Let's start from the beginning:

The state of a quantum particle is represented by a vector $|\psi\rangle$ in the Hilbert space, observables are represented by Hermitian operators which eigenvalues represent the possible outcomes of a measurement and which eigenvectors represent the possible states after the measurement. One fundamental postulate is that the eigenvectors of any observable form a complete basis for the possible states of the particle. At my current level of understanding I think this is simply a postulate and cannot be demonstrated. Suppose we are dealing with an observable $R$ with only two possible results for a measurement: $+r,-r$, then we can write that the generic state of a quantum particle is: $$|\psi\rangle = a_+|+\rangle +a_-|-\rangle \ \ \ \ \ \ (1)$$ where $|+\rangle,|-\rangle$ are the eigenvectors of $R$. The probabilities of one outcome or the other are respectively: $|a_+|^1,|a_-|^2$, this is also a postulate.
But what to do if our observable is, for example, the position, with an infinite number of possible measurement outcomes? We start by noticing that we can rewrite (1) as: $$|\psi\rangle = \langle +|\psi\rangle|+\rangle +\langle -|\psi\rangle|-\rangle \ \ \ \ \ \ (2)$$ this is just a mathematical property of the Hilbert space. We can then think to have an infinite number of basis vector with an infinite number of coefficients: $$\psi=\langle x_1 | \psi\rangle|x_1\rangle+\langle x_2|\psi\rangle|x_2\rangle+.....$$ this reasoning is not really rigorous, but in the limit of dense and infinite eigenstates allows us to think about the collection of coefficients as a function: $$\psi(x)=\langle x |\psi\rangle$$ then to find the probability of measuring $x$ in an interval $(a,b)$ we can think to sum all the probabilities of $x$ in that region, but since we are in the mentioned limit we integrate instead of summing: $$P(a<x<b)=\int_a^b|\psi(x)|^2dx$$ From what I currently understand this is what a wave function is!

Long story short: for me the wavefunction of a particle is a complex valued function that when integrated in its variable/s gives the probability of finding the particle under the interval of integration.

But I started to doubt my understanding, I think that the wave function may be more versatile than this. To explain why lets take an exercise as an example:

Given the wave function: $$\psi(r,\theta,\phi)=Ae^{-br}(1+2br\sin{\theta}\sin{\phi})$$ Find the possible outcomes of a measurement of $L^2,L_z$ with relative probabilities.

I strongly suspect that what I am about to say will appear ridiculous to a seasoned expert on this topic, but unfortunately I am not an expert, so hear me out:
Given my understanding of what a wave function is this question does not make sense! If you give me a wave function in the variables $r,\theta,\phi$ I can then give to you the probabilities for the particle to be in an area of $3D$ space. Because this is what a wave function allows us to do by definition, this is the information that it's carrying by construction! But how and why can I infer information about $L^2$ and $L_z$ with the wave function only?

This exercise of course has a standard procedure of resolution: write the wavefunction in terms of the eigenfunctions of $L^2,L_z$, so the Spherical Armonics $Y_{l,m}(\theta,\phi)$; if we do this we get: $$\psi=Ae^{-br}\sqrt{4\pi}\left[Y_{0,0}+\sqrt{\frac{2}{3}}ibr(Y_{1,1}+Y_{1,-1})\right]$$ and then we can somehow find the probabilities of $l=0,m=0$, ecc. by integration. Problem is I don't understand how or why this method works! I mean: the rewriting in terms of $Y(\theta,\phi)$ is mathematically ok, but why is it useful? The wave function, written in one way or the other, should still carry only the information about the probability of finding the particle in a region of space. How can we extract the information we need and why?

Is my definition/understanding of the wavefunction wrong? Is it more versatile than I thought? How?

