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 $H_2$

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

$$H_2=\left( \begin{array}{ccc} \text{t1} & 0 & 0 \\ 0 & \text{t2} & 0 \\ 0 & 0 & \text{t3} \\ \end{array} \right)$$

So now we have:

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

and

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

So now the new Hamiltonian is

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

Is that how it's done?

 

[Ravi Mohan]

You have done correctly with some typos.

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

and define identity operator on first space

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

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

$$H_1 \otimes I_2 = \left( \begin{array}{cccccccccccc} \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \text{s1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \text{s2} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s3} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{s4} \\ \end{array} \right),$$
as you have done correctly.
Second term is
$$I_1 \otimes H_2 = \left( \begin{array}{cccccccccccc} \text{t1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \text{t2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \text{t3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \text{t1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \text{t2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \text{t3} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \text{t1} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t2} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t3} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t1} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t2} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text{t3} \\ \end{array} \right).$$
(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 $|F=3;m_F=3 \rangle$

$$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)$$

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:

$$H^\prime=D^\dagger H D$$

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
$$\hat{H}^\prime=D^\dagger \hat{H} D.$$

 

[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.