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:


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 $\hat{H}=\hat{H}_r+\hat{H}_s$.

  • $\begingroup$ 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? $\endgroup$ – 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


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