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One of the postulates of quantum mechanics is that for every observable $A$, there corresponds a linear Hermitian operator $\hat A$, and when we measure the observable $A$, we get an eigenvalue of $\hat A$ as the result.

To me, this result seemed to come out of nowhere. While I could understand representing an observable $A$ by a linear operator $\hat A$, I can't understand why must the results of measuring $A$ have to be an eigenvalue of $\hat A$. Is it possible to better motivate this postulate?

Edit: Since my question might be a little vague, let me try to rephrase it - how can one motivate this postulate to a student first being introduced to Quantum Mechanics? Are there experimental results, for example, which can be used as motivation?

Ishan Deo
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    This is not quite true. A measurement will give a value in the spectrum of the operator, but not necessarily an eigenstate. – Slereah Oct 04 '20 at 15:57
  • What exactly is the difference between the two? – Ishan Deo Oct 04 '20 at 18:45
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    @IshanDeo Try reading here. – Charlie Oct 04 '20 at 22:58
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    Consider reading one of the standard textbooks on functional analysis. The spectrum of an operator is not the same as the set of eigenvalues of an operator. – Tom Oct 05 '20 at 04:44
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    @Slereah Was the last word of your comment supposed to be "eigenvalue"? – Acccumulation Oct 05 '20 at 05:25
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    @Tom well, but you can just look at the generalized eigenvectors (nuclear spectral theorem). So basically physicist got it right in the begining, just the formalization took a bit longer. – lalala Oct 05 '20 at 10:08
  • If you are new to QM do not trouble yourself with functional analysis before you have understood the basics. – Charlie Oct 05 '20 at 13:46
  • It didn't come out of nowhere: it came out of the lab. It's the math that happened to fit what was observed. – John Doty Oct 05 '20 at 19:12
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    And my question is - what experiments in the lab suggest an eigenvalue formulation – Ishan Deo Oct 05 '20 at 23:29
  • I'm surprised that none of these answers mentions what happens physically when you do a measurement on a quantum system. OP do you happen to know anything about the density matrix picture? – DanielSank Oct 10 '20 at 16:54

7 Answers7

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One of the postulates of quantum mechanics is that for every observable A, there corresponds a linear Hermitian operator A^, and when we measure the observable A, we get an eigenvalue of A^ as the result.

In spirit, yes. For technical reasons, this is not quite true. As mentioned by Slereah in the comments, the more precise statement is that a measurement of $A$ returns values which lie in the spectrum of $\hat A$. If the spectrum of $\hat A$ is purely continuous, as is the case for the position observable for a particle on a line, then $\hat A$ does not actually have any eigenvalues because there are no states $\psi$ in the Hilbert space such that $\hat A \psi = \lambda \psi$ for some complex number $\lambda$.

This introduces technical difficulties, but the takeaway which is relevant to this discussion is that to each observable $A$, there corresponds to a self-adjoint operator $\hat A$, and when we measure the observable $A$ we get a result which lies in the spectrum of $\hat A$.

This can be motivated in a few ways, but my favorite is the following. Note that this was not the historical route to quantum mechanics, which was filled with lots of twists and turns and dead ends.

If we view classical physics through the lens of Hamiltonian mechanics, we can define an observable as a continuous function from the phase space variables (the generalized coordinates and momenta) to the real numbers. With some extremely mild additional assumptions like the connectedness of the phase space, this immediately implies that the possible outcomes for measurements take the form of connected intervals in $\mathbb R$. For example, the possible positions of a point on an infinite line is given by $\mathbb R$, the possible kinetic energies for such a particle is the interval $[0,\infty)$, and the possible z-coordinates for a particle attached to a unit sphere is $[-1,1]$.

The results of the Stern-Gerlach experiment (in which the possible z-components of the spin angular momentum are $\{\frac{\hbar}{2},-\frac{\hbar}{2}\}$) and the emission spectra of hydrogen (in which the possible bound-state energy constitute the discrete set $\{-\frac{13.6\text{ eV}}{n^2}\}$) immediately fly in the face of this result. We also now understand that e.g. the energy spectra of solids lie in disconnected bands, which is once again incompatible with the previous line of reasoning.

There is no clear way to modify Hamiltonian mechanics to account for these possibilities, so we are motivated to seek out an entirely different framework which can. As it turns out, the spectral theory of linear operators on Hilbert spaces contains precisely the flexibility we need. A generic operator $\hat A$ on a Hilbert space has $\sigma(\hat A)\subseteq \mathbb C$, so in the context of observable quantities it is reasonable to ask which operators have spectra which lie entirely in $\mathbb R$; the answer is that $\sigma(\hat A)\subseteq \mathbb R \iff \hat A$ is self-adjoint$^\dagger$.

As a result, we say that to our system we associate a Hilbert space, which takes the place of the phase space from classical physics and whose elements (roughly) constitute the space of possible states of the system. Observable quantities are now represented by self-adjoint operators, and their spectra correspond to possible measurement outcomes.

