I have taken an introductory level course in QM and have covered some advanced topics by myself and don't really understand what it means to 'measure in a particular basis'.

A projective measurement {${P_{m}}$} is defined w.r.t. a physical observable $M = \sum_{m}mP_{m}$ where $P_{m}$ is the projection operator onto the eigenspace corresponding to the eigenvalue $m$.

Let the physical observable represented by the Hermitian operator $M$ be called 'mopentum'. Also, let there be a system which is represented by a state ket $|\psi\rangle$ and we measure the 'mopentum' of the system. The experiment should return only one of the eigenvalues $m$ as a result with a probability $p(m) = \langle\psi|P_{m}|\psi\rangle$ and the state of the system after measurement should be $$|\psi\rangle_{m} = \frac{P_{m}|\psi\rangle}{\sqrt{p(m)}}$$

Suppose there's a basis $\{|a\rangle\}$for the ket space of the system above and I want to 'measure in it'. I hope that means that we have to make some measurement that projects a state $|\psi\rangle$ into one of the kets in the basis.

I want to know what physical observable would correspond to making such an observation. In other words, if I define the set of projectors $P_{a} = |a\rangle\langle a|$, what is the physical observable (like 'mopentum') that I need to measure to perform the operation? What are the eigenvalues I will get when I do perform the measurement and what is the operator that will correspond to the measurement I make?

Will any $A = \sum_{a}a|a\rangle\langle a|$ (where $a$ is a real number) suit my purpose? But if I construct something so arbitrary, what quantity am I going to actually, physically measure?

  • 1
    $\begingroup$ You're not starting with a basis when you want to measure something. You're handed the physical observable as a self-adjoint operator and construct the basis through the spectral theorem. $\endgroup$
    – ACuriousMind
    Commented Aug 15, 2015 at 18:16
  • $\begingroup$ I get that that's the usual MO but what do things like 'measurement in the Bell basis' mean then? $\endgroup$
    – transistor
    Commented Aug 15, 2015 at 18:59
  • 3
    $\begingroup$ @transistor If you wonder what "measurement in the Bell basis" means, why don't you ask "What does measurement in a Bell basis mean?"? $\endgroup$ Commented Aug 15, 2015 at 22:00

2 Answers 2


I define the set of projectors $P_{a} = |a\rangle\langle a|$, what is the physical observable (like 'mopentum') that I need to measure to perform the operation?

Even if you have some projections, even if they are each self adjoint, and even if they are mutually orthogonal and their range spans the whole space they still don't linearly combine into a specific operator until you say what the eigenvalues are. To get $M=\sum_{a}a|a\rangle\langle a|$ you have to pick the correct values for $a$ to associate with each $P_{a} = |a\rangle\langle a|.$

That is the mathematics, there are many operators that have the same collection of projection operators. Which is interesting because the way a measurement changes the states (by projecting them) is not enough to determine the operator. So a positivist could be confused and a student could ask where and how you get the eigenvalue.

And you might ask

What are the eigenvalues I will get when I do perform the measurement and what is the operator that will correspond to the measurement I make?

You don't get the eigenvalues from the projection operators. And the set of projection operators is not the observable. And doing a measurement is not just projecting some states. A measurement also produces eigenvalues. And the projection is the orthogonal projection onto an eigenspace. So specifying the projections isn't telling us what the measurement of observable is.

Will any $A = \sum_{a}a|a\rangle\langle a|$ (where $a$ is a real number) suit my purpose?

Some collections of $\{a\}$ make $A = \sum_{a}a|a\rangle\langle a|$ be unbkunded (discontinuous) which make KT not be defined on all states. Which might not be what you want. So be careful saying "any $\{a\}$." You might not like the result.

But if I construct something so arbitrary, what quantity am I going to actually, physically measure?!

Measurement isn't magical it is an interaction that is dictated by the actual initial state, the actual Hamiltonian for the physical setup and the actual time you spend doing the measurement.

For instance if you write the Hamiltonian for a spin 1/2 particle travelling through an inhomogeneous magnetic field you see a beam with a fixed soon state split into two beams each with a spin state aligned or antialigned with the direction of the inhomogeneous magnetic field. And you see the size of the two beams determined by the relative size of the projection of the original spin onto the two basis states. All with the Schrödinger equation, without any special hypothesis or someone saying it is a measurement, busting looking at the actual Hamiltonian for the actual experimental set up.

