Finding energy Eigenvalue from two spin Hamiltonian

Let's consider a system of two spins, named spin 1 and spin 2.

Let's also consider, in a Hamiltonian, spin part has been defined as $$\sigma_1 \cdot \sigma_2$$.

For example:

$$H= E_0 + \sigma_1 \cdot \sigma_2$$

What I want to do is to apply the operator and find the energy Eigenvalue from it.

For example I see one has written the Eigenvalue for spin operator in this way.

$$\sigma_1 \cdot \sigma_2 |\uparrow \uparrow\rangle = 2 |\downarrow \uparrow\rangle - | \uparrow\downarrow \rangle \ \ \ \ \ \ \ \ \ (1)$$

where the first entry denotes the state of spin 1 and the second entry the state of spin 2.

Do you think I would have to decompose the spin operator to get that or is there a very simple way to do that?

What I'm saying is like:

$$\sigma_1 \cdot \sigma_2= \sigma_{1x} \sigma_{2x} + \sigma_{1y} \sigma_{2y} + \sigma_{1z} \sigma_{2z}$$

How did they write the right side of equation (1)?

Following the strategy I will work on other spin states as well.

Yes you can expand but of course the simplest way is to observe that $$\vec S^2:=\left(\vec \sigma_1+\vec\sigma_2\right)^2= \sigma_1^2+\sigma_2^2+2\vec\sigma_1\cdot\vec\sigma_2\, .$$ so that $$\vec\sigma_1\cdot\vec\sigma_2=\frac{1}{2} \left( \vec S^2-\sigma_1^2-\sigma_2^2\right)$$ and everything on the right is diagonal when acting on $$\vert \uparrow\uparrow\rangle$$ since this state is an eigenstate of the total spin $$\vec S$$ with $$S=1$$.

What I want to do is to apply the operator and find the energy Eigen value from it.

In your problem, the Hamiltonian is of the form $$\hat{H}=\hat{H}_r+\hat{H}_s$$, where the subscripts stand for spatial ($$r$$) and spin ($$s$$). In this case, the wavefunction describing the system can be written as $$\psi=\psi_r\chi_s$$, i.e. it separates into a spatial and a spin part.

We can apply separation of variables to the (time-dependent) Schroedinger equation to generate two decoupled ordinary differential equations. From these, we obtain the following time-independent eigenvalue equations:

$$\hat{H}_r\psi_r=E_r\psi_r\quad\text{and}\quad\hat{H}_s\chi_s=E_s\chi_s$$

To answer your question, then, the energy eigenvalue is given by $$E=E_r+E_s$$ as you might have expected. Note that this is only valid in the case of a Hamiltonian of the form $$\hat{H}=\hat{H}_r+\hat{H}_s$$.

• I think you misunderstood the question. What I wanted to do is to apply to spin operator $\sigma_1 \cdot \sigma_2$ on the spin state and want to know how I got the right side of the equation? Where did the factor 2 came from? – user193422 Aug 20 at 10:26

(Sorry, I'm not good at speaking english)

You are right. You have to use the define $$\sigma_1 \cdot \sigma_2 = \sigma_{1,x} \, \sigma_{2,x} + \sigma_{1,y} \, \sigma_{2,y} + \sigma_{1,z} \, \sigma_{2,z}$$ But you have the wrong answer, let's see.

Maybe it's hard using that components and easier using ladder operators, define as

$$\sigma_{1+}=\sigma_{1x}+i \, \sigma_{1y} \quad \quad \sigma_{1-}=\sigma_{1x}-i \, \sigma_{1y} \quad \quad \sigma_{2+}=\sigma_{2x}+i \, \sigma_{2y} \quad \quad \sigma_{2-}=\sigma_{2x}-i \, \sigma_{2y} \quad \quad j=1,2$$ or $$\sigma_{j,x}= \frac{1}{2} \, \left[ \sigma_{(j,+)}+\sigma_{(j,-)} \right] \quad \quad \sigma_{j,y}= \frac{1}{2 i} \, \left[ \sigma_{(j,+)}-\sigma_{(j,-)} \right]$$ whit this relations $$\sigma_{j,+} \left|\uparrow \, \right \rangle_j= 0 \quad \quad \sigma_{j,+} \left|\downarrow \, \right \rangle_j= \left|\uparrow \, \right \rangle_j \quad \quad \sigma_{j,-} \left|\uparrow \, \right \rangle_j= \left|\downarrow \, \right \rangle_j \quad \quad \sigma_{j,-} \left|\downarrow \, \right \rangle_j= 0$$ Then you know that $$\sigma_{1,x} \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 =\frac{1}{2} \, \left[ \sigma_{(1,+)}+\sigma_{(1,-)} \right] \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 \quad \Longrightarrow \quad \sigma_{1,x} \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 =\frac{1}{2} \, \left|\downarrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2$$ $$\sigma_{2,x} \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 =\frac{1}{2} \, \left[ \sigma_{(2,+)}+\sigma_{(2,-)} \right] \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 \quad \Longrightarrow \quad \sigma_{2,x} \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 =\frac{1}{2} \, \left|\uparrow \, \right \rangle_1 \, \left|\downarrow \, \right \rangle_2$$ $$\sigma_{1,y} \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 =\frac{1}{2 i} \, \left[ \sigma_{(1,+)}-\sigma_{(1,-)} \right] \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 \quad \Longrightarrow \quad \sigma_{1,y} \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 =-\frac{1}{2 i} \, \left|\downarrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2$$ $$\sigma_{2,y} \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 =\frac{1}{2 i} \, \left[ \sigma_{(2,+)}-\sigma_{(2,-)} \right] \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 \quad \Longrightarrow \quad \sigma_{2,y} \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 =-\frac{1}{2 i} \, \left|\uparrow \, \right \rangle_1 \, \left|\downarrow \, \right \rangle_2$$

$$\sigma_{1,z} =\left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 = \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 \quad \quad \sigma_{2,z} =\left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2 = \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2$$

So finally have $$\sigma_1 \cdot \sigma_2 \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2= \left|\uparrow \, \right \rangle_1 \, \left|\uparrow \, \right \rangle_2$$ Of course for anothers eingvectors it will be diferent