  • $\begingroup$ The wavefunction contains all of the information of the current state. This includes not only positional data but crucially also momentum-like data (which often shows up as the phase). I don't think you'd find it surprising that you could extract angular momentum data from the joint probability distribution $P(x,p)$ would you? $\endgroup$
    – jacob1729
    Dec 11, 2020 at 16:37

3 Answers 3


The wavefunction is certainly more versatile that you are giving it credit for. The wavefunction contains all of the information about the particle. This may be spatial information, as in the case you are considering, but it could be "internal" information about the particle, that has nothing to do with any spatial position or momentum. For example, quarks come in three "colors", and "which color" it is is encoded by a part of the wavefunction that, rather than being a function of $x$, is a function of a discrete index $i=\{1,2,3\}$.

This could be confusing if you are thinking of the wavefunction as a literal function of space, the end. The correct way to understand it is as you said in the quoted passage: it is a vector in an abstract space that we call Hilbert space. Now what is the most basic thing you know about vectors? "You can describe them in many different bases." Yes. So when you talk about the wavefunction as a function of $x$, you mean you are describing this vector in the basis of spatial points: this vector has some component along each of these basis directions, that is, at each point $x$. When you convert to spherical harmonics you are just changing the basis to the basis that corresponds to angular momentum. And in this basis the probabilities for being in different angular momentum states can just be read off, the way the probabilities for being in different spatial regions could be read off before.

So your question

The wave function, written in one way or the other, should still carry only the information about the probability of finding the particle in a region of space. How can we extract the information we need and why?

is equivalent to the following question about the vectors you are used to:

This vector is currently expressed in a certain basis, so it only carries information about its components along that basis. How can I ever extract information about its components along a different basis?

Except you know the answer to that. The vector in one basis still contains all the same information as it would if expressed in a different basis. To find one from the other you just do some trigonometry. It's the same thing with the wavefunction...

So how does this "trigonometry" look in quantum mechanics? I'll use Dirac notation because it looks like you are comfortable with that. In that notation the wavefunction as a function of space is $\psi(x) = \langle{x}|{\psi}\rangle$, that is, the vector $|\psi\rangle$ projected onto the basis $\langle x|$. Now we could have expressed the wavefunction instead in the basis of spherical harmonics. Let's write that as $\psi_{lm} = \langle lm|{\psi}\rangle$. These things would be the complex numbers you would square to get probabilities of measuring various angular momentum values! The remaining question is how to convert from one to the other. In order to do that, we will need to express the $\langle lm|$ in terms of the $\langle x|$. Let's call this $Y_{lm}(x) = \langle x|lm\rangle$. (I'm still writing $x$ here, but of course $x$ can be expressed in terms of $\theta$ and $\phi$.) The famous $Y_{lm}$s are the projections of the angular momentum basis vectors onto the position basis vectors. Lastly, we just need to insert the $\langle lm|$ into our starting expression $\langle x|\psi \rangle$. Because the $\langle lm|$ form a complete basis, if we sum over them it gives the identity operator in the following sense: \begin{equation}1=\sum_{lm} |lm\rangle \langle lm| \end{equation} where the sum runs over the appropriate values of $l$ and $m$. (I assume you are familiar with this kind of identity. It seems like you are based on your original post.) Since this is equal to 1, we are free to insert it in our original expression \begin{equation} \psi(x)=\langle x|\psi \rangle = \sum_{lm} \langle x|lm\rangle \langle lm|\psi\rangle = \sum_{lm} Y_{lm}(x) \langle lm|\psi\rangle. \end{equation} Recall that $\langle lm|\psi\rangle$ gives the probabilities that we are interested in. So by looking at this expression, we see that they are the coefficients of the $Y_{lm}$ which you can now read off from the expression in your post.

