Degenerate Perturbation Theory: Two Spin 1/2 Particles

[Xyius]

So I know this might be a lot to read but I am having a very hard time understanding how to use the formulas in degenerate perturbation theory. Here is the problem I am on.

Homework Statement

A system of two spin-1/2 particles is described by the following Hamiltonian.

$$\hat{H}=\hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2 + B \hat{S}_1^z$$

For simplicity, let $\hbar =1$ and the Hamiltonian be dimensionless.

(i) Compute the exact energy eigenvalues and eigenvectors.
(ii) Treating the term proportional to B as a perturbation, compute the corrections to the unperturbed energies through second order in B and to the eignkets through first order in B.
(iii) How do your results in (ii) compare to the Taylor expansion of the exact results for the energy eigenvalues and eigenvectors in (i)?

Homework Equations

In the following equations, superscripts represent orders of approximation (first order, second order, ect.). $|n^0,l_n>$ represents a ket in the degenerate subspace. S_n is the degenerate subspace. $|k^0>$ represents a ket that is NOT in the degenerate subspace.

(1): First order Energy Correction
$$\Delta^1_{n,l_n}=<n^0,l_n|\hat{V}|n^0,l_n>$$

(2): Second order Energy Correction
$$\Delta^2_{n,l_n}= \sum_{k \notin S_n}\frac{|<k^0|\hat{V}|n^0,l_n>|^2}{E_n^0-E_k^0}$$

(3): First order approximation for the eigenkets in the degenerate subspace
$$|n^1,l_n>=\sum_{l_n \ne l_n'}\frac{|n^0,l_n>}{\Delta_{n,l_n}^1-\Delta_{n,l_n'}^1}\sum_{k \notin S_n}<n^0,l_n|\hat{V}|k^0>\frac{1}{E_n^0-E_k^0}<k^0|\hat{V}|n^0,l_n>$$

The Attempt at a Solution

(i) So for this one, I first re-wrote the Hamiltonian by computing the dot product and writing the $S_x$ and $S_y$ terms in terms of the ladder operators. I then wrote all the possible kets I can have in a system with two spin 1/2 particles, then using clebch-gordon coefficients, re-wrote them in the $|s_1,s_2,m_1,m_2>$ basis. I then computed the entire matrix and calculated the eigenvalues. There was a LOT of room for error here but I got..

$$E_1=-\frac{1}{4}$$
$$E_2=\frac{1}{4}-\frac{1}{2}B$$
$$E_{3,4}=\frac{1}{3}B \pm \sqrt{\frac{1}{9}B^2+\frac{5}{6}}$$

First question: How is it possible I got a negative energy? I then plugged these into the matrix and found the eigevectors.

(ii) This is the part I am really stuck on. First I computed the matrix form of the perturbation V. Which was NOT diagonal, but only had TWO terms survive. I then diagonalized it and found the eigenvalues to be..

$$\text{Eigenvalues}=0,-\frac{B}{2}$$

Meaning that, these are the first order corrections to the energies via equation (1) above. This makes sense considering that in the first two energies I listen above for the total Hamiltonian, the term $-\frac{1}{4}$ is not being added to anything (eigenvalue of 0 in the perturbation) and the term $\frac{1}{4}-\frac{B}{2}$, has the original energy of 1/4, plus the perturbation of B/2. I then wanted to find the eigenenergies of the unperturbed Hamiltonian to confirm this, so I did and indeed the eigenenergies are $\pm \frac{1}{4}$ and $\pm \frac{1}{2}\sqrt{\frac{5}{2}}$.

So here is my second question: Once again I am getting negative energies, why? Also, is this system degenerate? I have four distinct eigenenergies in the unperturbed Hamiltonian, doesn't that mean it is non-degenerate and I can use non-degenerate perturbation theory?

So now my issue comes with calculating the second order energy level shift and the first order eigenkets. For both equations (2) and (3) above, I don't know how to evaluate the following term in the summation.

$$\frac{1}{E_n^0-E_k^0}$$

Both of the energy terms in the denominator represent unperturbed energies. But I am just having a really hard time figuring out how to evaluate the summations. I don't know where to go here and I keep trying to look up examples in second order but everything I find only goes to FIRST order.

