# Diagonalization Of $(\sigma_x+\sigma_y)$

Can this matrix $(\sigma_x\pm\sigma_y)$ be diagonalised? Clearly, if $\sigma_x$ is diagonalized by a similarity transformation $S_1\sigma_x{S_1}^{-1}$, then $\sigma_y$ can't be diagonalized by $S_1$, but by something else, say $S_2$, because they don't commute and therefore can't be diagonalized with the same similarity transformation. But is it possible to find some $S_3$ which makes a combination like $(\sigma_x\pm\sigma_y)$ diagonal? My second question is can a matrix have eigenvalues if it is not diagonalizable?

• You may want to ask your second question separately as it's quite different from the rest of your post. However, I recommend you have a search through math.se first, and probably post there. One key thing to look for is Jordan matrices. Feb 13, 2014 at 9:27

As the Pauli matrices are hermitian, every linear-combination (with real coefficients) of them is hermitian as well, in particular $(\sigma_x\pm\sigma_y)$. And because every hermitian operator can be diagonalized, the answer to your question is yes. Just write the corresponding 2x2-matrix and try to diagonalize it.

Linear combinations of Pauli matrices play particularly nicely with diagonalization. The reason for this is that a linear combination of the form $$\vec v\cdot\vec\sigma=\sum_j v_j\sigma_j =\begin{pmatrix}v_z&v_x-iv_y\\ v_x+iv_y&-v_z\end{pmatrix}\tag1$$ represent the density matrix $\rho=\tfrac12(1+\vec v\cdot\vec\sigma)$ of a state at the point $\vec v$ on the Bloch sphere. (Of course, normalization needs to be dealt with: it's a pure state on the sphere if $|\vec v|=1$, a mixed state inside the ball if $|\vec v|<1$, and a nonphysical 'density matrix' outside the sphere if $|\vec v|>1$.

To be more precise, consider a general pure state of a spin-1/2 particle, which can be written as $$|\psi\rangle=\cos\tfrac\theta2|\!\uparrow\rangle +e^{i\phi}\sin\tfrac\theta2|\!\uparrow\rangle =\begin{pmatrix}\cos\tfrac\theta2\\e^{i\phi}\sin\tfrac\theta2\end{pmatrix}.\tag2$$ The corresponding density matrix is \begin{align} \rho &=|\psi\rangle\langle\psi| = \begin{pmatrix}\cos\tfrac\theta2\\e^{i\phi}\sin\tfrac\theta2\end{pmatrix} \begin{pmatrix}\cos\tfrac\theta2 & e^{-i\phi}\sin\tfrac\theta2\end{pmatrix} \\ & =\begin{pmatrix} \cos^2\tfrac\theta2 & e^{-i\phi}\sin\tfrac\theta2\cos\tfrac\theta2\\ e^{i\phi}\sin\tfrac\theta2\cos\tfrac\theta2 & \sin^2\tfrac\theta2 \end{pmatrix} \\ & =\frac12\begin{pmatrix} 1+\cos\theta& e^{-i\phi}\sin\theta\\ e^{i\phi}\sin\theta & 1-\cos\theta \end{pmatrix}. \end{align} This is already in the form (1), if you set $\vec v=(\cos\phi\sin\theta,\sin\phi\sin\theta, \cos\theta)$, and that is precisely the standard polar coordinates representation (as used by physicists, of course). The upshot of this is that, modulo constants and the identity matrix (which affect eigenvalues but not the eigenvector structure), any linear combination of Pauli matrices can be easily written as a (displaced) projector on a known, easily describable state vector. Find the vector $\vec v$ from your linear combination, take its polar-coordinate representation, plug them into (2), and you get one eigenvector. The other one is, of course, its orthogonal complement, and is given exactly by changing $\vec v$ to $-\vec v$ (so changing $\theta$ to $\pi-\theta$ and $\phi$ to $\phi+\pi$).

Nice, right?

The matrix you consider is Hermitian because real linear combination of Hermitian matrices, thus it can be diagonalized (i.e. $S_3$ does exist) for a known general theorem.

If a matrix admits eigenvalues it may be non-diagonalizable. Consider the matrix $A$ with the form: $$A=\begin{pmatrix}0 & 1\\ 0& 0\end{pmatrix}$$

The eigenvalues are the complex solutions of $\det(A-\lambda I)=0$. There is only the solution $\lambda =0$. If the matrix were diagonalizable, its diagonal form would be$$\begin{pmatrix}0 & 0\\ 0& 0\end{pmatrix}$$ namely the zero matrix $O$. But this matrix has the same form in every basis, in particular in the initial one. Consequently, we would have $A=O$ that is false. So $A$ admits eigenvalues but cannot be diagonalized.