I Is the Wavefunction a Contravariant Component?

[JohnH]

I've heard that the wavefunction as a function of x has units of square root of inverse distance, but I haven't heard an intuitive description of why this is aside from that the math works out when you integrate to get the probability. But aside from the math working out, I'm hoping to get a more intuitive understanding of what this means for the wavefunction as a physical object. If it means that the wavefunction is a contravariant component of a vector, then I can understand why its units are of inverse distance. Is this a common understanding of it? Or is the intuition for the units just a mystery? In short, philosophically, what is to be made of the wavefunction having these oddly inverse units if it's a physical object that exists in the world?

 

[PeterDonis]

I've heard

Where? Please give a reference.

 

[atyy]

I've heard that the wavefunction as a function of x has units of square root of inverse distance, but I haven't heard an intuitive description of why this is aside from that the math works out when you integrate to get the probability. But aside from the math working out, I'm hoping to get a more intuitive understanding of what this means for the wavefunction as a physical object.

It is just because the maths works out. For a probability density, it has units of 1/x so that when you integrate against dx, you get a unitless probability. Similar reasoning gives the units for the wave function.

Regardless of units, it is a matter of interpretation whether the wave function is considered a physical object, or just something like a probability that may be taken to represent subjective knowledge.

 

[PeterDonis]

If it means that the wavefunction is a contravariant component of a vector

It isn't.

then I can understand why its units are of inverse distance.

Why? A vector can have all kinds of different units.

 

[haushofer]

Where? Please give a reference.

We should have a standard reply button with this phrase on PF.

 

[haushofer]

Or a new acronym: SMTR, "Show Me The Reference" :P

 

[JohnH]

Where? Please give a reference.

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."

 

[JohnH]

Why? A vector can have all kinds of different units.

That's just a statement of a fact. The question is what is the intuition for these particular units.

 

[JohnH]

It is just because the maths works out.

I think that pretty much answers the question.

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?

 

[PeterDonis]

The question is what is the intuition for these particular units.

But that intuition has nothing to do with whether or not the wave function is the contravariant component of a vector, which is what your statement that I quoted implied. It has to do with the wave function being a probability amplitude, i.e., its square is a probability density. So the units of the wave function have to be the square root of the inverse of whatever the probability density is a density over (in this case position, i.e., units of length).

 

[PeterDonis]

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?

You can't just choose to "interpret" the wave function (or anything else, for that matter) as a contravariant component of a vector. To be a component of a vector, it would have to be a member of a set of quantities that, taken together, transform as a contravariant vector. The wave function isn't.

 

[JohnH]

You can't just choose to "interpret" the wave function (or anything else, for that matter) as a contravariant component of a vector. To be a component of a vector, it would have to be a member of a set of quantities that, taken together, transform as a contravariant vector. The wave function isn't.

That's a better answer. But it still seems a bit Nhilistic to simply say that "the math just works out" is as intuitive as it gets.

 

[PeterDonis]

it still seems a bit Nhilistic to simply say that "the math just works out" is as intuitive as it gets.

That was the answer to why the units of the wave function are what they are, not to why it isn't a component of a contravariant vector. I expanded on that some in post #10.

 

[JohnH]

That was the answer to why the units of the wave function are what they are, not to why it isn't a component of a contravariant vector. I expanded on that some in post #10.

Yes, I think we both are agreeing on the basic facts (and I was alluding to what you were talking to in post 10 in my first post). But where there's some conflict is as to whether or not there's any more intuitive understanding of these facts than have yet been mentioned. I was trying to puzzle through the intuition and thought perhaps it had to do with contravariance, but apparently I was wrong about that.

So from here, I'd ask would you all agree that it is intuitive to say that the wavefunction has the units it has? If so, why? And if not, what does the literature say on this subject? Is there any deeper meaning/interpretation to be had here even if only theoretical? In other words, with classical physics there's always an intuition and an imaginable reasoning behind everything and I understand that's not always the case with quantum mechanics. If this is simply a case where we don't have an intuitive answer for it, then that's a fine and complete answer. Not complaining. Just wondering.

 

[weirdoguy]

I'd ask would you all agree that it is intuitive to say that the wavefunction has the units it has? If so, why?

For me it is, but I guess it's a matter of what intuition one has. Studying physics forces you to rebuild your way of thinking and what's "intuitive". As to why it is intuitive (and I'm repeating what PeterDonis said):
$P=\int_\mathcal{O}|\psi|^2\textsf{d}\mu$
$P$ has no unit, so $|\psi|^2$ has to have the inverse unit of measure. Then we take a square root and that's all. That's my intuition.

