6
$\begingroup$

Let a quantum system be described by Hilbert space $\mathscr{H}$ and let $|\psi\rangle$ be an arbitrary state. Define the operator

$$P=|\psi\rangle\langle \psi|$$

This is hermitian. It has two eigenvalues: $0$ and $1$ with two eigenspaces. The $1$ eigenspace is the subspace spanned by $|\psi\rangle$, in other words $$\mathscr{H}_1=\{\lambda |\psi\rangle : \lambda \in \mathbb{C}\}$$

while the eigenspace corresponding to zero is its orthogonal complement $\mathscr{H}_2 = \mathscr{H}_1^\perp$.

Since this is one observable, one would expect it could be measured. But how physically such measurement can be made?

The point is that $P$ doesn't correspond directly to a physical quantity like momentum, energy or angular momentum, which one experimentalist would know of a procedure to measure in the lab.

The point is that if $A$ is one physical quantity with eigenspaces $\mathscr{H}_\lambda$ corresponding to the values $\lambda\in \sigma(A)$ the postulates of quantum mechanics allows us to say "well the system's state lies in $\mathscr{H}_\lambda$" if when we measure $A$ we get $\lambda$.

This in particular allows us to preparate a system in any eigenstate of any physical quantity that we can measure. But preparing on arbitrary states still is somewhat weird to me.

Of course, if measuring $P$ is possible, a measure of $P$ yielding value $1$ would prepare a system in the state $|\psi\rangle$.

So, is there any "generalized way" to measure this observable?

$\endgroup$
  • 2
    $\begingroup$ Why do you think that just because you have defined an operator with eigenvalues means that these eigenvalues must be something that can be physically measured? We know that classical quantities like position or momentum have corresponding operators in QM, but the converse is not necessarily true. We can't just make an operator and say it must be physically measurable. $\endgroup$ – Aaron Stevens Jul 16 '18 at 17:35
  • $\begingroup$ I agree with @Aaron Stevens. I addressed that in my attempt at answering the question $\endgroup$ – InertialObserver Jul 16 '18 at 22:16
  • $\begingroup$ In principle, you can measure the expectation value of any operator, not only hermitian ones. To do it in practice can be tricky, though (just as many other things). $\endgroup$ – Norbert Schuch Jul 17 '18 at 21:14
3
$\begingroup$

Given a description of $|\psi\rangle$, there are two possibilities. Either you know how to apply a unitary operator $U$, that maps a standard basis state that you can measure, let's say $|0\rangle$, to it or not. In either case that unitary operator, $U$ exists; so let's assume you know it too. I should also mention that computational complexity-wise, for the general case, this is a hard problem.

Assuming that you know $U$, then you can apply $U^\dagger$ to your operator:

$$Q=U^\dagger P U \ .$$ And then you can do a measurement in the standard basis.

As a side note, let's assume in your lab or quantum computer, you can only apply a set of limited standard gates:$\left\{U_1,U_2,\cdots,U_N\right\}\in\mathcal G$. If $\mathcal G$ is universal, it is guaranteed that you can approximate any unitary operator with desired precision with its operators.

$\endgroup$
  • $\begingroup$ Would the downvoter care to share their thoughts on what they feel is incorrect or missing with my answer? $\endgroup$ – Ali Jul 26 '18 at 12:06
2
$\begingroup$

A measurement procedure is just a unitary operator acting on the tensor product of the system and the experimental device which entangles the two as much as possible. In the projector observable we're talking about, the outcome of measurement is either 0 or 1 so the device may consist of a single qubit with "classical" states $|0\rangle$ and $|1\rangle$ which show up on the screen of our device when we press the "big red button".

The Hamiltonian can be pretty much anything, it's a matter of engineering. We will choose it so that after time $T$, it induces the unitary evolution $$U =(1-P) \otimes (|0\rangle\langle x| + | x\rangle\langle0|) + P \otimes(|1\rangle\langle x| + |x \rangle\langle 1|),$$ where $|x\rangle$ is the initial state of the measurement device, a state that we can reliably prepare for our device by pressing the "reset" button. One can check that $U$ is unitary. The Hamiltonian that produces this operator after time $T$ is $H = -i\hbar \log U / T$.

To make the measurement, we prepare our device in state $|x\rangle$ by pressing the reset button. Then we bring it into contact with the unknown state $|\psi_0\rangle$ for time $T$, after which the combined state has evolved to $$U(|\psi_0\rangle \otimes |x\rangle) = (1-P)|\psi_0\rangle \otimes |0\rangle + P|\psi_0\rangle \otimes |1\rangle.$$ Then we press the big red button and now the machine either reads 0 and the state is $(1-P)|\psi_0\rangle$ (which doesn't tell us much) or the machine reads 1 and the state is $P|\psi_0\rangle \sim |\psi\rangle$, which tells us everything about the system.

You can think of this as a state preparation machine for $|\psi\rangle$ which displays 0 if unsuccessful and 1 if successful. The number of components used to implement $U$ and how many times you have to press the big red button are measures of the quantum complexity of the state $|\psi\rangle$.

