In quantum mechanics, why is it useful to apply operators to anything other than eigenfunctions, when only eigenfunctions correspond to observables?

For context, I was looking at the definition of the operator postulate from hyperphysics given here, which matches up pretty well with how I remember Griffiths explaining the definition of an operator in QM in his QM text.

The mathematical operator $$Q$$ extracts the observable value $$q_n$$ by operating upon the wavefunction which represents that particular state of the system.

If all an operator does is "extract the observable value", then it should always return a scalar, since we always just measure a scalar value, right? And that's presumably why the postulate is given in terms of eigenvalues. But what doesn't make sense is, if an operator always return scalar values, then isn't it just a scalar function in the first place and so wouldn't even have eigenvectors and eigenvalues?

The obvious answer is that operators don't just act on "wavefunctions which represent [a] particular state of [a] system." But then, what else do they act on? Wavefunctions that don't "represents [a] particular state of [a] system"? But what else could wavefunctions represent, if not particular states of a particle or system of particles?

Or, perhaps a better way to put it is, if only the eigenvalues of the wavefunction correspond to something observable, why do we care about the rest of the wavefunction? Like, what useful information, even just as an intermediate step in a calculation to allow us to do the rest of the calculation, do we get from applying an operator to a part of a wavefunction that isn't an eigenfunction?

• there is a relevant question and answer physics.stackexchange.com/q/693702 Commented Mar 27, 2023 at 4:38
• @annav, I don't see how that's relevant. It seems like a pretty different question to me. I'm not asking about what eigenfunctions are -- I already know what they are from linear algebra. I'm asking why we care about parts of the wavefunction that aren't eigenfunctions, since only the eigenfunctions correspond to observables. Commented Mar 27, 2023 at 4:41
• You state "then isn't it just a scalar function in the first place and so wouldn't even have eigenvectors and eigenvalues?" I do not think it is true, that scalar eigenvalues imply a scalar wavefunction. An operator can act on a complex function and give a scalar back. Commented Mar 27, 2023 at 5:00
• @annav, I suppose I was being a bit sloppy with my terminology. I simply meant a function that maps a vector space to it's field of scalars isn't normally considered a vector function (but rather a linear functional, which is probably the term I should have used) and doesn't have eigenvectors or eigenvalues. Commented Mar 27, 2023 at 5:08

Let me try to answer your questions. First a comment: operators do not always return eigenvalues. Sometimes they may return collections of eigenvalues (such as the position vector components, which can be obtained by letting the position operator, 2D or 3D etc, acting on a state), which if they are considered together, the result is a vector, i.e. $$\hat{\vec{r}}\psi(x)=\vec{r}\psi(x)$$ if $$\psi(x)$$ is an eigenfunction of the position operator.

There are wavefunctions that represent a particular state of the system, that is true. But there are also wavefunctions that represent another particular state of the system, and another and another, depending on what you wish to measure. An example, here would be a good idea: let $$|\alpha\rangle$$ be the ket corresponding to the wave function $$\psi_{\alpha}(x)=\langle x|\alpha\rangle$$. One can act with the operator $$A$$ on the ket $$|\alpha\rangle$$ and obtain an eigenvalue $$\alpha$$, whereas at the same time one can act with operator B on state $$|\beta\rangle$$ and obtain an eigenvalue $$\beta$$. $$\alpha$$ and $$\beta$$ are the results we get if we conduct measurements of different physical quantities on different states. For instance, $$\alpha$$ can be a position eigenvalue and $$\beta$$ a momentum eigenvalue.

At the same time, one might want to act with the position operator on a momentum eigenstate, namely of $$|\beta\rangle$$. So, one acts with the operator $$A$$ on $$|\beta\rangle$$. But $$|\beta\rangle$$ is not an eigenstate for the operator $$A$$, so the result is not of the form $$A|\beta\rangle=c |\beta\rangle$$ where $$c$$ is some eigenvalue (scalar if we restrict ourselves to 1 dimensional systems). So, one must expand $$|\beta\rangle$$ in the basis of $$|\alpha\rangle$$, namely write something like $$|\beta\rangle=\sum_{\alpha}c_{\alpha}|\alpha\rangle$$ where the $$c_{\alpha}$$s are constant coefficients, one for each possible state $$|\alpha\rangle$$. So, suppose for instance that $$A$$ again is the position operator and there are two possible positions labelled by the numbers $$1$$ and $$2$$, namely $$|1\rangle$$ and $$|2\rangle$$, with equal probability of measuring both. Then, one can write $$|\beta\rangle=\frac{1}{\sqrt{2}}(|1\rangle+|2\rangle)$$ and hence, $$A|\beta\rangle=\frac{1}{\sqrt{2}}(\alpha_1|1\rangle+\alpha_2|2\rangle)$$ which is not an eigenvalue of the operator $$A$$.

