We have some arbitrary quantum state, lets say $$\vert\Psi\rangle=\alpha_{1}\vert\uparrow\rangle+\alpha_{2}\vert\downarrow\rangle= \begin{pmatrix} \alpha_{1} \\ \alpha_{2} \\ \end{pmatrix}$$. And we act on it with some operator, whatever is appropriate math for it, lets say in this case particular case a linear combination of Pauli matrices $$ \sigma_{n} = \begin{pmatrix} n_z & n_x-in_y \\ n_x+in_y & -n_z \\ \end{pmatrix}$$ My understanding of physical process, that is happening here, is as follows. We have some arbitrary electron, with operator we put on magnetic field in arbitrary direction and eigenvalues make sure we calculate expectation value correctly. So eigenvalues are measurement result. (?)

On the other hand operator acts on a vector and we get a new state. (Rotates the state on a Bloch sphere?) So eigenvalues change basis vectors and we rotate a state.

How does the operators relate to rotations on Bloch sphere? In my interpretation eigenvalue existence mess up probabilities.

Addition: It seems I can't connect the idea of an observable (mathematical construction for expectation value - to get measurement results) and idea of an unitary operator which changes state of a system (due to magnetic field). How are they related? Are they one and the same?

  • $\begingroup$ I changed your "mixture" to "linear combination" - "mixture" is a dangerous word to use near pure quantum states as it means something else altogether. Use "linear combination" or "superposition". $\endgroup$ Mar 26, 2015 at 9:15
  • $\begingroup$ You state: "So eigenvalues change basis vectors and we rotate a state." I take you to mean that applying a rotation to the state changes the eigenbasis. In fact (as I note in my 1st answer) the only way to change the eigenbasis is to change the orientation of the polarizing B field. $\endgroup$ Apr 3, 2015 at 23:45
  • $\begingroup$ @hyperpolarizer I indeed made a mistake in reasoning here exactly. You understood me correctly. I thought hard on what does it mean - to do an operation, your explanation helped to set my knoweledge in right place, thanks! (And Messiah indeed has a great book.) $\endgroup$
    – Rena
    Apr 4, 2015 at 0:33
  • $\begingroup$ I'm glad I could help out; I think we can both agree that the subject of your query is beautiful, and that Messiah is a great book. $\endgroup$ Apr 4, 2015 at 4:45

3 Answers 3


I would re-state the problem slightly to conform to conventional notation. In any discussion of magnetic resonance -- of spin 1/2 particles, be they electrons, or say, protons-- the direction of the magnetic (polarizing) field is always chosen as +z in a laboratory coordinate frame. Then the two stationary states, which (for convenience) I write as |+> and |->, correspond to the spin aligned with its z-component of angular momentum parallel or anti-parallel to the polarizing field.

The stationary states are in fact eigenstates of the z component Pauli matrix, with respect to the current laboratory coordinate frame.

Then the operator you illustrate is a linear combination of all three Pauli matrices in this frame. Any Pauli matrix (or linear combination thereof) acts mathematically as the generator of an infinitesimal rotation of the spin. Your conventional x, y, and z Pauli matrices each individually generate infinitesimal rotations of the spin about these axes. The best discussion I know of this is Messiah's book (as noted in an earlier response, above); you may also want to consult M. E. Rose 'Theory of Angular Momentum,' which is available as a Dover reprint.

Rotating an eigenket gives you a new state, but it doesn't change your basis, which is defined essentially by the direction of the polarizing magnetic field. You can rotate the spins any way you like, but your result will still be expressed as a linear combination of those same eigenkets, as long as you don't physically change the direction of the field.

So far we have dealt exclusively in quantum mechanical rotation.

Once we bring in the 'Bloch sphere', we need to introduce the Bloch equations, which are classical. In fact, the Bloch equations without relaxation are exactly generators of infinitesimal rotations, but for 3 dimensional classical vectors, specifically (in the case of magnetic resonance) the magnetization. I usually (for convenience) try to keep the classical and quantum views separate in my mind-- so I think of the Pauli matrices as rotating an individual spin, and the Bloch matrices as rotating a bulk magnetization vector comprising the resultant of many millions of individual spins.

However, I emphasize that the there is no logical requirement to view things this way. The simple view is that the Pauli matrices give a quantum rotation and the Bloch matrices give a classical rotation. Nonetheless, you will see purely quantum discussions which refer also to the Bloch sphere-- I don't think this should cause you any confusion.

I hope this helps a bit; do not hesitate to request clarification.


In quantum mechanics an "observable" like the position of an atom, $\hat X$, is represented by a Hermitian operator. An "observable" is some physical quantity you can actually go out and measure, like the position. The eigenvalues of that operator are the possible measurement results.

In quantum mechanics a "system" (some physical entity that has properties that can be measured) like an atom is described as being in some "state", represented as a vector, e.g., $|\Psi\rangle$, in the same space as the above-mentioned operator.

