0
$\begingroup$

In wikipedia the definitions for pure and mixed state, are: " https://en.wikipedia.org/wiki/Purity_(quantum_mechanics)#:~:text=A%20pure%20quantum%20state,of%20the%20Hilbert%20space. "

A pure quantum state can be represented as a single vector $|\Psi\rangle$ in the Hilbert space. In the density matrix formulation, a pure state is represented by the matrix:

$\rho_{pure}= |\Psi\rangle\langle\Psi|$

However, a mixed state cannot be represented this way, and instead is represented by a linear combination of pure states:

$\rho_{mixed}=\Sigma_i p_i|\Psi_i\rangle\langle\Psi_i|$. Where $|\Psi_i\rangle$ are d orthonormal vectors that constitute a basis of the Hilbert space.

Clearly we can see the different notations here:

For a pure state we use $|\Psi\rangle$and we can write $|\Psi\rangle=\Sigma_ic_i|\Psi_i\rangle$. For the mixed state we use $|\Psi_i\rangle$ and it is also mentioned that $|\Psi_i\rangle$ constitute a basis of the Hilbert space., meaning they are the eigenvectors. Which means that $|\Psi\rangle$ must be a linear combination of the basis {$|\Psi_i\rangle$}. If I understood the notations correctly, does it mean that the mixed density matrix can be expressed also as a weighted sum of superpositions of the eigenstates:

$\rho_{mixed}=\Sigma_k p_k|\Psi_k\rangle\langle\Psi_k|$, where as $|\Psi_k\rangle = \Sigma_ic_i^k|\Psi_i\rangle$ is a superposition of the eigenstates $|\Psi_i\rangle$

Or it's always weighted sum of the eigenstates of the hilbert space?

$\endgroup$
4
  • $\begingroup$ If the $|\Psi_i\rangle$ are not orthogonal, then that means that they do not represent a basis, which means is possible to do what I asked, or no? $\endgroup$
    – imbAF
    Aug 10 at 19:29
  • $\begingroup$ Eigenstates = ket basis of the hilbert space. The eigenstates are the eigenstates of the hamiltonian $\endgroup$
    – imbAF
    Aug 10 at 19:38
  • $\begingroup$ My claim is that for the pure state you use the notation $|\Psi\rangle$, and for the mixed state you use $|\Psi_i\rangle$, and $|\Psi_i\rangle$ are basis of the the hilbert space. Then $|\Psi\rangle$ without the index i must be a linear combination of $|\Psi_i\rangle$. My question is if we can write a mixed state using $|\Psi\rangle$ vectors, which are superposition of the basis kets {$|\Psi_i\rangle$} $\endgroup$
    – imbAF
    Aug 10 at 19:42
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – imbAF
    Aug 10 at 19:42
0
$\begingroup$

Pure state can be written as the superposition of eigenstates of the system. However, a mixed state is statistical ensemble of pure states, which is different from superposition. If you write, $|\psi_k\rangle=\sum_ic_{i}^k|\psi_i\rangle$, then $|\psi_k\rangle$ becomes a pure state by definition. This contradicts you assumption.

$\endgroup$
6
  • $\begingroup$ Please clarify what you mean with 'eigenstates of the system'. $\endgroup$
    – Jakob
    Aug 10 at 19:31
  • $\begingroup$ If the pure state can be as the superposition of eigenstates of the system and if a mixed state is statistical ensemble of pure states, can't you say that the mixed state is a statistical ensemble of superposition of eigenstates (pure states) ?? $\endgroup$
    – imbAF
    Aug 10 at 19:31
  • $\begingroup$ basis of the hilbert space. Which means, every other vector which is not an eigenstate (eigenvector) can be written as a linear combination of them $\endgroup$
    – imbAF
    Aug 10 at 19:33
  • 1
    $\begingroup$ @imbAF Yes it implies that, but you cannot write the statistical ensemble as some linear combination of states. That's why you statement is correct, however mathematically, it is not correct to write the last equation you wrote. $\endgroup$
    – sslucifer
    Aug 10 at 19:34
  • $\begingroup$ why not? if logically my statement is correct $\endgroup$
    – imbAF
    Aug 10 at 19:39
0
$\begingroup$