Can anyone help??

 

[maajdl]

For question (i), I didn't get the sames eigenvalues as you.
I got these:

(-1 - 2*Sqrt(1 + B²))/4
(+1 - 2*Sqrt(1 + B²))/4
(-1 + 2*Sqrt(1 + B²))/4
(+1 + 2*Sqrt(1 + B²))/4

I don't see why negative values would be impossible.
The unperturbed Hamiltonian represents the the scalar product of the spins: it can be positive or negative equally well. The same for the perturbation. You can see the perfect symmetry between positive and negative eigenvalues.

The unperturbed eigenvalues, which you need for the perturbation calculations, are:

-3/4
-1/4
-1/4
+3/4

This unperturbed spectrum shows no degeneracy at all.

Am I wrong?

 

[vela]

So I know this might be a lot to read but I am having a very hard time understanding how to use the formulas in degenerate perturbation theory. Here is the problem I am on.

Homework Statement

A system of two spin-1/2 particles is described by the following Hamiltonian.

$$\hat{H}=\hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2 + B \hat{S}_1^z$$

For simplicity, let $\hbar =1$ and the Hamiltonian be dimensionless.

(i) Compute the exact energy eigenvalues and eigenvectors.
(ii) Treating the term proportional to B as a perturbation, compute the corrections to the unperturbed energies through second order in B and to the eignkets through first order in B.
(iii) How do your results in (ii) compare to the Taylor expansion of the exact results for the energy eigenvalues and eigenvectors in (i)?

Homework Equations

In the following equations, superscripts represent orders of approximation (first order, second order, ect.). $|n^0,l_n>$ represents a ket in the degenerate subspace. S_n is the degenerate subspace. $|k^0>$ represents a ket that is NOT in the degenerate subspace.

(1): First order Energy Correction
$$\Delta^1_{n,l_n}=<n^0,l_n|\hat{V}|n^0,l_n>$$

(2): Second order Energy Correction
$$\Delta^2_{n,l_n}= \sum_{k \notin S_n}\frac{|<k^0|\hat{V}|n^0,l_n>|^2}{E_n^0-E_k^0}$$

(3): First order approximation for the eigenkets in the degenerate subspace
$$|n^1,l_n>=\sum_{l_n \ne l_n'}\frac{|n^0,l_n>}{\Delta_{n,l_n}^1-\Delta_{n,l_n'}^1}\sum_{k \notin S_n}<n^0,l_n|\hat{V}|k^0>\frac{1}{E_n^0-E_k^0}<k^0|\hat{V}|n^0,l_n>$$

The Attempt at a Solution

(i) So for this one, I first re-wrote the Hamiltonian by computing the dot product and writing the $S_x$ and $S_y$ terms in terms of the ladder operators. I then wrote all the possible kets I can have in a system with two spin 1/2 particles, then using clebch-gordon coefficients, re-wrote them in the $|s_1,s_2,m_1,m_2>$ basis. I then computed the entire matrix and calculated the eigenvalues. There was a LOT of room for error here but I got..

$$E_1=-\frac{1}{4}$$
$$E_2=\frac{1}{4}-\frac{1}{2}B$$
$$E_{3,4}=\frac{1}{3}B \pm \sqrt{\frac{1}{9}B^2+\frac{5}{6}}$$

First question: How is it possible I got a negative energy?

If the two spins are anti-parallel, it's not unreasonable to expect their dot product to be negative.

I got different energies than you did. You should try this part again except use the $\lvert s, m_s \rangle$ basis, where $\vec{S} = \vec{S}_1 + \vec{S}_2$. The trick is to write
$$\vec{S}_1\cdot\vec{S}_2 = \frac{\vec{S}^2 - \vec{S}_1^2 - \vec{S}_2^2}{2}.$$ In this basis, the unperturbed Hamiltonian will be diagonal, which is what you want before you start in on the perturbation theory for part (b).

I then plugged these into the matrix and found the eigevectors.