I recommend John Preskill's notes (pdf) to learn (much much) more.

$\endgroup$
  • 1
    $\begingroup$ This confuses me. Which state are we using to define the projection operator? Also, this example an a "state preparation machine which displays 0 if unsuccessful" are two different things. You can have a state that is not $\psi$ but it can still have a projection along $\psi$ $\endgroup$ – Aaron Stevens Jul 17 '18 at 13:52
  • $\begingroup$ $P = |\psi \rangle \langle \psi |$ and the initial state is $|\psi_0\rangle$. Once the measurement is made and the outcome is 1, then the state is $P|\psi_0\rangle \sim |\psi\rangle$, as I said. This is the usual "wavefunction collapse" people talk about. Measurement and state preparation are the same thing if you think about it. $\endgroup$ – Ryan Thorngren Jul 17 '18 at 19:48
1
$\begingroup$

Let's say the system is in a state $|\phi\rangle$, and you want to measure said operator $P=|\psi\rangle\langle\psi|$, i.e., $$ \begin{align} \langle\phi|P|\phi\rangle&=\langle\phi|\psi\rangle\langle\psi|\phi\rangle \\ &=|\langle\phi|\psi\rangle|^2\ . \end{align} $$ This is, what you need to measure is the overlap of the two states. There are different ways to do that, but the "canonical" way would be that you prepare $|\psi\rangle$ (in addition to $|\phi\rangle$, which is the input to your measurement scheme) and then let them interfere. The degree of interference you see (when you carry out the experiment many times) will exactly correspond to $$ |\langle\phi|\psi\rangle|^2= \langle\phi|P|\phi\rangle \ . $$ (More formally, in Quantum Information, there is the concept of a "swap test" which allows you to measure said overlap.)

$\endgroup$
  • 1
    $\begingroup$ Thanks for your answer @NorbertSchuch. Now, what bothers me is really how would one go in preparing $|\psi\rangle$. The reason is: in a spin 1/2 system, to prepare the state $|\uparrow\rangle$ one would exactly measure $S_z$. Now if $|\psi\rangle$ doesn't correspond to a simple physical quantity, how can it be prepared? If this is too complicated for a post here, references on the matter would be appreciated. Thanks! $\endgroup$ – user1620696 Jul 17 '18 at 21:24
  • $\begingroup$ @user1620696 That's a good point. But you agree that for any state you can write down, there should be a preparation procedure? Otherwise, why don't we kick all states we cannot prepare out of quantum theory, which would render it much simpler (as all we need are, say, position eigenstates to describe all of quantum physics)? $\endgroup$ – Norbert Schuch Jul 17 '18 at 21:27
  • $\begingroup$ I agree in the sense that it seems intuitive to me that all states should be attainable by a suitable time evolution. In other words, intuitively to me, it seems obvious that for arbitrary $|\psi\rangle$ there is $|\psi_0\rangle$ an eigenstate of some simple physical quantity and a Hamiltonian $H$ such that if $U(t)$ is the corresponding time evolution there is $t_0$ such that $U(t_0)|\psi_0\rangle=|\psi\rangle$. Albeit I have this intuition, I'm unsure if this is rigorously true. $\endgroup$ – user1620696 Jul 17 '18 at 21:35
  • 2
    $\begingroup$ @user1620696 This is a central question in quantum information: Which set of Hamiltonians or unitaries is universal, i.e. allows us to construct arbitrary unitaries, and how to we actually construct (or well approximate) such a unitary given a universal set. $\endgroup$ – Norbert Schuch Jul 19 '18 at 12:17
-1
$\begingroup$

Given $|\psi\rangle$, You can always make a machine that produces as many copies of $|\psi\rangle\langle \psi|$ pure states as required. By measuring the state copies you would be measuring the eigenspace of value $\lambda = 1$ for your observable

Measuring the eigenspace of $\lambda = 0$ is less straightforward, as it requires to determine the much bigger subspace of states orthogonal to your $|\psi\rangle$

The above procedure can be done as long as $|\psi\rangle$ is known. No-cloning theorem would forbid you from building above machine if you don't have enough information about it, and all you have is a physical instance of the quantum state

$\endgroup$
  • $\begingroup$ Thanks for the answer! So the point is that if we know $|\psi\rangle$ we can actualy make a machine that produces copies of this state. Now could you elaborate further on how that would be done, or if it is too involved, provide a reference expanding further? Furthermore, how would one "measure the state copies"? Because again, in my naive view it seems all we can measure are physical quantities like energy, momentum, angular momentum, etc. But I'm probably clearly missing something quite basic here. Thanks again! $\endgroup$ – user1620696 Jul 16 '18 at 14:53
  • 2
    $\begingroup$ What do you mean by "measuring state copies"? $\endgroup$ – Aaron Stevens Jul 16 '18 at 18:02
  • $\begingroup$ How does this explain how to measure the expectation value of $P$ on some arbitrary state $|\phi\rangle$? $\endgroup$ – Norbert Schuch Jul 17 '18 at 21:15

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.