I Is the Wavefunction a Contravariant Component?

[Nugatory]

So my problem with the current explanation is that it treats the square integral and the resulting probability distribution as if it's all that mattered and as if all the innerworkings don't need to make any sense.

That's how quantum mechanics works - it predicts measurement results but does not tell us anything about what's "really" going on behind the scenes to make those outcomes happen.

You may find this frustrating and unsatisfying, and if so you are in good company. However there's no reason why the universe should care whether we like the way it works, and whether we like it or not quantum mechanics is the way the universe works.

It's worth noting that this problem with "what's 'really' going on behind the scenes" isn't (as your scare-quotes suggest) unique to QM. Consider Newtonian gravity: We all learn in high school that masses attract with a force proportional to their masses and inversely proportional to the square of the distance between them and we're generally quite happy with this statement of physical law. But if someone were to ask what's going on behond the scenes to create this force, we'd be somewhat embarassed and would eventually fall back on the answer "Because observation tells us that's how the universe works". The situation with QM differs only in that we don't like that answer as much, and that's our problem not the universe's.

 

[JohnH]

The square of the wave function is the probability distribution; that's the definition of the wave function. What else would you expect?

No, the wavefunction is the wavefunction. The ingetration of the wavefunction multiplied by its complex conjugate is the probability distribution. When you multiply the wavefunction by its complex conjugate it's real valued. In order to extract the information of the momentum or the energy, for example, you wouldn't be able to do that by applying the operators to the square of the wavefunction. You have to apply those operators to the wavefunction (or its complex conjugate--if it's Hermitian) on its own and then multiply it by the complex conjugate (or wavefunction).

What "innerworkings" are you talking about?

See above answer.

Those units aren't units of the eigenfunctions; the eigenfunctions are wave functions, whose units are "square root of probability density".

The units you are talking about--momentum, energy, etc.--are units of observables. Observables are operators, not wave functions.

But the operators act upon the wavefunction to get the eigenfunctions, so you're scaling the wavefunction by something with units to get the eigenfunction which has an eigenvalue. The eigenvalues can be observed. I think I just used the word eigenfunction a bit flippantly is all, but correct me if I've missed something more than that. At any rate I think the point still remains, if operators involve extracting information from the wavefunction, you'd expect some more consistency between the units of the wavefunction and the units of the observables. Why does one use inverse units of length and the other uninverted? I don't think this is an unreasonable point to ponder over, do you? But perhaps we're just getting into something too theoretical, and perhaps this is not the place for that.

What textbooks on QM have you read?

Mostly Griffith's and the MIT lectures I linked to among others. I've just not been super concise with my language and in some cases, admittedly, my understanding as I've tried to come up with my own intution for what I've been reading, wrongfully thinking the science community might have come to the same conclusion. But it seems that what we're talking about is fairly basic: bras and kets with Hermitian operators and the linear algebra involved there. I don't think we're disagreeing on the what quantum mechanics says. Where we're diverging is how it is and can be interpreted.

 

[PeterDonis]

the operators act upon the wavefunction to get the eigenfunctions

No, the operators, when acting on their eigenfunctions, give the same eigenfunctions back (times whatever the corresponding eigenvalue is). But an operator acting on a wave function that is not an eigenfunction does not give an eigenfunction.

if operators involve extracting information from the wavefunction, you'd expect some more consistency between the units of the wavefunction and the units of the observables

There is: the wave function's units, in the representation that corresponds to a particular observable, are the inverse square root of that observable. So if the observable you're interested in is position, the wave function in position representation has units of inverse square root of position. If the observable you're interested in is momentum, the wave function in momentum representation has units of inverse square root of momentum. And so on for other observables.

Perhaps that's one of the missing pieces you hadn't realized: the wave function can have different units in different representations.

I don't think we're disagreeing on the what quantum mechanics says. Where we're diverging is how it is and can be interpreted.

Discussions of interpretations of QM do not belong in this forum. They belong in the subforum on QM interpretations and foundations.

 

[JohnH]

Perhaps that's one of the missing pieces you hadn't realized: the wave function can have different units in different representations.

Yeah, any links to more on this?

 

[atyy]

I think that pretty much answers the question

One way to think about it is that if one took the square root of a classical probability density, one would also get something with the same units as the wave function. So the question is what does squaring the wave function give you that you would not otherwise get? What you get is interference.

As a follow up, would it be a theoretical possibility to interpret the wavefunction as a contravariant component of a vector, or would there be some logical reason for necessarily disqualifying that interpretation?

