Past exam of the mathematical methods I course of the University of Oxford / 2014 / 7

Table of contents
- i 7
  - Solution i
- ii 7
  - Solution ii
- iii 7
  - Solution iii
- iv 7
  - Solution iv

i

Solution

The finite-dimensional spectral theorem diagonalizes the Hermitian matrix in its orthonormal eigenbasis. The determinant is the product of the eigenvalues:

det H = λ_{1} λ_{2} .

ii

Solution

The trace is unchanged by similarity, so in the same eigenbasis

tr H = λ_{1} + λ_{2} .

iii

Solution

Every eigenvalue of a Hermitian matrix is real: if

He = λ e

with

e^{†} e = 1

, then

λ = e^{†} He = \overline{e^{†} He}

. Therefore

Im (λ_{1} λ_{2}) = 0.

iv

Solution

Distinct eigenvalues of a Hermitian matrix have orthogonal eigenvectors, since

λ_{2} e_{1}^{†} e_{2} = e_{1}^{†} H e_{2} = λ_{1} e_{1}^{†} e_{2}

. The given normalization then gives

e_{1}^{†} (e_{1} + e_{2}) = 1 + 0 = 1.

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