Given an infinite collection of identically prepared systems, all in the same state, the average value you will find if you measure the observable on each system is $$ \langle\Psi |\hat X|\Psi\rangle $$

The result of any individual measurement you perform must end up actually being one of the eigenvalues of the operator representing the measurement.

For example, if you measure the position, you might find the value $x_0$, which is an eigenvalue of $\hat X$. And then, in this case, the state of the system after the measurement is $$ |x_0\rangle\;. $$

The case of the position operator is actually a little more complicated than the case of spin 1/2 (because of the dimensionality and the non-normalizability).

For the case of spin

For more information, check out a reputable Quantum Mechanics book that covers the basics. For example "Quantum Mechanics" by Cohen-Tannoudji et al. or "Quantum Mechanics" by Messiah.


For your first questions:

Observables correspond to hermitian (or self-adjoint) operators.

As such, the eigenvalues are real, and these values are the possible outcomes.

Also because of the hermitian/self-adjointness of the operator, when you have eigenvectors with different eigenvalues, the eigenvectors are orthogonal.

So you know the values, but what happens? The shortest story is that a measurement projects the vector onto an eigenspace, and you get the eigenvalue corresponding to that eigenspace, and the relative frequency of getting different results is the relative squared-length of the projections of the original vector onto the orthogonal eigenspaces. And any book will tell you this. To visualize this, you can imagine projecting onto an eigenspace, and then rescaling to be unit length again (so this is not a rotation). And imagine that when you do this you get the eigenvalue (somehow) as a result, and then you can imagine that it happens probabilistically in some way with a frequency equal to the ratio of the post-projection squared length over the pre-projection squared length.

But you wanted more. How does measurement relate to unitary evolution? Two parts remain the same when you look at the process realistically. First, the ends results are indeed orthogonal. Second, the experimentally observed frequency is equal to what the post-projection squared length over the pre-projection squared length would have been had there been a projection. So now let's look at what happens.

What happens is that the system evolves according to the Schrödinger equation. And we call something a measurement when it evolves into a sum of orthogonal states, that will always and forever more remain orthogonal. A common way to achieve orthogonality is to have no spatial overlap, for instance a stern-gerlach device can deflect the single beam into two nonoverlapping beams. Wavefunctions are in configuration space, and configuration space is staggeringly vast, so once those beams start to interact with large numbers of different particle to make them move differently, the wavepackets are incredibly unlikely to ever overlap again. This is a prerequisite to calling an evolution a measurement.

The other thing you need, is to have those wavepackets be (at some moment after they become forever orthogonal) eigenvectors to the observable. So for instance, in the stern-gerlach, the spin (bi)vector for the two spatially nonoverlapping beams needs to become polarized, all spin up in one beam and all spin down in the other beam. How does this happen? Well, the Hamiltonian for an inhomogeneous magnetic field does this naturally, an example is available at [http://dx.doi.org/10.1119/1.4848217](this nice article in the American Journal of Physics), [http://arxiv.org/abs/1305.1280](arxiv version). If you don't want to read the article, the punchline is that the single beam bifurcates, and a volume goes one way and a volume goes the other way, and the relative volume depends on how much the original state had of the up and down, and all the volume that goes one way has the spin polarize one way and every part of the volume that goes the other way has the spin polarize the other way. This is literally how you get the repeatability of identical measurements, and the relative fractions. All from the Schrödinger equation.

This is how a measurement happens, and this allows weak measurements too, which are often what actually happens, and sometimes is what is desired. Plus it's what is described by the actual evolution equations. And is what is observed by the lab, and it requires that you actually interact with the system to do a measurement rather than wave your hands and hope a measurement happens.

But what about the probabilities? When the beams deflected and polarized, the squared norm of each deflected wave is equal to what the post-projection squared length over the pre-projection squared length would have been had there been a projection. So that part is, again, already taken care of by the Schrödinger equation.

But since all the beams exist according to the Schrödinger equation it might look like a measurement hasn't happened. After all, some of the beam went left, and some went right. But not only are the beams orthogonal, they must always and forever more remain orthogonal, which actually requires that each now acts like the other one doesn't exist. Each wave is in configuration space, the configuration space of every particle in the entire universe, so the universe now has two parts, each of which acts as if it were all by itself, it is exactly that sense in which the measurement has "happened" with different outcomes. There are now parts of the wavefunction that all act independently. And there is no harm done if the people in each part (whose particles are part of the configuration space) each decide to act as if the other parts don't matter. So at any point they can disbelieve the existence of the other parts, and it will not contradict anything about the evolution of their part of the wave.

This is exactly why I described ratios of post-projection and pre-projection. If your want to rescale your part because it will never be influenced by the other parts, it changes nothing. The overall norm affects nothing, only relative magnitudes matter, and even then it only affects the math if the waves are not orthogonal. So at some point you can rescale (or not), and at some point you can act like a measurement happened (as long as they are now orthogonal, will remain orthogonal forever more, and were at one time in separate eigenspaces).


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.