The quantum state is not a vector or covector that is localized in spacetime (it is a vector in the Hilbert space). Most usually, we make a choice of simultaneity before specifying the quantum state. After making that choice, the quantum state at a given time is assigned to a plane of simultaneity.

 

[PeterDonis]

any links to more on this?

It's not something that is discussed much in textbooks, but it should be obvious by the same line of reasoning we already used in this thread. For example, just consider the wave function in the momentum representation and what its squared modulus means as a probability density.

 

[JohnH]

it should be obvious by the same line of reasoning we already used in this thread

What had been unclear was that the wavefunction took on varying units based on its Fourier transform. I'm still going to have to work on my intuition of exactly why this happens and also on my understanding of eigenfunctions, but it's starting to come into focus, so thank you.

 

[vanhees71]

The wave function in position representation is, expressed in terms of bras and kets,
$$\psi(x)=\langle x|\psi \rangle.$$
Usually one also normalizes the state ket,
$$\langle \psi|\psi \rangle=1. \qquad (*)$$
Further $|x \rangle$ are "generalized eigenvectors" of the position operator, which has entire $\mathbb{R}$ as its spectrum, i.e., the "eigenvalues" form a continuum, and that's why these "eigenvectors" cannot be normalized in the usual way but as a Dirac $\delta$ distribution.
$$\langle x|x' \rangle=\delta(x-x'). \qquad (**)$$
This makes the dimensional analysis very simple: The normalized ket $|\psi \rangle$ is dimensionless, because of Eq. (*). From (**) it then follows that the dimension of $|x \rangle$ and $\langle x|$ must be $1/\sqrt{\text{length}}$. The wave function has thus dimension $1/\sqrt{\text{length}}$.

This must also be so because of the physical meaning of the wave function. With a normalized state ket and the position "eigenvectors" normalized as given by (**), $|\psi(x)|^2$ is the probability density distribution of the particle's position, and thus this quantity must have the dimension $1/\text{length}$, as already mentioned above.

 

[Sunil]

Here's a reference: https://ocw.mit.edu/courses/physics...pring-2013/lecture-notes/MIT8_04S13_Lec03.pdf

A few lines into page 3 it says "so the wavefunction has units of reciprocal square root of length."

The part before this is essential: "which in this case is position, so the wavefunction has units of
reciprocal square root of length."

So the claim is only about one-dimensional position x. In general we have a configuration space Q, which already for two particles in 3D is six-dimensional, so that the unit would be the reciprocal square root of that corresponding six-dimensional volume. Not only the configuration space can vary a lot, becoming infinite-dimensional in field theory, but we can measure also other things like momentum or energy or whatever. Each will give some other unit, but it will be always a reverse square root of the volume of the space of measurement results.

Let's also be careful about spacetime. There is no notion of a wave function on relativistic spacetime, only on the configuration space at a given moment of absolute time. In classical QM we have absolute time. In relativistic QM one tries to hide this, simply by not talking about those things which depend on absolute time but focusing on those things where this dependence is hidden. So, the volume which matters is the volume of of the configuration space at a fixed moment of time. (And there is not even such an animal like a configuration spacetime.)

Once time is fixed: Probability distributions are covariant: If you have a probability distribution on $X$ and a map $X\to Y$ you get a probability distribution on $Y$, which is defined by the probability that the preimage. Functions of probability distributions are covariant too (it does not matter where you compute the inverse square root of some object).

 

[stevendaryl]

But the operators act upon the wavefunction to get the eigenfunctions, so you're scaling the wavefunction by something with units to get the eigenfunction which has an eigenvalue. The eigenvalues can be observed. I think I just used the word eigenfunction a bit flippantly is all, but correct me if I've missed something more than that. At any rate I think the point still remains, if operators involve extracting information from the wavefunction, you'd expect some more consistency between the units of the wavefunction and the units of the observables. Why does one use inverse units of length and the other uninverted? I don't think this is an unreasonable point to ponder over, do you? But perhaps we're just getting into something too theoretical, and perhaps this is not the place for that.

I don't understand exactly what you're driving at. When you solve the Schrodinger equation to get the wave function, the units are arbitrary. It's a linear equation, so multiplying by a constant doesn't change whether it solves the Schrodinger equation. Also, multiplying by a constant doesn't change the fact that a wave function is an eigenstate of momentum or energy or angular momentum. So the normalization doesn't change anything about the physics of working with wave functions.

