (\beta )-1)+1\right) & -\frac{1}{2} \sqrt{\frac{5}{2}} e^{i \alpha +2 i \gamma } \cos ^3\left(\frac{\beta }{2}\right) (6 \cos (\beta )-2) \sin \left(\frac{\beta }{2}\right) & \sqrt{15} e^{i \alpha +3 i \gamma } \cos ^4\left(\frac{\beta }{2}\right) \sin ^2\left(\frac{\beta }{2}\right) \\
    \sqrt{6} e^{2 i \alpha -3 i \gamma } \cos \left(\frac{\beta }{2}\right) \sin ^5\left(\frac{\beta }{2}\right) & \frac{1}{2} e^{2 i \alpha -2 i \gamma } (6 \cos (\beta )+4) \sin ^4\left(\frac{\beta }{2}\right) & \frac{1}{2} \sqrt{\frac{5}{2}} e^{2 i \alpha -i \gamma } \cos \left(\frac{\beta }{2}\right) (6 \cos (\beta )+2) \sin ^3\left(\frac{\beta }{2}\right) & \sqrt{30} e^{2 i \alpha } \cos ^2\left(\frac{\beta }{2}\right) \cos (\beta ) \sin ^2\left(\frac{\beta }{2}\right) & \frac{1}{2} \sqrt{\frac{5}{2}} e^{2 i \alpha +i \gamma } \cos ^3\left(\frac{\beta }{2}\right) (6 \cos (\beta )-2) \sin \left(\frac{\beta }{2}\right) & \frac{1}{2} e^{2 i \alpha +2 i \gamma } \cos ^4\left(\frac{\beta }{2}\right) (6 \cos (\beta )-4) & -\sqrt{6} e^{2 i \alpha +3 i \gamma } \cos ^5\left(\frac{\beta }{2}\right) \sin \left(\frac{\beta }{2}\right) \\
    e^{3 i \alpha -3 i \gamma } \sin ^6\left(\frac{\beta }{2}\right) & \sqrt{6} e^{3 i \alpha -2 i \gamma } \cos \left(\frac{\beta }{2}\right) \sin ^5\left(\frac{\beta }{2}\right) & \sqrt{15} e^{3 i \alpha -i \gamma } \cos ^2\left(\frac{\beta }{2}\right) \sin ^4\left(\frac{\beta }{2}\right) & 2 \sqrt{5} e^{3 i \alpha } \cos ^3\left(\frac{\beta }{2}\right) \sin ^3\left(\frac{\beta }{2}\right) & \sqrt{15} e^{3 i \alpha +i \gamma } \cos ^4\left(\frac{\beta }{2}\right) \sin ^2\left(\frac{\beta }{2}\right) & \sqrt{6} e^{3 i \alpha +2 i \gamma } \cos ^5\left(\frac{\beta }{2}\right) \sin \left(\frac{\beta }{2}\right) & e^{3 i \alpha +3 i \gamma } \cos ^6\left(\frac{\beta }{2}\right) \\
    \end{array}
    \right)
    [/tex]

    where the angles are the Euler angles, and the matrix is a unitary matrix, so I can use it to transform any Hamiltonian in the F=3 basis as follows:

    [tex]
    H^\prime=D^\dagger H D
    [/tex]

    The calculation of that matrix is another story, I got that from Mathematica. So my question is: assume I know that matrix for F=3 and F=4, and I know the Hamiltonians for both F=3 and F=4. How can I calculate the combined Hamiltonian after applying an arbitrary (but equal for each) transformations?

    In other words: Say I use the matrix to rotate the F=3 Hamiltonian and F=4 Hamiltonian by some angles (a,b,c). Now I have 2 new Hamiltonians for each case. How can I combine them both?
     
    Last edited: Oct 30, 2013
  11. Oct 31, 2013 #10
    The hamiltonian matrix is
    [itex] \hat{H} = \hat{H}_1 \otimes I_2 + I_1 \otimes \hat{H}_2 + \hat{H}_{int}[/itex]. The rotation matrix in the new space should be defined as
    [itex] D = D^{F=3} \otimes I_2 + I_1 \otimes D^{F=4} [/itex]. And now you can use your formula
    [tex]
    \hat{H}^\prime=D^\dagger \hat{H} D.
    [/tex]
     
    Last edited: Oct 31, 2013
  12. Oct 31, 2013 #11
    If I do this product the result's side length will become 7*9=63 elements... while my combined system from F=3 and F=4 has only 7+9=16 elements, since F=3 has side-length 2*3+1=7 and F=4 has 2*4+1=9.

    So here's the problem right now.
     
  13. Oct 31, 2013 #12

    kith

    User Avatar
    Science Advisor

    This seems to be about the difference between the direct product and the direct sum. The dimension of the product is n*m while the dimension of the sum is n+m. You get the direct sum if you write the initial matrices on the diagonal of the resulting matrix and fill the off-diagonals with zeros.

    If you combine two systems, you need the direct product. But what you want to do is to combine different possible states of a single system, don't you? There, the direct sum seems to be the right thing to me.

    I've never used Wigner's matrix but I remember that I found Sakurai really insightful on it at the time. Maybe it helps you.

    /edit: typo
     
    Last edited: Oct 31, 2013
  14. Oct 31, 2013 #13
    Your combined system also has 69 elements. Consider the first term of Hamiltonian. H (7*7 matrix) of system 1 tensored with Identity of system 2 (9*9 matrix) giving you a total of 63*63 matrix. Similarly the second term is 63*63 matrix.
     
  15. Oct 31, 2013 #14
    Thank you for your answers, guys.

    Like Kith said, I need to add components and not have the product. A 63*63 matrix is redundant, where it gives me, for example, a state [itex]| F=3,F=4,m_F=3,m_F=-4 \rangle[/itex], which is definitely wrong. What I'm doing is adding components... please consider re-assessing the situation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook