1
$\begingroup$

If a quantum system is described to be only in Quantum states $s_1$ and $s_2$ (described by computational basis $\vert {0} \rangle$ and $\vert {1}\rangle$) or a combination of both (for a qubit system for example), should not all the operators, which we interpret as representations of measurable properties of the system, be only diagonal matrices? What happens if we have an operator whose basis is not the computational basis?

If we measure a property of the quantum system (operator O) and measure an eigenvalue for an eigenvector which is not associated to $s_1$ nor $s_2$, but an eigenvector which is a linear combination of both, does this mean the quantum state collapsed to such eigenstate? Would not that be contradictory to the fact that the only observed states from the Quantum System are $s_1$ and $s_2$?

$\endgroup$
6
  • 1
    $\begingroup$ Any linear combination of basis states is an allowed (observable) state. $\endgroup$ – kludg Jul 8 '20 at 7:32
  • $\begingroup$ I do not understand the second paragraph. You are essentially saying "if I measure a third state, is this not contradictory with only two different states being observable?". Well, sure, it is a contradiction: if you find a third state then clearly there were not only two possible observable states to begin with. I don't see what's the point $\endgroup$ – glS Jul 8 '20 at 11:00
  • $\begingroup$ Klug, so based on your comment I would assume that each property has its own set of possible measurable states? So the computational basis is just one frame of reference but it does not necessarily mean that $\left|0\right\rangle$ and $\left|1\right\rangle$ are the only possible "states of the system"? $\endgroup$ – César Leonardo Clemente López Jul 8 '20 at 20:09
  • $\begingroup$ @CésarLeonardoClementeLópez $\vert 0 \rangle$ and $\vert 1 \rangle$ are basis vectors in a Hilbert space, very similar to how $x$ and $y$ are basis vectors in a Euclidean plane. You wouldn't say that $x$ and $y$ are the only possible values in the plane, it's the same for $\vert 0 \rangle$ and $\vert 1 \rangle$. $\endgroup$ – Jonathan Trousdale Jul 8 '20 at 23:52
  • $\begingroup$ Yes, I am aware of that, and that is why I specifically asked this question :) My confusion arised from the physical interpretation of what this meant. If i got it correctly, then, the correct way of interpreting if a Quantum System is in state $\ket{0}$ and $\ket{1}$ is always with respect to an operator $\endgroup$ – César Leonardo Clemente López Jul 9 '20 at 0:15
1
$\begingroup$

You seem to be referring specifically to measurement operators ("operator" without qualification also refers to unitary operators, which evolve the system). In the case of projective measurements, the measurement operator decomposes into a set of projection operators, say $M = \sum m \, P_m$, where $m$ is an eigenvalue of $M$ and $P_m$ is a projection operator (POVM is slightly more complicated, but conceptually similar).

The projection operator $P_m$ projects every vector in the relevant vector space into the eigenspace associated with $m$. So if the state vector, say $\vert \psi \rangle$, is an eigenvector of $M$ associated with the eigenvalue $k$, then the probabilities of the measurement will be $p(m=k)=1$ and $p(m \ne k)=0$.

For the general case when $\vert \psi \rangle$ is not necessarily an eigenvector of $M$, then $$p(m)=\langle \psi \vert P_m \vert \psi \rangle = \sum P_{m,ij} \, \psi_i^\ast \psi_j.$$ So each $P_m$ is playing the role of an Hermitian form (the complex version of a quadratic form) that gives the probability of the measurement observing $\vert \psi \rangle$ in the $P_m$ eigenspace.

The $P_m$ are always complete, $\sum P_m = I$, positive semi-definite, and, in this case (projective measurements), also orthogonal, $P_i \, P_j = \delta_{ij} P_i$. Two $M$'s with distinct eigenspaces are simply measuring $\vert \psi \rangle$ over distinct bases that are related by some coordinate transform.

If $M$ is diagonal, the $P_m$ are permutations of $\text{diag}(1,0,...,0)$. This makes the calculations easy, but it's certainly not necessary for a valid measurement operator.

$\endgroup$

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.