  • $\begingroup$ I used the indent block to separate arguments, the text is not a direct citation; it is merely my current understanding of the definition of wave function. But of course I derived the definition by studying multiple books on the topic. $\endgroup$
    – Noumeno
    Dec 10, 2020 at 18:11
  • $\begingroup$ Then you were completely right :) $\endgroup$
    – kaylimekay
    Dec 10, 2020 at 18:14
  • $\begingroup$ The problem with your argument is that the wave function is still written in the basis of the angular coordinates, there is no change of basis whatsoever. I informally kinda understand what you are trying to say but I would like a formal explanation. The base remains angular coordinates in the first form and also in the spherical armonics one since the spherical armonics are function of the angular coordinates. $\endgroup$
    – Noumeno
    Dec 10, 2020 at 18:15
  • $\begingroup$ Also you said nothing about my example and the math behind it, and how to find the probabilities, which is one of the main points of my question. $\endgroup$
    – Noumeno
    Dec 10, 2020 at 18:17
  • $\begingroup$ OK I answered the 'why' questions and hoped that 'how' would be clear from that, but I'll update my answer to try and be clearer. $\endgroup$
    – kaylimekay
    Dec 10, 2020 at 18:25

First, a nice review of vectors. If we have a vector $\mathbf v$, we can express this vector as a linear combination of basis vectors. And there is not a unique way to do this. The typical example we first see is something like $\mathbf v=v_x\,\hat x+v_y\,\hat y+v_z\,\hat z$, where the basis vectors align with the coordinate axes. But we can choose some other set linearly independent basis vectors that are not aligned with the axes if we want to. In general we just have $\mathbf v=v_1\,\hat e_1+v_2\,\hat e_2+v_3\,\hat e_2$.

Even more important is the fact that if we know how the basis vectors $\hat x,\hat y, \hat z$ can be expressed in terms of $\hat e_1,\hat e_2,\hat e_3$, then we can easily move from describing $\mathbf v$ in the first basis to describing $\mathbf v$ in the second bases (or vice versa).

Now, onto your question

But how and why can I infer information about $L^2$ and $L_z$ with the wave function only?

The information of a quantum system are mathematically encoded in the state vector $|\psi\rangle$. What we say by the "wave function" $\psi(\mathbf r)$ is really just a function describing the components of the state vector $|\psi\rangle$ in the position basis (i.e. the basis formed by the continuous set of position eignstates $|x\rangle$).

Furthermore, if we know how the position basis vectors $|x\rangle$ relate to angular momentum basis functions $|\ell,m\rangle$ (through the spherical harmonic functions), then we instantly know the angular momentum information of $|\psi\rangle$ by relating $\psi(\mathbf r)$ (the information of the "position components") to information about the "angular momentum components".

But in order to use the wave function in this way, to find results about the angular momentum, the wave function should be expressed in the base of the eigenstates of $L_z$, right? We don't do this! Instead we rewrite it with the eigenfunctions, which is not the same thing, or at least is not obvious.

So here is what you are doing. You can express your state as a linear combination of angular momentum states: $$|\psi\rangle=\sum_{\ell,m}\psi_{\ell,m}|{\ell,m}\rangle$$

and then you are looking at the position basis components of this $$\langle \mathbf r|\psi\rangle=\psi(\mathbf r)=\sum_{\ell,m}\psi_{\ell,m}\langle\mathbf r|{\ell,m}\rangle$$

and these $\langle\mathbf r|{\ell,m}\rangle$ are your spherical harmonics. Therefore, if you can express $\psi(\mathbf r)$ as a linear combination of spherical harmonics, you know $\psi_{\ell,m}$. If you know $\psi_{\ell,m}$ then you know what $|\psi\rangle$ is in terms of angular momentum eigenstates. And, of course, $|\psi_{\ell,m}|^2$ tells you the probability of measuring the state $|\ell,m\rangle$ given the initial state $|\psi\rangle$.