The normalization only comes into play when you interpret $\psi^* \psi$ as a probability density. For that interpretation to make sense, you have to choose the normalization so that $\int \psi^* \psi = 1$.

I really don't see that there is anything deeper to the choice of the normalization than that. I don't see how there could be anything deeper, since the normalization is irrelevant for everything else.

 

[Jazzdude]

I really don't understand how the misconception that it's meaningful to assign units to wavefunctions is still around. I remember having the exact same argument on this board maybe 10 years ago.

Quantum state wavefunctions are agnostic regarding any choice of unit or dimension. The dimensional analysis must be done for the full expression for the dimensionless probability $$P = \frac{\int_{\Omega \subset S} \Psi \Psi^* }{\int_S \Psi \Psi^*}$$ for which any non-zero factor to $\Psi$, including units, cancels.

So it doesn't matter what units you choose.

Cheers,

Jazz

 

[JohnH]

I really don't understand how the misconception that it's meaningful to assign units to wavefunctions is still around.

Let's assume, for a momement, that the wavefunction exists in physical reality--physics, after all, being the study of physical reality. If it's true that it doesn't matter what the units of the wavefunction are, how are we supposed to make sense of that? It exists in some dimensionless, purely numerical void?

It's a linear equation, so multiplying by a constant doesn't change whether it solves the Schrodinger equation.

To raise the Schrodinger equation to such a level of definitive importance that its solutions define reality seems like more of an interpretation than a verifiable fact. In other words, it may be that all of reality can be described by the Schrodinger equation, but that not all solutions to the Schrodinger equation describe reality (i.e. solutions with arbitrarily chosen units).

 

[PeterDonis]

Let's assume, for a momement, that the wavefunction exists in physical reality

What does "exists in physical reality" mean?

We know what it can't mean, at least as regards the wave function: it can't mean that the wave function exists in ordinary 3-dimensional space, because the wave function is a function on configuration space, and for any quantum system other than a single, spinless, non-interacting particle, the configuration space is not ordinary 3-dimensional space.

 

[JohnH]

What does "exists in physical reality" mean?

If it doesn't "exist in physical reality" then why are we trying to describe it with physics? Are you a spiritualist? I am a physicalist and I believe that if we're not describing physical reality then we're not describing anything at all. But quantum mechanics might push out the boundaries of how physical reality is defined.

 

[PeterDonis]

If it doesn't "exist in physical reality" then why are we trying to describe it with physics? Are you a spiritualist? I am a physicalist and I believe that if we're not describing physical reality then we're not describing anything at all.

None of this answers my question or addresses the issue I raised. If you think physics is about describing "what exists in physical reality" then you should be able to explain what you mean by that.

In fact, many physicists do not think physics is about describing "what exists in physical reality". They think that's a matter of philosophy or metaphysics. To them, physics is about building models that make accurate predictions about what will happen if we do various experiments or more generally take various actions. Perhaps that is what you mean by "physical reality" (certainly the things that happen when we do experiments or take actions would seem to qualify as part of "physical reality"). But models are not the same thing as reality, and even models that make good predictions in some domain don't necessarily "describe what exists in physical reality" unless you define that phrase to mean whatever it is the models are doing.

In short, this matter of "existing in physical reality" is nowhere near as simple and obvious as you appear to think it is.

 

[JohnH]

If you think physics is about describing "what exists in physical reality" then you should be able to explain what you mean by that.

I believe that the world is composed of an entirely numerical substance (Wolfram and many others also believe this). Furthermore, I believe that this is extremely well defined and also struggle to imagine there being any other explanation that doesn't resort to abstraction and/or spiritualism.

 

[PeterDonis]

I believe that the world is composed of an entirely numerical substance (Wolfram and many others also believe this). Furthermore, I believe that this is extremely well defined and also struggle to imagine there being any other explanation that doesn't resort to abstraction and/or spiritualism.

I understand that this is your opinion, but it's out of bounds for discussion here, since this forum is for discussion of quantum physics as a physical theory, based on how that theory is presented in textbooks and peer-reviewed papers, not people's opinions about "physical reality".

Even in the quantum interpretations forum, where the rules are somewhat looser because of the nature of the subject matter, discussion still needs to be based on interpretations as they are presented in textbooks and peer-reviewed papers.

 

[PeterDonis]

The OP's question has been addressed, and personal opinion/speculation is off limits for this forum. Thread closed.