(ii) This is the part I am really stuck on. First I computed the matrix form of the perturbation V. Which was NOT diagonal, but only had TWO terms survive. I then diagonalized it and found the eigenvalues to be..

$$\text{Eigenvalues}=0,-\frac{B}{2}$$

Meaning that, these are the first order corrections to the energies via equation (1) above. This makes sense considering that in the first two energies I listen above for the total Hamiltonian, the term $-\frac{1}{4}$ is not being added to anything (eigenvalue of 0 in the perturbation) and the term $\frac{1}{4}-\frac{B}{2}$, has the original energy of 1/4, plus the perturbation of B/2. I then wanted to find the eigenenergies of the unperturbed Hamiltonian to confirm this, so I did and indeed the eigenenergies are $\pm \frac{1}{4}$ and $\pm \frac{1}{2}\sqrt{\frac{5}{2}}$.

So here is my second question: Once again I am getting negative energies, why? Also, is this system degenerate? I have four distinct eigenenergies in the unperturbed Hamiltonian, doesn't that mean it is non-degenerate and I can use non-degenerate perturbation theory?

So now my issue comes with calculating the second order energy level shift and the first order eigenkets. For both equations (2) and (3) above, I don't know how to evaluate the following term in the summation.

$$\frac{1}{E_n^0-E_k^0}$$

Both of the energy terms in the denominator represent unperturbed energies. But I am just having a really hard time figuring out how to evaluate the summations. I don't know where to go here and I keep trying to look up examples in second order but everything I find only goes to FIRST order.

Can anyone help??

 

[Xyius]

Thank you everyone for your comments so far and taking the time to help me out.

If the two spins are anti-parallel, it's not unreasonable to expect their dot product to be negative.

I got different energies than you did. You should try this part again except use the $\lvert s, m_s \rangle$ basis, where $\vec{S} = \vec{S}_1 + \vec{S}_2$. The trick is to write
$$\vec{S}_1\cdot\vec{S}_2 = \frac{\vec{S}^2 - \vec{S}_1^2 - \vec{S}_2^2}{2}.$$ In this basis, the unperturbed Hamiltonian will be diagonal, which is what you want before you start in on the perturbation theory for part (b).

But how would I apply $|s,m_s>$ ket to, say, $\vec{S}_1$? Say for example I had the vector $|0,0>$ in this basis. $s_1$ can either be +1/2 or -1/2 since the sum of the two spins must equal zero. Wouldn't I need to convert this basis to the $|m_1,m_2>$ basis?

 

[vela]

Yes, you want to express $\lvert 0, 0 \rangle$ in terms of the $\lvert m_1, m_2 \rangle$ basis, apply $S_1$, and then express the result in terms of the $\lvert s, m_s \rangle$ basis. It's very straightforward to do. Do you know what the $\lvert s, m_s \rangle$ kets are in terms of the $\lvert m_1, m_2 \rangle$ ket?

 

[Xyius]

Yes, I got the kets to be..

$$|0,0>=\frac{1}{\sqrt(2)}\left[ |-1/2,1/2>-|1/2,-1/2> \right]$$
$$|1,-1>=|-1/2,-1/2>$$
$$|1,0>=\frac{1}{\sqrt{2}}\left[ |-1/2,1/2>+|1/2,-1/2> \right]$$
$$|1,1>=|1/2,1/2>$$

So say for example, I want to find <1,1|H|1,1>
The $S^2$ term would be 1(1+1)=2
The $S_1^2$ term would be (1/2)(1/2+1)=3/4
The $S_2^2$ term would be the same, 3/4
But this would mean that $S_1 \cdot S_2$ would be 2+3/4+3/4=7/2??

(EDIT: Oops I forgot to subtract, it would be 2-3/4-3/4=1, but then divide by 2 to get 1/2. And then there is a factor of 1/2 out in front from the normalization constants making the total result 1/4.)

The other two terms are more straight forward for me since $S_1^z$ on |1,1> would simply be 1/2 (setting h-bar =1).

Am I on the right track?