So what makes something a measurement. Note that the relative size of those two splits was determined by the relative size of the two projections. That means if you send one of those split beams into a similarly constructed device the beam is merely deflected in the same direction (no additional splitting) and it is that repeatability that is associated with the projection.

How about the eigenvalue? In one sense they are just labels for the differently separated portions of the wave as it evolves according to the Schrödinger equation. In another sense the operator in question appears in the Hamiltonian. For instance, in our case the Hamiltonian has a term proportional to the dot product of the spin vector and the external magnetic field vector. And so the eigenvalues literally affect how quickly and how severely the different eigenspaces become entangled with the degree of freedom that is determining the measurement (in our spin example that degree of freedom is the direction of deflect of the two beams).

Measurements create entanglement in a way that splits general states but doesn't split something that was just split. Different eigenvalues affect how strongly and quickly you can separate the different eigenspaces every place where the operator appears in the Hamiltonian will have an effect proportional to the eigenvalue's magnitude.

Finally. What you really have is first you have no measurement then later you have a measurement. So you have dynamics so you have time evolution so you have a Hamiltonian. You have Hamiltonians and when you see how they evolve factorizatized states into a state that entangles a collection of eigenspaces with another collection of orthogonal states then you can call it a measurement if doing it twice doesn't split it a second time. The result (eigenvalue) you get is actually related to how and how quickly the state becomes entangled.

And experimentally it is actually related to the calibration of the device. A device that deflects spin up right and deflects spin down is just as effective as one that does the reverse. But they have different Hamiltonians because the direction the magnetic field points versus the direction in which it gets stronger affects which goes which way. But they are just as good as separating the spin eigenstates.

The fact that differently calibrated devices literally evolve states differently actually confuses professionals that should know better because we strangely call the process an observation or measurement which makes it sound like the results (such as which portion of the beam was deflected where under the flow determined by the probability current) don't depend in these calibrations. They do. So how you associate the eigenvalues to the entanglement does depend on the calibration of the device. The relative size of the various split beams does not depend on those details.

So sometimes people don't care about how you calibrated your device. If you want to do the experiment in the lab and compare those results to the mathematics then you do have to pay attention to that.

  • $\begingroup$ What makes "something" a measurement is one requirement and one fact: we require that the measurement leaves a permanent change in a physical system, i.e. it has to be an irreversible process. The third law of thermodynamics states that such a "recording system" has to have a thermodynamic temperature T>0, which means that we have to transfer at least E>kT of energy from the quantum system to the measurement system, otherwise we won't be able to distinguish the measurement from the thermodynamic background. It follows that every physical measurement is an irreversible energy transfer. $\endgroup$ Commented Mar 27, 2023 at 3:58

I'll just pick up on one of your follow-up comments:

I get that that's the usual MO but what do things like 'measurement in the Bell basis' mean then?

If I understand correctly, the problem is that you're trying to bridge the gap between

  1. the abstract algebra of quantum measurement (i.e., Born's rule) on the one hand and
  2. the physical meaning (i.e., the experimental implementation) of the measurement on the other hand.

A quantum measurement as per (1) is straightfoward. From your OP, it seems that you understand what measurement operator $\hat{O}$ is and how it leads to eigenvalues etc.

The problem is (2), namely, how does one implement, or rather---approximate---, $\hat{O}$ "physically". For trivial cases, such as photon number detection for example, the measurement bases are given by the number operator $\hat{O} = \hat{n}$. Such devices actually exist, namely photon number resolving detectors (PNRD). Similarly, if your goal is to measure polarization you'd set up a polarizing filter followed by a PNRD. A less trivial case is if you're trying to measure more complex operators such as the quadrature of light $\hat{O} = \hat{x}$. You'd then have to resort to the so-called homodyne detection, which involves the need for much more elaborate setups.

Note that, most often, the measurement operators that one builds in the lab are often approximations of the ideal, algebraic $\hat{O}$. Depending on the problem at hand, such as your Bell state measurement, finding out how the operator is approximated in the lab is as much of an art as a science.


Your Answer

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

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