  • $\begingroup$ But in order to use the wave function in this way, to find results about the angular momentum, the wave function should be expressed in the base of the eigenstates of $L_z$, right? We don't do this! Instead we rewrite it with the eigenfunctions, which is not the same thing, or at least is not obvious. $\endgroup$
    – Noumeno
    Dec 10, 2020 at 18:22
  • $\begingroup$ Also how do we go about finding the probabilities in my exercise? And why? What is the formula for them? Can you expand on those issues in your answer? $\endgroup$
    – Noumeno
    Dec 10, 2020 at 18:24
  • $\begingroup$ @Noumeno I will update when I can :) $\endgroup$ Dec 10, 2020 at 18:24
  • $\begingroup$ @Noumeno It is updated $\endgroup$ Dec 10, 2020 at 18:34

Your statement about what a wave function is is not at all incorrect and your second example does not in any way contradict your first statement! The utility of the wave function is accurately and succinctly explained. Your second example merely expresses a wave function in a different set of coordinates, spherical. But you are still justified in saying that the integral of this function is done correctly (and your statement does have a mistake in it) it measures the probability of finding the particle within the interval of integration. In 3D this interval would be a 3D cell. Your spherical coordinate example does not violate your expectations.

To correct your mistake it is NOT the wave function that is integrated but its magnitude squared, $|\psi|^2$, which measures the probability density.

As for getting measurements of other quantities, it is not the utility of the wave function that you are asking about but the utility of the mathematical abstraction of linear operators acting on a function space. This is present even in classical field theory, acoustic, electromagnetics, etc.

The ability to "infer" information about other quantities is built into the postulates of quantum theory. Interpreting $|\psi|^2$ as a probability density then the expectation value of any variable, $v$, denoted $<v>$ , related to the particle would be the integral of $v |\psi|^2$. This is just classical probability theory applied to the measurable quantities related to the state of a particle. One major difference, expressed in the postulates of QM is that an observable is represented as an operator which acts on the functions in Hilbert space. So the QM version of the expectation value is $<\psi|v\psi>$ where $v$ is now an operator. Based on this one can get an estimate for any well defined quantity using the wave function and an operator for the quantity.

As for the last example which discusses a wave function that is written as a sum over other functions, again this is more of a math principle that has been mapped to a physics postulate. Hilbert space is an abstraction of a linear vector space and like all linear spaces any function can be expressed as a sum over an orthonormal basis, as other answers point out. This is not necessary for the wave function to be well defined or useful. In contrast, this is a way to use math to get more out of it. Each observable quantity is expressed as an operator (specifically a Hermitian operator) and the eigenvalues of that operator represent the allowed observed values. The solutions to the eigenvalue equation for each and every such operator for a basis for the space of functions. Any function in the space can be written as a linear combination of any one of the bases. So what you have provided as an example is one such linear combination of specific eigenfunctions of the angular momentum operators. But this could have just as easily been the linear momentum operator, or the position operator, or any other "physical measurement" operator. Given any arbitrary wave function its projection onto the basis of a specific measurement eigenfunction will give you the probability to measure the eigenvalue corresponding to that eigenfunction.

I could give you the wave function $sin(kr)/kr$ which is well defined everywhere and normalizable. If you want to determine the probability to measure an angular momentum with $l=6$, you express the function as a infinite series in $L^2$ eigenfunctions and pull the coefficient of the $l=6$ term. It's a little easier than that since that coefficient is just $<\phi_6|\psi>$, where $\phi_6$ is the $l=6$ eigenfunction. Then the probability to observe this value is just $|<\phi_6|\psi>|^2$. Same for linear momentum, just find the eigenfunctions for that operator and calculate the projection onto it, $|<\phi_k|\psi>|^2$.

If one were to construct a large statistical ensemble of experiments to measure this variable with identical starting wave functions then the expectation value of that measurement operator, $<\psi|M\psi>$, would equal the average value observed. Once you take the measurement you "collapse the wave function" and the particle is not in an eigenstate of $M$, all future measurements will yield the exact same quantity. The only way to verify the statistical nature of this is to reset the wave function and perform the measurement again. You will most likely get a different result each time, some repetition may occur as with flipping a coin and getting {H, T, H, H, T, H, T, T, ...}.

To understand the "value" of the wave function you really need to know the complete set of postulates for QM and the currently accepted interpretation, which is always subject to change.


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