A generic element of a finite-dimensional Hilbert space can be decomposed into a linear combination of eigenvectors of whatever self-adjoint operator you wish. If $\hat A$ has $\lambda$ as an eigenvalue, then it does not seem unreasonable to guess that the corresponding eigenstate is one for which measurement of $A$ returns precisely $\lambda$. The situation is more complex when the spectrum of the operator is continuous, but the spirit of the argument remains the same.

Of course, none of this is mathematical proof that we're making the right choices - indeed, no such proof could possibly exist. All we can do is to throw these ideas together into a coherent framework, make predictions, and compare with experiment. As it happens, this particular recipe is enormously successful - though that does not rule out the possibility that it will someday be replaced with something better.

$^\dagger$This is not entirely true - see here for a generalization. However, it is a good starting point for the standard formulation of QM, which can then be extended.

J. Murray
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  • Interesting point about the outcomes being in a connected set in Hamiltonian Mechanics. Thanks! – Marc Oct 05 '20 at 05:16
  • So this basically boils down to continuity: In the Hamiltonian framework observables need to be continuous, but they don't have to be in quantum mechanics. – Jannik Pitt Oct 05 '20 at 08:54
  • It's been a long time I did not study physics but if $\hat{A}$ is the position operator, don't we have $\hat{A} \psi(x) = x \psi(x)$? The $x$ are eigenvalues and $\psi(x)$ eigenstates, aren't they? – Evariste Oct 10 '20 at 17:18
  • @Evariste Eigenvalues are constants, not functions. You’d need to have that $\hat A\psi(x)=\lambda \psi(x)$ for some constant $\lambda$. Delta functions like $\delta(x-\lambda)$ have this property for the position operator, but they are not elements of the Hilbert space because they’re not square-integrable. As a result, the position operator has no eigenfunctions/values. These technical details are cleared up by developing the full machinery of the spectral theorem, but it requires a substantial amount of work to do so. – J. Murray Oct 10 '20 at 17:38
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The Stern-Gerlach experiment and similar experiments show

  • A system has a state.
  • The states of a system form a Hilbert space. You can choose a set of basis states and represent the system's current state as a sum of those basis states. E.G. For electron spin, your choice of up/down, left/right, or states at some other angle.
  • A measurement is a physical interaction that changes the state of the system and produces a measured value. In general, the measured value is probabilistic even if the state is known. E.G. A spin left state in a spin up/down measurement produces spin up and spin down results with equal probability.
  • A measurement changes the state to one consistent with the measured value. E.G. If a spin up/down measurement produces a spin up result, the system is in a spin up state. Another measurement will also produce a spin up value.

Measurements transform one state in a Hilbert space to another. This is just what operators on the Hilbert space do.

A measurement leaves some states unchanged and produce a predicable value. Some operators leave states unchanged. Those states are called the eigenstates of the operator.

A very similar operator takes the state to a scalar multiple of itself. This operator can represent both the effect of the measurement on the state of the system and the measured value. The scalar multiple/measured value is call an eigenvalue of the operator. This gives us $\hat{A} \left|a\right> = \lambda \left|a\right>$

Measured values are real. The eigenvalue is real when the operator is self-adjoint.

The magnitude of $\left|a\right>$ isn't important to us, so we can require that $\left<a|a\right> = 1$ for all states. This normalization works well when we work with basis states and probabilities.

mmesser314
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By claiming that the observable $A$ is representable by an operator which has specific eigenvalues, you are claiming that the only possible outcome of measuring $A$ are those eigenvalues. After you have measured the system and you have made sure that the outcome is some eigenvalue $a_n$ then you are also sure that the system is in a state that is attributed to the eigenvalue $a_n$, hence a specific eigenstate. For example, you can try to measure if a particle in a box is on the left part of the box or on right part. This measurement is described by an operator which only has two eigenvalues and eigenstates, because the outcomes we are looking for are only two distinct outcomes.

Alphy
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The measurement postulate of quantum mechanics can be formulated like this:

Measurement of observable $A$ is modeled as a probabilistic process: With probability $p_i$ it will give the result $a_i$ (an eigenvalue of $A$) while throwing the state from $|\Psi\rangle$ (a normalized vector) to $|a_i\rangle$ (a normalized eigenvector of $\hat{A}$). The probabilities are given by $p_i=|\langle a_i|\Psi\rangle|^2$.

A measurement needs to meet the following physical requirement.

When a measurement of observable $A$ on a state has given the result $a_i$, then repeating the measurement immediately again will give the same result $a_i$ again. This is a basic requirement for any measurement (otherwise we would not even call it a measurement). Countless experiments (the Stern-Gerlach experiments being the first ones) confirmed this requirement.

