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Theme 2 — Hilbert Spaces and Dirac Notation

Diagonalisation of a two-by-two Hermitian matrix

Keywords: diagonalisation · Hermitian matrix · two-level system · mixing angle · avoided crossing · level repulsion

Exercise 1 : Diagonalisation of a two-by-two Hermitian matrix

Consider the most general $2\times2$ Hermitian matrix,

\begin{aligned} H=\begin{pmatrix}a&b\\ b^*&d\end{pmatrix},\qquad a,d\in\R,\quad b=|b|e^{i\varphi}\in\C . \end{aligned}

For example, it represents the Hamiltonian of a two-level system in a basis in which $a$ and $d$ are the energies of the two states and $b$ is their coupling. Let

m=\frac{a+d}{2},\qquad\Delta=\sqrt{\Bigl(\frac{a-d}{2}\Bigr)^2+|b|^2}.

Question 1

Determine the eigenvalues of

H

Solution

The characteristic polynomial is

\det(H-\lambda\mathbf{1})=(a-\lambda)(d-\lambda)-|b|^2=\lambda^2-(a+d)\lambda+ad-|b|^2 .

Its discriminant is $(a+d)^2-4ad+4|b|^2=(a-d)^2+4|b|^2=4\Delta^2\geq0$ , and the eigenvalues are

\lambda_\pm=\frac{a+d}{2}\pm\frac12\sqrt{(a-d)^2+4|b|^2}=m\pm\Delta .

They are real, as they must be for a Hermitian matrix. They are equal if and only if $\Delta=0$ , that is, if $a=d$ and $b=0$ : the matrix is then proportional to the identity.

Question 2

Suppose that

\Delta>0

, and define the angle

\theta\in[0,\pi]

\cos\theta=\frac{a-d}{2\Delta}

and

\sin\theta=\frac{|b|}{\Delta}

. Show that

H=m\,\mathbf{1}+\Delta\,K

with

\begin{aligned} K=\begin{pmatrix}\cos\theta&e^{i\varphi}\sin\theta\\ e^{-i\varphi}\sin\theta&-\cos\theta\end{pmatrix}, \end{aligned}

then show that the vectors

\ket{+}=\begin{pmatrix}\cos\frac\theta2\\ e^{-i\varphi}\sin\frac\theta2\end{pmatrix},\qquad \ket{-}=\begin{pmatrix}-e^{i\varphi}\sin\frac\theta2\\ \cos\frac\theta2\end{pmatrix}

form an orthonormal basis of eigenvectors of $H$ .

Solution

Direct calculation gives

m+\Delta\cos\theta=\frac{a+d}{2}+\frac{a-d}{2}=a

m-\Delta\cos\theta=d

, and

\Delta e^{i\varphi}\sin\theta=|b|e^{i\varphi}=b

. The eigenvectors of

H

are those of

K

, with eigenvalues

m+\Delta k

k

is an eigenvalue of

K

. Let us calculate

K\ket+

, using the identities

\cos\theta=\cos^2\frac\theta2-\sin^2\frac\theta2

and

\sin\theta=2\sin\frac\theta2\cos\frac\theta2

. The first component is

\cos\theta\cos\tfrac\theta2+e^{i\varphi}\sin\theta\,e^{-i\varphi}\sin\tfrac\theta2=\cos\theta\cos\tfrac\theta2+\sin\theta\sin\tfrac\theta2=\cos\bigl(\theta-\tfrac\theta2\bigr)=\cos\tfrac\theta2,

and the second is

e^{-i\varphi}\sin\theta\cos\tfrac\theta2-\cos\theta\,e^{-i\varphi}\sin\tfrac\theta2=e^{-i\varphi}\sin\bigl(\theta-\tfrac\theta2\bigr)=e^{-i\varphi}\sin\tfrac\theta2 .

Thus $K\ket+=\ket+$ , and one similarly shows that $K\ket-=-\ket-$ . The two vectors are normalised, and $\braket+-=-\cos\frac\theta2e^{i\varphi}\sin\frac\theta2+e^{i\varphi}\sin\frac\theta2\cos\frac\theta2=0$ . They are therefore eigenvectors of $H$ , with eigenvalues $m\pm\Delta$ .

Question 3

Apply these results to

H=\begin{pmatrix}1&i\\-i&-1\end{pmatrix}

. Verify the first eigenvector by direct calculation.

Solution

Here

a=1

d=-1

b=i

, so

m=0

|b|=1

\varphi=\pi/2

, and

\Delta=\sqrt{1+1}=\sqrt2

. The eigenvalues are

\pm\sqrt2

. The angle satisfies

\cos\theta=\sin\theta=\frac{1}{\sqrt2}

, hence

\theta=\pi/4

, and

e^{-i\varphi}=-i

\ket+=\begin{pmatrix}\cos\frac\pi8\\-i\sin\frac\pi8\end{pmatrix},\qquad \ket-=\begin{pmatrix}-i\sin\frac\pi8\\ \cos\frac\pi8\end{pmatrix}.

Let us check the first component of $H\ket+$ : $1\cdot\cos\frac\pi8+i\cdot(-i)\sin\frac\pi8=\cos\frac\pi8+\sin\frac\pi8$ . It must equal $\sqrt2\cos\frac\pi8$ , which it does because $\cos\frac\pi8+\sin\frac\pi8=\sqrt2\cos\bigl(\frac\pi8-\frac\pi4\bigr)=\sqrt2\cos\frac\pi8$ . Numerically, $0{,}924+0{,}383=1{,}307=\sqrt2\times0{,}924$ .

Question 4

Consider the case

a\neq d

with weak coupling,

|b|\ll|a-d|

. Give an expansion of the eigenvalues to the first non-zero order in

|b|

, and describe the eigenvectors. What happens instead when

a=d

? Sketch qualitatively the two eigenvalues as functions of

a-d

, with

m

and

b

fixed.

Solution

Suppose

a>d

for definiteness. Expanding the square root gives

\Delta=\frac{a-d}{2}\sqrt{1+\frac{4|b|^2}{(a-d)^2}}\simeq\frac{a-d}{2}+\frac{|b|^2}{a-d}, \qquad \lambda_+\simeq a+\frac{|b|^2}{a-d},\quad\lambda_-\simeq d-\frac{|b|^2}{a-d}.

The coupling pushes the two levels apart: the upper level rises and the lower level falls, by an amount proportional to the square of the coupling and inversely proportional to the initial separation. The angle $\theta$ is small, $\theta\simeq2|b|/(a-d)$ , and the eigenvectors are close to the basis vectors: each is only slightly mixed with the other. When $a=d$ , by contrast, $\theta=\pi/2$ , and the eigenvectors $\frac{1}{\sqrt2}(1,\pm e^{-i\varphi})$ are equal-weight superpositions of the two basis states: the mixing is maximal, even for weak coupling, and the separation between the levels is $2|b|$ . As $a-d$ varies, the two eigenvalues $m\pm\sqrt{(a-d)^2/4+|b|^2}$ trace the two branches of a hyperbola: they never cross, unlike the uncoupled energies $a$ and $d$ , which cross at $a=d$ . This is the phenomenon of an avoided crossing, or level repulsion, which occurs throughout quantum physics, from molecules to qubits.