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Question

Consider the matrix $$ \left[ \matrix { 0&0&-1+i \\ 0&3&0 \\ -1-i&0&0 } \right] $$

(a) Find the eigenvalues and normalized eigenvectors of $A$. Denote the eigenvectors of $A$ by $|a_1\rangle$, $|a_2\rangle$, $|a_3\rangle$. Any degenerate eigenvalues?

(b) Show that the eigenvectors $|a_1\rangle$, $|a_2\rangle$, $|a_3\rangle$ form an orthonormal and complete basis ; $|a_1\rangle\langle a_1| + |a_2\rangle \langle a_2| + |a_3\rangle\langle a_3|= I$, where $I$ is the $3\times 3$ unit matrix, and that $\langle a_j|a_k\rangle$ is the Kronecker delta function

(c) Find the matrix corresponding to the operator obtained from the ket-bra product of the first eigenvector $P=|a_1\rangle\langle a_1|$. Is $P$ a projection operator?

My attempt

I have done part (a). I got the eigenvalues as $3,\sqrt{2},\sqrt{2}$ with corresponding eigenvectors

$(0, 1, 0)$ , $( (1-i)/\sqrt{2}, 0, 1 )$ , $( -(1-i)/\sqrt{2}, 0, 1 )$

Even after normalizing the vectors, I still can't work out part (b). I just don't get the $3\times3$ unit matrix. Any help would be greatly appreciated.

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  • $\begingroup$ you need to find appropriate linear combinations of $|a_2>$ and $|a_3>$ which are mutually orthogonal. $\endgroup$ – vnd Feb 22 '16 at 16:59
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    $\begingroup$ @vnx No need for that: eigenvectors belonging to different eigenvalues of a Hermitian matrix are mutually orthogonal. They only need to be normalized $\endgroup$ – DonAntonio Feb 22 '16 at 17:26
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After normalization, theory tells us that they must be orthonormal, since they correspond to different eigenvalues (note that you've written $\sqrt 2$ twice in your question, one of them should be $- \sqrt 2$; nevertheless, the eigenvectors are correct).

I suspect that your error stems from having used a Euclidean inner product instead of a Hermitian one (i.e. forgetting a complex conjugation). The Hermitian product to use should be

$$\langle u, v \rangle = u_1 \overline v_1 + u_2 \overline v_2 + u_3 \overline v_3 .$$

Then,

$$\left\| \left( \frac {1 - \textrm i} {\sqrt 2}, 0, 1 \right) \right\| = \sqrt {\frac {1 - \textrm i} {\sqrt 2} \cdot \overline {\frac {1 - \textrm i} {\sqrt 2}} + 0 \cdot \overline 0 + 1 \cdot \overline 1} = \sqrt 2 = \left\| \left(\frac {-1 + \textrm i} {\sqrt 2}, 0, 1 \right) \right\| .$$

Similarly, with this Hermitian product

$$\left< \left( \frac {1 - \textrm i} {\sqrt 2}, 0, 1 \right), \left( \frac {-1 + \textrm i} {\sqrt 2}, 0, 1 \right) \right> = \frac {1 - \textrm i} {\sqrt 2} \cdot \overline {\frac {-1 + \textrm i} {\sqrt 2}} + 0 \cdot \overline 0 + 1 \cdot \overline 1 = \frac {-2} 2 + 0 + 1 = 0 .$$

The other products are easily seen to be $0$.

The truth is that I should have written all these vectors vertically, this is the usual convention, but this would have made my typing more difficult. Keeping this in mind, though, place these column vectors one by the other to form the matrix (don't forget the normalization of the 2nd and 3rd vectors!)

$$\begin{pmatrix} 0 & \frac {1 - \textrm i} 2 & \frac {-1 + \textrm i} 2 \\ 1 & 0 & 0 \\ 0 & \frac 1 {\sqrt 2} & \frac 1 {\sqrt 2} \end{pmatrix} .$$

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  • $\begingroup$ The issue was that I used the Euclidean inner product instead of the Hermitian inner product. Managed to solve the question, thanks a lot $\endgroup$ – Sammy Williamson Feb 23 '16 at 8:06

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