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\heading{E148: A system of two qubits, perturbation theory} 
\auname{Konstantin Yavilberg and Ronen Hershman}


{\bf The problem:}

\Dn Qubit is a device with two basis states ${|x=0\rangle , |x=1\rangle}$. Assume that both states have the same energy and that the jumping amplitude between them is ${c}$. We are given a system of two qubits with basis states ${|x_a,x_b\rangle}$. It turns out that when the two qubits are in the state ${x=1}$, there is an energetic payment ${u}$. The hamiltonian of the system can be written as ${\mathcal{H}=\mathcal{H}_0+uV}$.


\Dn



(1) Write the matrix of ${V}$ in the standard basis (no algebraic work is needed).



The hamiltonian ${\mathcal{H}_0}$ is diagonal in a certain basis. Indicate the basis states as ${|SS\rangle,|SA\rangle,|AS\rangle,|AA\rangle}$.

(2) Write column vectors which represent the new basis states in the standard basis.

(3) Write the diagonal matrix which represents ${\mathcal{H}_0}$ in the basis you defined (no algebraic work is needed).

(4) Write in the new basis the perturbation matrix ${V}$.



(5) Find the energy eigenvalues of the system up to the first order in the interaction. 
  


(6) Find the ground state energy up to the second order in the interaction.

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) The states of a single qubit are:
\[|0\rangle = \left(\amatrix {1 \cr 0} \right) \quad
|1\rangle = \left(\amatrix {0 \cr 1} \right)\]
With the appropriate hamiltonian:
   
\[
\tilde{\mathcal{H}} 
=\left( \amatrix { 0 & c \cr c & 0}\right)
\]



We construct the basis of the composite system using the tensor product ${|x_a,x_b\rangle = |x_a\rangle\otimes|x_b\rangle}$ on the single qubit states:
\[|0,0\rangle = \left( \amatrix {1 \cr 0 \cr 0 \cr 0} \right) \quad
 |0,1\rangle = \left( \amatrix {0 \cr 1 \cr 0 \cr 0} \right) \quad 
 |1,0\rangle = \left( \amatrix {0 \cr 0 \cr 1 \cr 0} \right) \quad
 |1,1\rangle = \left( \amatrix {0 \cr 0 \cr 0 \cr 1} \right) \quad \]


The effect of perturbation is only on the ${|1,1\rangle}$ state, meaning a change in this state's energy by ${u}$. Therefore the perturbation matrix in the standard basis is:


\[
V 
= u|1,1\rangle \langle 1,1| = \left( \amatrix { 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & u}\right)
\]



%%%%%%%%%%%%%%%
\Dn

(2) The unperturbed hamiltonian for a single qubit is diagonal in the symmetric and antisymmetric basis:
\[|S\rangle = \frac{1}{\sqrt{2}}\left(\amatrix {1 \cr 1} \right) \quad
|A\rangle = \frac{1}{\sqrt{2}}\left(\amatrix {1 \cr -1} \right)\]


From here we obtain the basis for the composite system, in the same method as in section (1):

\[|SS\rangle = \frac{1}{2}\left( \amatrix {1 \cr 1 \cr 1 \cr 1} \right) \quad
|SA\rangle = \frac{1}{2}\left( \amatrix {1 \cr -1 \cr 1 \cr -1} \right) \quad
|AS\rangle = \frac{1}{2}\left( \amatrix {1 \cr 1 \cr -1 \cr -1} \right) \quad
|AA\rangle = \frac{1}{2}\left( \amatrix {1 \cr -1 \cr -1 \cr 1} \right)
  \]



%%%%%%%%%%%%%%
\Dn

(3)
The unperturbed hamiltonian in the new basis is: 

\[\mathcal{H}_0 = \tilde{\mathcal{H}}\otimes 1_{2\times 2} +1_{2\times 2} \otimes\tilde{\mathcal{H}} = \left( \amatrix { c & 0 \cr 0 & -c}\right)\otimes \left(\amatrix { 1 & 0 \cr 0 & 1}\right) + \left(\amatrix { 1 & 0 \cr 0 & 1}\right)\otimes \left( \amatrix { c & 0 \cr 0 & -c}\right)
\]
\[
\Rightarrow \mathcal{H}_0 = \left( \amatrix { 2c & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & -2c}\right)
\]


%%%%%%%%%%%%%
\Dn

(4)
The perturbation matrix is easily obtained by noticing that ${V}$ operates only on the last component of the state vectors:
\[V|SS\rangle = V|AA\rangle = \frac{u}{2}\left( \amatrix {0 \cr 0 \cr 0 \cr 1}\right) \quad V|SA\rangle = V|AS\rangle = \frac{u}{2}\left( \amatrix {0 \cr 0 \cr 0 \cr -1}\right)\]
\[
\Rightarrow V 
=\frac{u}{4}  \left( \amatrix { 1 & -1 & -1 & 1 \cr -1 & 1 & 1 & -1 \cr -1 & 1 & 1 & -1 \cr 1 & -1 & -1 & 1}\right)
\]
%%%%%%%%%%%%%
\Dn

(5) We can see that the central block of the hamiltonian is degenerate, and the perturbation couples these degenerate states, so before using perturbation theory we must remove the degeneracy:
\[
\mathcal{H}^{(deg)} 
=\left( \amatrix { 0 & 0 \cr 0 & 0}\right)+\frac{u}{4}\left( \amatrix { 1 & 1 \cr 1 & 1}\right)
\]


After diagonalization, the central block is:

\[
\mathcal{H}^{(deg)} 
=\frac{u}{4}\left( \amatrix { 2 & 0 \cr 0 & 0}\right)
\]

Where we used the symmetric and antisymmetric states. We are only interested in the diagonal of the hamiltonian for the first order perturbation, meaning: 
\[E_n = E^{(0)}_n + V_{n,n} + \mathcal{O}(u^2)\]
So we don't have to worry about off-diagonal values that were changed by this diagonalization. The energy values are (from the ground state up):

\[ E_0 \approx -2c + \frac{u}{4}\]
\[ E_1 \approx 0\]
\[ E_2 \approx \frac{u}{2}\]
\[ E_3 \approx 2c + \frac{u}{4}\]

%%%%%%%%%%%%%
\Dn

(6) The ground state energy up to the second order is of the form:
\[E_0 = E^{(0)}_n + V_{0,0} + \sum_{m \neq 0} \frac{|V_{m,0}|^2}{E^{(0)}_0 - E^{(0)}_m} + \mathcal{O}(u^3)\]
\[E^{(2)}_0 = \frac{u^2}{16}\left(\frac{1}{-2c}+\frac{1}{-2c}+\frac{1}{-4c}\right) = -\frac{5u^2}{64c}\]
So the ground state energy is:
\[E_0 \approx -2c+\frac{u}{4}-\frac{5u^2}{64c}\]
 


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