4
$\begingroup$

The statement in the question comes from the answer to the following question on Reddit, if more context is needed:

And how does a perfect quantum computer ensure that we know for sure that the state is |0> in one measurement?

There is no way to ensure this. Strictly speaking, quantum states cannot be measured. You are only allowed to project the state onto some measurement basis. If the state belongs to this basis (|0> and |1> in your case) then you "measure" it. Otherwise your measurement is as good as your basis choice. Moreover, there is an infinite number of possible bases but only a finite number of measurement outcomes. So preparing your measurement is kinda more informative than actually doing it.

But if you can prepare many copies of the same quantum state you kinda can measure it by sampling your measurements. For a two-sized state you just imagine the Bloch sphere and perform many measurement along all 3 axes to sample the projections. For entangled states you need (exponentially) many more measurements.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

This comes down to the difference between the dimension of a vector space and the number of elements of a vector space (here, the vector space is a Hilbert space, but that doesn't really matter).

For example, given just two orthonormal basis vectors $|0\rangle=\begin{pmatrix}1\\0\end{pmatrix}$ and $|1\rangle=\begin{pmatrix}0\\1\end{pmatrix}$, we can conceive of infinitely many states of the form $$|\psi\rangle=\cos\frac{\theta}{2}|0\rangle+\mathrm{e}^{\mathrm{i}\phi}\sin\frac{\theta}{2}|1\rangle=\begin{pmatrix}\cos\frac{\theta}{2}\\\mathrm{e}^{\mathrm{i}\phi}\sin\frac{\theta}{2}\end{pmatrix}$$ because the parameters $\theta\in[0,\pi]$ and $\phi\in[0,2\pi)$ are continuous. Then, we can always consider measurements to be of the form "is the state equal to $|\psi\rangle$ or equal to some other state $|\psi_\perp\rangle$ that is orthogonal to $|\psi\rangle$ (i.e., $\langle \psi_\perp |\psi\rangle=0$)?" Without loss of generality, we can consider $$|\psi_\perp\rangle=\sin\frac{\theta}{2}|0\rangle-\mathrm{e}^{\mathrm{i}\phi}\cos\frac{\theta}{2}|1\rangle=\begin{pmatrix}\sin\frac{\theta}{2}\\-\mathrm{e}^{\mathrm{i}\phi}\cos\frac{\theta}{2}\end{pmatrix}.$$ Each possible measurement in this space boils down to probing which of some pair $|\psi\rangle$ and $|\psi_\perp\rangle$ the system is in, for a specific value of $\theta$ and $\phi$. This means we have an infinite number of measurements ($\theta$ and $\phi$ are continuous) but only two possible outcomes (the state is $|\psi\rangle$ or the state is $|\psi_\perp\rangle$).

An analogy: If you stand in one spot, there are an infinite number of directions in which you can walk. However, your ultimate final point can always be reached by walking a certain distance along the left/right axis and a certain distance along the forward/backward axis. This is because the finite number of independent directions (two) can be combined in an infinite number of ways to let you walk in an infinite number of directions.

In words: The number of possible measurement outcomes is the number of completely unique possibilities. The number of possible measurement bases is much higher because there are many ways of combining the unique possibilities.

Improve this answer
$\endgroup$
3
$\begingroup$

There is always an infinite number of measurement bases because there is an infinite number of orthonormal bases in any (more than one-dimensional) complex vector space. In fact, the set of all such bases in $\mathbb C^n$ is isomorphic to the unitary group $\mathbf U(n)$. This, however, overcounts the possible projective measurements, as some bases correspond to the same one (e.g. $\{|0\rangle,|1\rangle\}$ and $\{|0\rangle,-|1\rangle\}$ correspond to the same measurement). This is fixed considering the projective unitary group: $\mathbf{PU}(n)$.

Regarding the number of possible measurement outcomes, the statement is not really true.

For infinite-dimensions/continuous-variable systems you can clearly have infinitely many outcomes.

In finite-dimensional systems, a projective measurement does indeed only have no more than $n$ measurement outcome, if the space is $n$-dimensional. However, you can consider more general types of measurements: POVMs can have as many measurement outcomes as you want. Granted, POVMs with more than $n^2$ components will necessarily be "redundant", meaning that its components won't be linearly independent.

Improve this answer
$\endgroup$
4
  • $\begingroup$ We can uniquely identify a unitary matrix $U$ by specifying an ordered basis $B$ and an $n$-tuple of eigenvalues. We can write this as $U=B[\lambda_1, \dots, \lambda_n]$. The freedom in the choice of eigenvalues confirms that $U(n)$ does indeed overcount the bases (and thus projective measurements). However, note that we can uniquely identify an equivalence class $V$ of unitary matrices in the projective unitary group by specifying an ordered basis $B$ and an $n$-tuple of eigenvalues that begins with a unit $V=B[1, \mu_2, \dots, \mu_n]$. $\endgroup$
    – Adam Zalcman
    Oct 14, 2021 at 18:36
  • $\begingroup$ The numbers $\lambda_1, \dots, \lambda_n$ can be thought of as absolute phases applied by $U$ to each basis element. Similarly, $\mu_2, \dots, \mu_n$ can be thought of as relative phases introduced by $V$ between the first and the other basis vectors. Thus, both $U(n)$ and $PU(n)$ overcount bases. The preimage of a given basis $B$ in $U(n)$ is $n$-dimensional, corresponding to global and relative phases. The preimage of $B$ in $PU(n)$ is $(n-1)$-dimensional corresponding to the relative phases. $\endgroup$
    – Adam Zalcman
    Oct 14, 2021 at 18:36
  • $\begingroup$ Also, it is not clear what the operation is on the set of bases, so I'm not sure it is appropriate to talk about isomorphism. $\endgroup$
    – Adam Zalcman
    Oct 14, 2021 at 18:40
  • 1
    $\begingroup$ @AdamZalcman uhm, yes, I think you are right, thanks. Do you know of a better way to classify the actual set of orthogonal unit-trace projections? Googling I found discussions on the "Stiefel manifold", en.wikipedia.org/wiki/Stiefel_manifold, but it seems like this doesn't take into account that we also only care about orthonormal vectors each one defined up to phase $\endgroup$
    – glS ♦
    Oct 15, 2021 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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