2
$\begingroup$

So, I was studying quantum computation using the book Nielsen and Chuang and it stated a theorem known as "Spectral Decomposition theorem"

$$A=\sum _{i}\lambda _{i} | i \rangle \langle i|$$

I infer from this theorem that any normal operator can be diagonalized in the basis set $ \left\{ |i \rangle \right\} $ which should be the eigen vectors of the operator matrix A with $ \lambda _{i}$ as the eigenvalues.

Now when I started studying about the density operator with the definition $$\rho = \sum _{i}p_{i} |\psi _{i} \rangle \langle \psi _{i}|$$ I got a little confused. Since $\rho$ is a normal operator and it can be written as this decomposition, it must mean that the vectors $| \psi _{i} \rangle$ must be orthonormal to each other according to the spectral decomposition theorem. This seems totally absurd to me since there is no reason for the pure states (combining to make a mixed state) to have orthonormality as a prerequisite. I am sorry if my question is very trivial as this is my first time studying quantum information and I would be glad if someone could help me with this problem.

$\endgroup$
3
$\begingroup$

The vectors $|\psi_i\rangle$ out of which you build the density matrix don't have to be orthornormal. They don't even have to be a basis: you can have more or less of them than your dimension. However, the decomposition theorem tells you that you can always find an orthonormal basis in which the density matrix can be written as in your first equation.

As an example, suppose you have spin-1/2 particles in an equal mixture of eigenstates in the three positive cartesian directions:

$$|x\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad |y\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix}, \quad |z\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

From this we can build the density matrix

$$\rho = \frac13 |x\rangle \langle x | + \frac13 |y\rangle \langle y | + \frac13 |z\rangle \langle z |,$$

and the theorem assures us that there is an orthonormal basis (the eigenbasis of $\rho$) $\{|1\rangle, |2\rangle\}$ such that

$$\rho = p_1 |1\rangle \langle 1 | + p_2 |2\rangle \langle 2 |.$$

$\endgroup$
  • $\begingroup$ So this means that the density operator can always be written in an orthonormal basis but that doesn't necessarily mean that those orthonormal basis vectors are the pure states. Am I right in this statement? $\endgroup$ – Tachyon209 Jul 4 at 14:03
  • $\begingroup$ Hmm, according to what you say the following would be a density matrix: $$\rho = 0.5|\zeta\rangle\langle\zeta|+0.5|z\rangle\langle z|$$ with $|\zeta\rangle = \frac{1}{\sqrt{3}}(|x\rangle+|y\rangle)$. Now I expect $1 = Tr(\rho) = \sum\limits_\pm\langle \pm z|\rho|\pm z\rangle$. How ever, this doesn't seem to add up... $\endgroup$ – denklo Jul 4 at 14:11
  • $\begingroup$ I think there must be some mistake in your calculation. The density matrix which I am getting has Trace=1. $\endgroup$ – Tachyon209 Jul 4 at 14:25
  • $\begingroup$ @Tachyon209 I am not sure what you mean in your first comment. Any vector is a pure state. However, the pure states you used to build the density matrix are not necessarily the ones in the orthonormal basis. A given density matrix can represent many different mixtures. $\endgroup$ – Javier Jul 4 at 14:41
  • $\begingroup$ Found my mistake, thanks! $\endgroup$ – denklo Jul 4 at 14:42
0
$\begingroup$

Lets say we have a set of states $|\psi_i\rangle = \sum_j\alpha_{ij}|j\rangle$, which are not necessarilly orthonormal, but normalized: $\sum_j|\alpha_{ij}|^2 = 1$. Here $|j\rangle$ is some orthonormal basis.

We create some putative density matrix out of them: $\rho = \sum_i p_i|\psi_i\rangle\langle\psi_i|$ where $\sum_{i}p_i = 1$. Now to check, wether it really is a density-matrix we need to calculate the trace and make sure it's 1. $$ Tr(\rho) = \sum_{ji}p_i\langle j|\psi_i\rangle\langle\psi_i|j\rangle = \sum_{ij}p_i|\alpha_{ij}|^2 = \sum_{i}p_i\sum_j|\alpha_{ij}|^2 = \sum_{i}p_i = 1 $$ Thus the states which span a density matrix need not to be orthonormal necessarily. Infact we even never needed to calculate the overlaps $\langle \psi_j|\psi_i\rangle$ during the proof...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.