So, to answer to your last question, we do care about the remaining wavefunction, as it gives us information for about other physical quantities. If I only cared about position measurements in my example, for instance, I would only care about the set of eigenfunctions $$\{|\alpha\rangle\}$$, but instead, one may have a more general wavefunction and obtain different kind of information each time one acts on the wavefunction with different operators.

I hope this helps. If there are any questions, please do let me know.

• One important thing that I really think that you should emphasize (and also change the language in your post): you say "At the same time, one might want to measure the position of a momentum eigenstate... So, one acts with the operator A on..." This seems to imply that acting with the operator on the state corresponds to the physical act of measurement of the operator, which it does not, and this is such a common misconception (and can cause such problems with learners of QM), that that should be addressed. Commented Mar 27, 2023 at 19:59
• No, acting with some operator on a vector does not correspond to a measurement. Acting with an operator on a vector is a mathematical operation which yields a new vector. That the expectation value is defined like this has nothing to do with the point mentioned by @march. Commented Mar 28, 2023 at 8:00
• You should use the @username function to notify others. Anyway, I don't understand your query here, to be honest. It is just wrong that the action of an operator corresponds to some measurement... Let me ask you instead: Elaborate how or why do you think this corresponds to a measurement?! Again: if you have some operator/observable $A$ , then $(A\psi)$ is another vector. However, in the simplest case of a non-deg. observable, the state after the measurement is $\psi_n$, where this is the eigenstate of $A$ corresponding to the measured value $a_n$, i.e. $A\psi_n=a_n\psi_n$. Commented Mar 28, 2023 at 10:27
• Again: IMHO it is useless and in general wrong to say that the action of an operator corresponds to a measurement. For example, even if your system is in an eigenstate $\psi_n$, then if you say the action of $A$ corresponds to a measurement, then the new state is $a_n \psi_n$, or what? This is not normalized and can even be the $0$ vector... Commented Mar 28, 2023 at 11:13
• @MikaylaEckelCifrese ...The problem in QM is that there is a large inferential gap between the mathematical operations and the physical interpretations, and you just shouldn't get in the habit of trying to understand the "physical interpretation" of every mathematical operation that you do. Commented Mar 29, 2023 at 3:38

When it comes to observable operators (there are also operators used for different things in QM, for which we almost never solve the eigenvalue problem) - the logic kind of goes the other way around. We need the operator in order to even know what the eigenstates (and corresponding) are. That's really the main purpose of those observable operators. Sure, if you already had the outputs of the solved eigenvalue problem you wouldn't need the observable operators much anymore, except for $$H$$ because it has the additional role that it specifies time evolution in the Schrödinger equation.

Specifically with the momentum operator here's how it goes. We are given an observable operator, let's say (this is the 3d version, to get 1d replace $$\nabla \to \frac{\partial}{\partial x}$$).

$$\hat{p} = -i \hbar \vec{\nabla}$$

From there we want to know what its eigenvectors and eigenvalues are, so we solve the eigenvalue problem for $$\psi_p$$ such that $$\hat{p_i} \psi_p = p_i \psi_p$$ Note that $$p_i$$ without the hat is the eigenvalue for a component of momentum, it is a number. The $$\hat{p}_i$$ is the operator for the $$i$$th component of momentum. We find, in this case, that

$$\psi_p = e^{i\vec{p}\cdot \vec{r}/\hbar}$$

Is the eigenvector, simultaneously an eigenstate of the operators $$\hat{p}_x, \hat{p}_y, \hat{p}_z$$ corresponding to eigenvalues which we can write as a vector $$\vec{p}$$. We therefore call that state $$|\vec{p}\rangle$$, just labeling it by its momentum. This allows us to take the inner product $$|\langle \vec{p} | \phi \rangle|^2$$ to get probabilities for momentum measurements when the system is in a state $$\phi$$.