 

[vela]

Yes, I got the kets to be..

$$|0,0>=\frac{1}{\sqrt(2)}\left[ |-1/2,1/2>-|1/2,-1/2> \right]$$
$$|1,-1>=|-1/2,-1/2>$$
$$|1,0>=\frac{1}{\sqrt{2}}\left[ |-1/2,1/2>+|1/2,-1/2> \right]$$
$$|1,1>=|1/2,1/2>$$

So say for example, I want to find <1,1|H|1,1>
The $S^2$ term would be 1(1+1)=2
The $S_1^2$ term would be (1/2)(1/2+1)=3/4
The $S_2^2$ term would be the same, 3/4
But this would mean that $S_1 \cdot S_2$ would be 2+3/4+3/4=7/2??

(EDIT: Oops I forgot to subtract, it would be 2-3/4-3/4=1, but then divide by 2 to get 1/2. And then there is a factor of 1/2 out in front from the normalization constants making the total result 1/4.)

You should check your arithmetic, e.g., 2 - 3/4 - 3/4 = 1/2, not 1.

The other two terms are more straight forward for me since $S_1^z$ on |1,1> would simply be 1/2 (setting h-bar =1).

You mean $S_{1z} \lvert 1, 1 \rangle = \frac{1}{2} \lvert 1, 1 \rangle$.

Am I on the right track?

Looks like you have a handle on it. What do you get when you apply $S_{1z}$ to $\lvert 1, 0 \rangle$?

 

[maajdl]

Yes, you want to express $\lvert 0, 0 \rangle$ in terms of the $\lvert m_1, m_2 \rangle$ basis, apply $S_1$, and then express the result in terms of the $\lvert s, m_s \rangle$ basis. It's very straightforward to do. Do you know what the $\lvert s, m_s \rangle$ kets are in terms of the $\lvert m_1, m_2 \rangle$ ket?

Vela,

I am missing something.
Why not using the |m1,m2> basis?
This is how I did it, and this was extremely simple.
Is the use of the total spin basis implicitely required somehow by this exercise?

Thanks to clarify,

maajdl

 

[vela]

You didn't do it correctly because the energies you mentioned above aren't correct. $\vec{S}_1\cdot \vec{S}_2$ doesn't commute with $S_{1z}$ or $S_{2z}$, so it's not going to be diagonal in the $\lvert m_1, m_2 \rangle$ basis. To use perturbation theory, however, you want the basis that diagonalizes $\vec{S}_1\cdot \vec{S}_2$.

 

[Xyius]

What do you get when you apply $S_{1z}$ to $\lvert 1, 0 \rangle$?

This would be 0 correct? Since one ket has +1/2 and the other has -1/2.

 

[vela]

No. That's why it's important to keep the kets around. Applying an operator to a ket results in another ket, not a number. For example, $S_{1z} \lvert + + \rangle = 1/2 \lvert + + \rangle$. It's not $S_{1z} \lvert + + \rangle = 1/2$.

 

[Xyius]

Sorry I was having expectation values stuck in my head and I guess I got confused.

Just for clarity couldn't I also just re-write $S_1\cdot S_2$ in terms of ladder operators and compute it that way, like I originally did? This just seems like another way

 

[vela]

If it's mathematically valid, you can do anything you want. Some ways will be more efficient than others. If you use the $\lvert s, m_s \rangle$ basis, the first term of the Hamiltonian is already diagonalized. Finding the matrix for the second term is like four lines of math. Diagonalizing the full Hamiltonian reduces to solving a 2x2 system. Or you can do it the way you tried, which is more complicated and presents the opportunity for more errors.

 

[Xyius]

Thank you so much Vela for your help so far.

I just calculated the diagonal components of the unperturbed Hamiltonian using the basis you suggest and I got.. (starting from the top left corner)

0,5/4,3/4,1,4

Is this correct? So the next step would be to find V in this basis, and then add them both together to get the total Hamiltonian matrix? Then find the eigenvalues and eigenvectors.

EDIT: One thing I am confused about. You say it is easy to find the perturbation matrix using this basis, but is V diagonal in this basis? If it is not diagonal, don't I need to find each term?

