Combine Hamiltonians of two different bases

[TheDestroyer]

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 | \psi \rangle to be H_\psi, and another system with bases | \phi \rangle with Hamiltonian H_\phi.

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

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 |F_1 m_{F_1} \rangle and |F_2 m_{F_2}\rangle). If the Wigner-D rotation matrix for the first Hamiltonian/basis is \mathcal{D}^{F_1}_{m_{F_1} m_{F_1}^\prime}, and for the second system is \mathcal{D}^{F_2}_{m_{F_2} m_{F_2}^\prime}. How can I rotate each system and then combine them to introduce interactions between them?

Thank you for any efforts.

 

[Ravi Mohan]

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

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

You suppose correct. For this situation we create a Hilbert Space by the tensor product of basis of the two spaces containing Hamiltonians \hat{H}_\psi and \hat{H}_\phi(which are spanned by | \psi \rangle and | \phi \rangle respectively).

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

So the new Hilbert Space is now spanned by basis | \psi_i \phi_j\rangle or | \psi_i\rangle \otimes |\phi_j\rangle (sum over i and j). The Hamiltonian of the system now becomes \hat{H}_\psi \otimes I + I \otimes \hat{H}_\phi + \hat{H}_{int}. If you are not aware of this algebra, I would recommend Quantum Computation and Quantum Information by Neilsen and Chaung, Chapter 2.

 

[TheDestroyer]

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 H_1 and H_2

H_1=\left(<br /> \begin{array}{cccc}<br /> \text{s1} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \text{s2} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \text{s3} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \text{s4} \\<br /> \end{array}<br /> \right)

H_2=\left(<br /> \begin{array}{ccc}<br /> \text{t1} &amp; 0 &amp; 0 \\<br /> 0 &amp; \text{t2} &amp; 0 \\<br /> 0 &amp; 0 &amp; \text{t3} \\<br /> \end{array}<br /> \right)

So now we have:

H_1 \otimes I_3 = \left(<br /> \begin{array}{cccccccccccc}<br /> \text{s1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \text{s1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \text{s1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \text{s2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4} \\<br /> \end{array}<br /> \right)

and

H_2 \otimes I_3 = \left(<br /> \begin{array}{cccccccccccc}<br /> \text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \text{t3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t3} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t1} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t2} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t3} \\<br /> \end{array}<br /> \right)

So now the new Hamiltonian is

H_1 \otimes I_3 + I_4 \otimes H_2 = \left(<br /> \begin{array}{cccccccccccc}<br /> \text{s1}+\text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \text{s1}+\text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \text{s1}+\text{t3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \text{s2}+\text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s2}+\text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s2}+\text{t3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3}+\text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3}+\text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3}+\text{t3} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4}+\text{t1} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4}+\text{t2} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4}+\text{t3} \\<br /> \end{array}<br /> \right)

Is that how it's done?

 

[Ravi Mohan]

You have done correctly with some typos.

H_1=\left(<br /> \begin{array}{cccc}<br /> \text{s1} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \text{s2} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \text{s3} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \text{s4} \\<br /> \end{array}<br /> \right),

and define identity operator on first space

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

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

H_1 \otimes I_2 = \left(<br /> \begin{array}{cccccccccccc}<br /> \text{s1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \text{s1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \text{s1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \text{s2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s3} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{s4} \\<br /> \end{array}<br /> \right),
as you have done correctly.
Second term is
I_1 \otimes H_2 = \left(<br /> \begin{array}{cccccccccccc}<br /> \text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \text{t3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t1} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t2} &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t3} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t1} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t2} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \text{t3} \\<br /> \end{array}<br /> \right).
(you did correctly but it seems lot of typesetting made you overlook the order I_1 \otimes H_2 .
And you can add these matrices to get total Hamiltonian matrix (without interaction).

 

[TheDestroyer]

Hmmmmmmm... Thank you so much. Though I thought this approach would solve my problem, but apparently it doesn't.

I have Hyperfine states with F=3,4 (where F=I+J 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?

 

[Ravi Mohan]

I have Hyperfine states with F=3,4 (where F=I+J 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?

I am sorry, I don't know what Wigner-D matrices are.

 

[TheDestroyer]

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 F=3 and m_F=-3,...,3, the Wigner-D matrix will be D_{m_F m_F^\prime}^{F=3}, where m_F, m_F^\prime will be the matrix elements, and we'll have 7\times 7=49 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?

 

[Ravi Mohan]

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?

Please give one demonstration. Say I need to rotate basis F=3. How will I proceed?

 

[TheDestroyer]

Thank you for your time. Wigner D-Matrix for that case will be the following, starting from the eigen-vector |F=3;m_F=-3 \rangle to |F=3;m_F=3 \rangle

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

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:

<br /> H^\prime=D^\dagger H D<br />

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?

 

[Ravi Mohan]

The hamiltonian matrix is
\hat{H} = \hat{H}_1 \otimes I_2 + I_1 \otimes \hat{H}_2 + \hat{H}_{int}. The rotation matrix in the new space should be defined as
D = D^{F=3} \otimes I_2 + I_1 \otimes D^{F=4}. And now you can use your formula
<br /> \hat{H}^\prime=D^\dagger \hat{H} D.<br />

 

[TheDestroyer]

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.

 

[kith]

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.

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

 

[Ravi Mohan]

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.

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.

 

[TheDestroyer]

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 | F=3,F=4,m_F=3,m_F=-4 \rangle, which is definitely wrong. What I'm doing is adding components... please consider re-assessing the situation.