• "there are also operators used for different things in QM, for which we almost never solve the eigenvalue problem" what else are they used for? Commented Mar 27, 2023 at 4:57
• @MikaylaEckelCifrese So 1. there is the Hamiltonian in the Schrödinger equation. I guess you don't have to think of it as an operator any more than in any other differential equation. 2. Then you can express the time evolution of the state as an operator $\psi(t)=U(t) \psi(0)$, where it turns out $U=e^{-itH/\hbar}$. This ends up often being useful. 3Then there are "projection operators", these become most useful when a measurement is done and the eigenvalue that results from it has more than one corresponding eigenvector. Then the probability of getting that outcome is $|P|\psi\rangle|$ rather Commented Mar 27, 2023 at 5:23
• than, for example, $|\langle x|\psi \rangle|^2$ in the case of position. Those above are the most "foundational" operators in QM, but along the way there are operators we use for miscellaneous purposes. One common one is the raising/lowering operators, which come up when solving the Schrödinger equation for a particle in a potential well $V(x)=kx^2$. This example, called the "harmonic oscillator", is an important example because so many real situations can be approximated by it. Applying the raising operator to one solution in this potential gets you another solution of a higher energy. Commented Mar 27, 2023 at 5:28
• I would also like to mention that, while many of these operators are here to stay, there is an alternative formulation of non-relativistic quantum mechanics called Bohmian Mechanics, which doesn't use observable operators at all (at least not in its foundations, but they can be derived in limiting cases). It also doesn't use projection operators. So it really just has the Schrödinger equation, and some of those miscellaneous operators. It's a nearly operator-free formulation of QM. Commented Mar 27, 2023 at 5:30

First, some background, in which I'll ignore subtleties and just address the main ideas. Solving the eigenvalue equation for a normal operator (of which self-adjoint operators are a special case) is essentially just a route to the so-called spectral decomposition of the operator, which takes the form $$\hat A = \sum_n \lambda_n \hat P_n$$ where $$\lambda_n$$ is the $$n^{th}$$ eigenvalue of $$\hat A$$ and $$\hat P_n$$ is the orthogonal projection operator which acts on a vector by projecting it into the eigenspace corresponding to $$\lambda_n$$. Assuming that $$\lambda_n$$ is non-degenerate, $$\hat P_n$$ takes the familiar form $$|\phi_n\rangle\langle \phi_n|$$ where $$|\phi_n\rangle$$ is the (normalized) eigenvector with eigenvalue $$\lambda_n$$.

This decomposition is important for a number of reasons. First, if $$\hat A$$ is a self-adjoint operator representing some observable, then the $$\mathbb R$$-valued eigenvalues correspond to possible measurement outcomes. If a system is in a pure state with (normalized) state vector $$|\psi\rangle$$, then the probability of measuring $$\hat A$$ to take the value $$\lambda_n$$ is given by $$\langle \psi|\hat P_n |\psi\rangle$$, and the (unnormalized) post-measurement state is given by $$\hat P_n|\psi\rangle$$. Therefore, the spectral decomposition of an operator allows you to compute the probability of all possible measurement outcomes, as well as the post-measurement state obtained in each case.

The decomposition is also important if we wish to compute a function of the operator. Given an ordinary function $$f:\mathbb C\rightarrow \mathbb C$$, we can define the operator $$f(\hat A)$$ as $$f(\hat A) = \sum_n f(\lambda_n) \hat P_n$$ Of particular importance is the time evolution operator which tells us how to evolve our state vectors forward in time; for a time-independent Hamiltonian $$\hat H$$, the time-evolution operator is given by $$\hat U(t) = \exp\big[-i\hat H t/\hbar\big] = \sum_n e^{-iE_nt/\hbar} \hat P_n$$ Other examples include the rotation operators, which are obtained by exponentiating the angular momentum operators, and inverse operators of the (formal) form $$\hat A^{-1} \equiv 1/\hat A$$ which are of great utility in e.g. perturbation theory.

But what doesn't make sense is, if an operator always return scalar values, then isn't it just a scalar function in the first place and so wouldn't even have eigenvectors and eigenvalues?

When people say that operating on an eigenvector turns the corresponding eigenvalue, what they mean is that the eigenvalue can be determined simply by looking at the resulting expression. The operator $$-\frac{d^2}{dx^2}$$ acting on $$\sin(2x)$$ yields $$\underbrace{-\frac{d^2}{dx^2}}_{\text{operator}} \underbrace{\sin(2x)}_\text{vector} = \underbrace{4}_\text{eigenvalue} \underbrace{\sin(2x)}_{\text{vector}}$$

In other words, the operator does not literally return $$4$$ as an output, but by acting on the vector with the operator we can read off the eigenvalue from the result.

Or, perhaps a better way to put it is, if only the eigenvalues of the wavefunction correspond to something observable, why do we care about the rest of the wavefunction? Like, what useful information, even just as an intermediate step in a calculation to allow us to do the rest of the calculation, do we get from applying an operator to a part of a wavefunction that isn't an eigenfunction?

I'm not really sure what to make of this question. The last part is answered in part by my background section above; there are many operators (e.g. symmetry transformations like rotation operators or translation operators, or the time evolution operator) which are interesting and useful because of the transformation they impose on the state vector of the system, not because they correspond to observables.