 

[Xyius]

Oh wait. Is the second term also diagonal? Because I notice that it commutes with both S and S_z if I did my math right.

 

[vela]

I'm not sure what you're doing. Say your first basis vector is $\lvert 1,1 \rangle$. Apply the unperturbed Hamiltonian to it. You get, using some of your results from above,
$$\vec{S}_1\cdot \vec{S}_2 \lvert 1,1 \rangle = \frac{\vec{S}^2 - \vec{S}_1^2 - \vec{S}_2^2}{2} \lvert 1,1 \rangle = \frac{2-\frac{3}{4}-\frac{3}{4}}{2} \lvert 1,1 \rangle = \frac{1}{4}\lvert 1,1 \rangle.$$ So the first column of the matrix would be
$$\begin{pmatrix}1/4\\0\\0\\0\end{pmatrix}.$$ What do you get for the other columns?

 

[Xyius]

I'm not sure what you're doing. Say your first basis vector is $\lvert 1,1 \rangle$. Apply the unperturbed Hamiltonian to it. You get, using some of your results from above,
$$\vec{S}_1\cdot \vec{S}_2 \lvert 1,1 \rangle = \frac{\vec{S}^2 - \vec{S}_1^2 - \vec{S}_2^2}{2} \lvert 1,1 \rangle = \frac{2-\frac{3}{4}-\frac{3}{4}}{2} \lvert 1,1 \rangle = \frac{1}{4}\lvert 1,1 \rangle.$$ So the first column of the matrix would be
$$\begin{pmatrix}1/4\\0\\0\\0\end{pmatrix}.$$ What do you get for the other columns?

Yes I did get that. Except I am writing my hamiltonian with the m=0,l=0 state in the top left and the m=1,l=1 in the bottom right, so I got the same thing, except the 1/4 is on the bottom. All the columns I got were..

$$\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1/4 \end{pmatrix}$$
$$\begin{pmatrix} 0 \\ 5/4 \\ 0 \\ 0 \end{pmatrix}$$

$$\begin{pmatrix} 0 \\ 0 \\ 3/4 \\ 0 \end{pmatrix}$$
$$\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

I then found all the entries to the second expression in the hamiltonian and found its matrix to be.

$$\begin{pmatrix} 0 & 0 & -\frac{B}{2} & 0 \\ 0 & -\frac{B}{2} & 0 & 0 \\ -\frac{B}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{B}{2} \end{pmatrix}$$

I then combined both the matrices of the first and second term in the Hamiltonian, and found the eigenvalues to be..

$$E_1=\frac{1}{4}\left( 5-2B \right)$$
$$E_2=\frac{1}{4}\left( 1+2B \right)$$
$$E_{3,4}=\frac{3}{8}\left( 1 \pm \sqrt{1-\left( \frac{4}{3}B \right)^2} \right)$$

 

[vela]

I don't see where you're getting the values like 5/4 in the unperturbed Hamiltonian.

 

[Xyius]

Hmm. Perhaps I just made another arithmetic mistake.

5/4 is from the term..

$$<1,-1|H_1|1,-1>$$

Where $H_1$ represents the first term in the Hamiltonian.
This becomes..

$$\frac{1}{2}<-\frac{1}{2},-\frac{1}{2}|S^2-S_1^z-S_2^z|-\frac{1}{2},-\frac{1}{2}>$$

The term involving $S^2$ is 1(1+1)=2

The term involving $S_1^z$ is, (-1/2)(-1/2+1)=-1/4
The term involving $S_2^z$ is, (-1/2)(-1/2+1)=-1/4

So all together I have. (1/2)[2+1/4+1/4]=5/4

Where did I go wrong?

 

[vela]

Take $\vec{S} = \vec{S}_1 + \vec{S}_2$, square both sides, and then solve for $\vec{S}_1 \cdot \vec{S}_2$.

 

[Xyius]

Oh oops in doing this I realized that in my previous post I wrote $S_1^2$ as $S_1^z$, but I still evaluated them as $S_1^2$.