The simplest way to understand a pure state (in a finite dimensional space) is to recognize that a pure state is always the eigenstate of some hermitian operator. Maybe the canonical example is the arbitrary qubit state \begin{align} \vert\psi\rangle = \cos\theta/2 \vert +\rangle + e^{i\phi}\sin\theta/2 \vert -\rangle\, . \end{align} (If I did this right) $\vert\psi\rangle$ is an eigenstate of (hermitian operator) $n_x\sigma_x+n_y\sigma_y+n_z\sigma_z$, where $n_x=\sin\theta\cos\phi$, $n_y=\sin\theta\sin\phi$, $n_z=\cos\theta$.

On the other hand, a mixed state will NOT be an eigenstate of any combination of the $\sigma_i$'s.

In a pure state, the coefficients $\alpha$ and $\beta$ in $\vert\psi\rangle =\alpha \vert +\rangle +\beta\vert-\rangle$ can be complex and indeed it is the relative phase of the complex coefficients that controls the interference between the terms in the evaluation of various quantities. In addition, $\vert\alpha\vert^2+\vert\beta\vert^2=1$.

In a mixed state, the coefficients $c_{ij},c_{ij}$ in $\rho=\sum_{i,j=+,-} c_{ij}\vert i\rangle\langle j\vert$ must be such that $\rho$ is hermitian. In a basis where $\rho$ is diagonal, $c_{12}=c_{21}=0$ and $c_{11}+c_{12}=1$ (rather than the sum of their modulus squared). The mixed state represents a partially coherent mixture of pure states, and (just like in optics) partially coherent mixtures do not completely interfere.

Physically, the mixed states account for classical uncertainty in the preparation of the system: your Stern-Gerlach oven will spit out a particle with spin-up half the time, and with down half the time so the oven gives you \begin{align} \rho=\frac{1}{2}\vert +\rangle\langle + \vert +\frac{1}{2}\vert -\rangle\langle -\vert\, . \end{align} A system described by this $\rho$ will give you spin up and down (along any direction) with equal probability. This doesn't depend on the orientation because as a matrix $\rho=\frac{1}{2}\mathbb{I}$, and this is invariant under rotation.

On the other hand, you can filter your mixed state by separating the component using a Stern-Gerlach magnet: the particles with spin up will go up and the ones with spin down will go down. You then just place a stopper to prevent the down beam from going any further, and you're left (after the filter) with a beam of pure states $\vert +\rangle\langle +\vert$ having 1/2 of the intensity of the output of the oven. The pure states can (individually) also be represented by the ket $\vert +\rangle$.

You can certainly use these pure states to measure $\sigma_x$, $\sigma_y$ or $\sigma_z$, but if you measure $\sigma_z$, you will get only one outcome: all your particles will have spin up. If you measure $\sigma_x$, you will get up and down equally probably. If you choose to measure an intermediate combination of $\sigma_x$ and $\sigma_y$ - something like $\cos\chi\sigma_x+\sin\chi\sigma_y$, you will get a beam polarized along $\cos\chi\hat x+\sin\chi\hat y$.

In the example I gave, I supposed the output of the oven was up/down with 50-50 probability, resulting in $\rho=\frac{1}{2}\vert +\rangle +\frac{1}{2}\vert -\rangle$ but there's nothing to suggest the outputs must be orthogonal: some device may in fact produce a different mixture of states - say $\frac{1}{3}\vert +;z\rangle\langle +;z\rangle +\frac{2}{3}\vert -;x\rangle \langle -;x\vert$, in which case the device spits out $\vert +\rangle$ along $\hat z$ 1/3 of the time and $\vert -\rangle$ along $\hat x$ 2/3 of the time.

$\endgroup$

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.