The postulate from above is motivated by the fact that it meets this requirement:

When you measure $A$ on the state $$|\Psi\rangle=\sum_i c_i |a_i\rangle$$ then with probability $|c_i|^2$ you will get result $a_i$ and throw the state to $$|\Psi'\rangle=|a_i\rangle.$$ When you now repeat the same measurement of $A$ on that new state, the postulate predicts that you certainly (with probability $100$%) get this same result $a_i$ and again the state $$|\Psi''\rangle=|a_i\rangle.$$

Thomas Fritsch
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In physics it is not really meaningful to ask why nature happens to be the way it is. If your question is essentially why does the numerical outcome of a measurement have to be an eigenvalue (or more strictly, as Slereah has said, an element in the spectrum) of the operator then the only admissible answer is "because that is what makes accurate predictions".

Charlie
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  • @JacopoTissino I'm not sure what you mean, there are pre-existing experiments that the theory must agree with but quantum mechanics isn't designed to predict any particular "kind of experiment". – Charlie Oct 04 '20 at 16:40
  • Perhaps the question should be rephrased. What is it about operators that make them model nature better than other mathematical constructs? If I were trying to construct a theory of quantum mechanics, what would lead me to consider operators? – mmesser314 Oct 04 '20 at 17:36
  • @mmesser314 I honestly don't think that question is satisfyingly answerable either, it's a philosophical question. "Why operators instead of X other mathematical object?" is only really answerable with "because that's what works". At a certain point you're asking why nature seems to have rigid, predictable structure, which is a question for a philosopher not a physicist. Perhaps I am misunderstanding what you've written. – Charlie Oct 04 '20 at 17:43
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    Ok then let me rephrase this as such - what experimental results motivated the founders of QM to consider eigenvalues of operators as the results of measurements on a system. – Ishan Deo Oct 04 '20 at 18:34
  • If that's the question you're asking then this may be better suited to the History of Science and Mathematics stackexchange. I don't know what exactly led the founders of quantum mechanics to use this specific branch of linear algebra. – Charlie Oct 04 '20 at 18:36
  • We don't actually know that nature is this way, rather there are a series of experiments which are consistent with the eigenvalue postulate. A proper answer would IMO give justification for the postulate, I think it is a reasonable question. Everything worthwhile in physics has a justification, this included. – doublefelix Oct 04 '20 at 22:07
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    @doublefelix Sure, I would go even further and say that "nature is this way" is not even a meaningful statement. The purpose of theories in physics is not to show that the mathematical objects used in physics are real things, rather that there seem to exist similar patterns in nature. Non-relativistic QM stops being predictive under certain conditions, it's not clear whether there even exist mathematical models that are arbitrarily accurate. – Charlie Oct 04 '20 at 22:55
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    My point wasn't that it's not a good question, rather that it's not a question we have an answer to. The justification for QM existing in its current form in physics is simply "because it works", that is the one and only reason. If you want to press harder and ask why nature has chosen to be this way you're asking a philosophical question. If OP is asking what historical discoveries led to QM existing in its current form that is a question for HSM. – Charlie Oct 04 '20 at 22:57
  • We do have an answer to the question "What experimental evidence is there which can be used to justify the eigenvalue postulate?", which is a reasonable interpretation of OP's question given his comments here. The answer to the above question is also not necessarily a historical one, as it need not be the same method of inference used when the postulate was first come up with. For this reason I cannot sympathize with your argument. – doublefelix Oct 05 '20 at 13:04
  • @doublefelix What OP has now added as an "edit" to the original equation has different connotations to the original post. If OP just wants a list of experiments for which non-rel QM is accurate then that of course has an answer. The original question seemed (to me) to be asking why measurement has to specifically return an eigenvalue and not, say, the square root of the eigenvalue. This would be asking why the postulates of QM were chosen the way they were, to which the correct and only answer is "because that it what makes QM predictive within its experimental scope". – Charlie Oct 05 '20 at 13:49
  • I am being semantic, and perhaps OP did not intend the question to have these connotations, but that is how the question read at the time of posting. – Charlie Oct 05 '20 at 13:50
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I think that the best example to motivate a student is the spin.

Particles prepared so that their spins are +1/2 in a given lab z-direction, if measured by an apparatus tilted arbitrarily with respect to the prepared orientation, have spins sometimes +1/2 and sometimes -1/2.

But the expected value of the average of a great number of measurements can be calculated by the eigenvectors $|S\rangle$ of the matrix resulting from the linear combination of the Pauli matrix: $\sigma_k = n_x\sigma_x + n_y\sigma_y + n_z\sigma_z$, where $n_i$ are the componentes of the unit vector of the new orientation.

$$E_k = \frac{1}{2}\langle S|σ_k|S\rangle$$

At least when that mathematical procedure was developed, it only happened to match the empirical data. The matrices resulting form the linear combination have always the same 2 eingenvalues.

Claudio Saspinski
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You might want to have a look at the ideas of Quantum Darwinism. I am not sure how popular this thoughts are, so decide for yourself.

As far as I understand there an attempt is made to explain why certain states are measured, based on how "stable" they are compared to other states when interacting with the measurement device and the environment.

amh23
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