EDIT:

So it should have been
$$\frac{1}{2}(S^2-S_1^2-S_2^2)$$

So I am still confused as to why I got 5/4 if it is incorrect :\

 

[vela]

The state $\lvert 1,-1\rangle$ corresponds to $s=1$, $m_s = -1$, $s_1 = 1/2$, and $s_2 = 1/2$.

 

[Xyius]

Oh! I accidentally used the m values in the expressions for S1 and S2 instead of using the s values!

So doing this again I would get.

For the S term: 1(1+1)=2
For the S1 term AND the S2 term: (1/2)(1/2+1)=3/4
So I get in total.

(1/2)(2-3/4-3/4)=1/4

 

[maajdl]

You didn't do it correctly because the energies you mentioned above aren't correct.

Yes, I was too fast in expanding the Cartesian product of the Pauli matrices.
Being more careful, here is the Hamiltonian I found in the {|++>,|+->,|-+>,|-->} basis:

\begin{pmatrix}
\frac{B}{2}+\frac{1}{4} & 0 & 0 & 0 \\
0 & \frac{B}{2}-\frac{1}{4} & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & -\frac{B}{2}-\frac{1}{4} & 0 \\
0 & 0 & 0 & \frac{1}{4}-\frac{B}{2} \\
\end{pmatrix}

And those are the eigenvalues of this matrix:

1/4 - B/2
1/4 + B/2
-1/4 - Sqrt[1 + B^2]/2
-1/4 +Sqrt[1 + B^2]/2

There is indeed a degeneracy of the unperturbed Hamiltonian!
Quite logical, since 1 and 2 can be exchanged before perturbation.

 

[Xyius]

SO i think I got the answer! I re-did the problem with this new insight and got every diagonal term in the first term of the Hamiltonian to be 1/4. I then found the matrix representation of the second term (which was NOT diagonal) and added them together. Then found the eigenvalues to be..

$$E_n=\frac{1}{4}+\frac{B}{2},\frac{1}{4}-\frac{B}{2},\frac{1}{4}+\frac{B}{4}(\sqrt{5}-1),\frac{1}{4}-\frac{B}{4}(\sqrt{5}+1)$$

I then found the first order corrections to the energies of the perturbation and got..

$$\Delta^1= \frac{B}{2},-\frac{B}{2},\frac{B}{4}(\sqrt{5}-1), -\frac{B}{4}(\sqrt{5}+1)$$

Which means that after only finding the first order energies, I get the exact solution!

But one question, does this mean I do not have to find the second order? Or should I show that the second order is zero?

 

[vela]

Recheck the unperturbed Hamiltonian. One of the diagonal elements should be -3/4.

 

[Xyius]

Hmm I check it a few times and I keep getting 5/4. Namely the terms involving the vectors |1,0> and |0,0>.

The $S^2$ term gives me 2, the $-S_1^2$ term gives me -3/8 and the $-S_2^2$ term gives me another -3/8. All together I get..

2-3/8-3/8=5/4

 

[vela]

I have to say I am baffled as to how you're finding calculating four numbers (really just two) so difficult to do correctly. In your previous post, you said you got all 1/4. Now you're saying you're getting a 5/4.

For the $\lvert 1, m \rangle$ states, what are $s$, $s_1$, and $s_2$ equal to?

For the $\lvert 0, 0 \rangle$ state, what are $s$, $s_1$, and $s_2$ equal to?

 

[Xyius]

I have to say I am baffled as to how you're finding calculating four numbers (really just two) so difficult to do correctly. In your previous post, you said you got all 1/4. Now you're saying you're getting a 5/4.

For the $\lvert 1, m \rangle$ states, what are $s$, $s_1$, and $s_2$ equal to?

For the $\lvert 0, 0 \rangle$ state, what are $s$, $s_1$, and $s_2$ equal to?

Sorry, for some reason I keep using the $m_s$ values instead of the $s_1$ or $s_2$ values and it is completely screwing up my calculations. Doing things again, I do indeed get -3/4 for the |0,0> expectation value. I am going to go through finding the eigenvalues and eigenvectors again and hopefully finish this problem.