I am currently self studying many-body physics from Introduction to Many Body Physics and am trying to derive (3.153) from (3.152). The Bogoliubov transformation for fermions $a_1$ and $a_2$ are given by the equations:

$c_1 = ua_1 + va_2^{+}$ and $c_2^+ = -va_1 + ua_2^+$, where $u,v \in \mathbb{R}$ and $u^2 + v^2 = 1$. From this one can see that that

$a_1 = uc_1 - vc_2^{+}$ and $a_2^+ = vc_1 + uc_2^+$. From this we see that $a_1^+ = uc_1^+ - vc_2$ and $a_2 = vc_1^+ + uc_2$.

Now my goal is to show that the Hamiltonian (3.152) $H = \epsilon(a_1^+a_1 - a_2a_2^+) + \Delta(a_1^+a_2^+ + h.c)$ diagonalizes to (3.153)$H = \sqrt{\epsilon^2 + \Delta^2}(c_1^+c_1 + c_2^+c_2 - 1)$.

where h.c. means Hermitian complex

Using the formulae from the previous system of linear equations I found that $a_1^+ a_1 - a_2a_2^+ = (u^2 - v^2)(c_1^+ c_1 - c_2c_2^+) - 2uv(c_1^+ c_2^+ + c_2c_1)$

$a_1^+a_2^+ = uvc_1^+c_1 + u^2 c_1^+c_2^+ - v^2c_2c_1 - uvc_2c_2^+$

However, I cannot see how to go from one Hamiltonian to another.

My Question: Am I doing something wrong here? Can anyone show me how to diagonalize this Hamiltonian?

  • 1
    $\begingroup$ I don't see any arithmetic mistakes. One suggestion is to check that the two Hamiltonians have the same eigenvalues and eigenstates. They are easy to see with the second form of the Hamiltonian; the four eigenstates are $|00\rangle$, $|01\rangle$, $|10\rangle$, and $|11\rangle$ in the transformed ($c_i$) basis. Since we know how $c_1$ and $c_2$ act on each of these eigenstates, we can try acting with the first Hamiltonian on them using the expressions for $a_1$ and $a_2$ that you've calculated, and verify that the states are eigenstates with the correct eigenvalues. $\endgroup$ May 20 at 21:19
  • $\begingroup$ @QuantumMechanic Is this a general way to diagonalize a Hamiltonian with a transformation? If there are other ways, what are they? $\endgroup$
    – Amatuer
    May 20 at 22:46
  • 1
    $\begingroup$ Essentially the idea is to find the eigenstates and their eigenvalues. So once somebody promises that they've given you the diagonalized Hamiltonian, you can check and see. Sometimes you can easily recognize eigenstates and eigenvalues, such as from the second Hamiltonian. When you can't, the easiest way for me is often to write the Hamiltonian as a matrix in some basis (ie the original Fock basis associated with the operators $a_i$) and then diagonalize the matrix using the standard eigenvalue/eigenvector procedure $\endgroup$ May 21 at 0:32
  • $\begingroup$ Maybe realizing that $u$ and $v$ need to depend on $\epsilon$ and $\Delta$ will help? $\endgroup$ May 21 at 15:57

We can write the initial Hamiltonian in the Fock basis of the $a_i$ operators. Realizing that $a_i=\vphantom{a}_i|0\rangle\langle 1|_i$ lets us write $a_1^+ a_1=|1\rangle\langle1|\otimes\mathbb{I}$, $a_2 a_2^+=\mathbb{I}\otimes|0\rangle\langle0|$, $a_1^+ a_2^+=|1\rangle\langle0|\otimes|1\rangle\langle0|$, etc., we can write the Hamiltonian in the $|i\rangle_1\otimes|j\rangle_2$ basis as $$H=\begin{pmatrix}-\epsilon&0&0&\Delta\\0&0&0&0\\0&0&0&0\\\Delta&0&0&\epsilon\end{pmatrix}.$$ This matrix is easy to diagonalize: there are two eigenvectors $|1\rangle\otimes|0\rangle$ and $|0\rangle\otimes|1\rangle$ with eigenvalue $0$, then there are two eigenvectors proportional to $$\left(-\epsilon\pm\sqrt{\epsilon^2+\Delta^2}\right)|0\rangle\otimes|0\rangle+ \Delta|1\rangle\otimes|1\rangle$$ with eigenvalues $\pm\sqrt{\epsilon^2+\Delta^2}$.

As for what the textbook does, I prefer to do it the other way around. We can express the second Hamiltonian as \begin{align} c_1^+c_1+c_2^+c_2-1=&(ua_1^++va_2)(ua_1+va_2^+)+(-va_1+ua_2^+)(-va_1^++ua_2)-1\\ =&u^2 a_1^+a_1 + v^2 a_2 a_2^++uv(a_1^+a_2^++a_2a_1)+v^2 a_1a_1^++u^2a_2^+a_2-uv(a_1a_2+a_2^+a_1^+)-1\\ =&u^2 a_1^+a_1 + v^2 a_2 a_2^++uv(a_1^+a_2^++a_2a_1)+v^2 (1-a_1^+a_1)+u^2(1-a_2a_2^+)+uv(a_2a_1+a_1^+a_2^+)-1\\ =&a_1^+a_1(u^2-v^2)+a_2a_2^+(v^2-u^2)+v^2+u^2+2uv(a_1^+a_2^++a_2a_1)-1\\ =&(u^2-v^2)(a_1^+a_1-a_2a_2^+)+2uv(a_1^+a_2^++h.c.). \end{align} The crucial step is the third equality, where I used the ferminonic properties $a_i a_i^+=1-a_i^+a_i$ and $a_i a_j=-a_ja_i$ for $i\neq j$. We can immediately read off the required relationships $u^2-v^2=\epsilon$ and $2uv=\Delta$.


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