 

[JohnH]

For me it is, but I guess it's a matter of what intuition one has. Studying physics forces you to rebuild your way of thinking and what's "intuitive".

So I take it you don't interpret the wavefunction to be a physical object then.

 

[weirdoguy]

I would say no one interprets it as a "physical object"

 

[dextercioby]

By the way, in order to properly speak about "covariance" and "contravariance" of mathematical objects, one first needs the full machinery of (usually finite-dimensional) differential geometry (i.e. topological/differential manifolds, points, charts, tangent spaces, tangent bundles, parametrized curves on the manifold, differentiability of curves, etc.).

The mathematical interpretation of wavefunctions: vectors from the infinite-dimensional Hilbert space $L^2 (\Omega\subseteq \mathbb{R}^n, d\mu)$, so yeah, quite unrelated to the paragraph above.
If you wish, you can perceive wavefunctions as contravariant vectors, but there is no solid mathematical background for it...

 

[weirdoguy]

Some interpret wave functions as $\frac{1}{2}$-densities.

 

[dextercioby]

Some interpret wave functions as $\frac{1}{2}$-densities.

That is correct, it's the "diff. geom." interpretation made popular by the relativist Robert Geroch, but I've not seen it formalized in terms of infinite-dim manifolds.

 

[JohnH]

Thank you all for the replies.

 

[stevendaryl]

So from here, I'd ask would you all agree that it is intuitive to say that the wavefunction has the units it has? If so, why? And if not, what does the literature say on this subject? Is there any deeper meaning/interpretation to be had here even if only theoretical? In other words, with classical physics there's always an intuition and an imaginable reasoning behind everything and I understand that's not always the case with quantum mechanics. If this is simply a case where we don't have an intuitive answer for it, then that's a fine and complete answer. Not complaining. Just wondering.

Just for clarity of the discussion, do you agree that there is no connection between being the two issues of the units for the wave function and whether it's a component of a contravariant vector? A vector can have any units you like. Momentum, for example, has units $\dfrac{\text{mass} \cdot \text{length}}{\text{time}}$, while current density has units $\dfrac{\text{charge}}{\text{length}^2 \cdot \text{time}}$

 

[JohnH]

do you agree that there is no connection between being the two issues of the units for the wave function and whether it's a component of a contravariant vector?

As I said right above what you quoted, I was wrong about it having anything to do with contravariance.

 

[JohnH]

Momentum, for example, has units , while current density has units

Yes, that's a good example, the charge density. Are you suggesting an analog--that the wavefunction is divided out over space like a charge? Because an answer like that is very intuitive.

 

[PeterDonis]

Are you suggesting an analog--that the wavefunction is divided out over space like a charge?

Schrodinger originally thought this about the wave function for an electron. Unfortunately, this analogy breaks down as soon as you consider a quantum system of more than one particle, since the configuration space--the space that the wave function "lives" in--is no longer ordinary 3-dimensional space. In fact, this is true even for a single electron since it also has spin as well as position; Schrodinger's original wave equation ignored spin.

 

[PeroK]

Dirac spinors have four components that transform according to a representation of the Lorentz group:

https://www.damtp.cam.ac.uk/user/tong/qft/four.pdf

 

[PeterDonis]

Dirac spinors have four components that transform according to a representation of the Lorentz group:

Yes, but it's not the vector representation.

 

[JohnH]

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. It's as if we built a car that works and when we open the engine, nothing makes any sense, but we say, but no look--the car drives just fine when I turn the wheel so "seems pretty intuitive to me."

And I'm fully prepared to accept that the answer might not be "intuitive" based on my macro-world associations, but I still believe the mathematical descriptions should be imaginable--at least as far as the fundamentals are concerned.

 

[JohnH]

Another thing that's been on my mind is the units of the eigenfunctions. If the wavefunction started out with these inverted units, how then might it arrive at units of momentum, energy, etc for the eigenfunctions? Perhaps I just don't have a popular way of thinking about things, but personally I need to be able to imaginatively reconstruct some narrative explaining these fundamentals.

As an aside, it was this issue that led me to believe that the wavefunction was interpreted as a contravariant component because then the eigenfunctions could be covariant and they would seem to map out right.

 

[PeterDonis]

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

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

as if all the innerworkings don't need to make any sense

What "innerworkings" are you talking about?

If the wavefunction started out with these inverted units, how then might it arrive at units of momentum, energy, etc for the eigenfunctions?

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.

Perhaps I just don't have a popular way of thinking about things,

What textbooks on QM have you read? How much of the actual math of QM are you familiar with?

If the answers are "none" and "not much", then I would strongly recommend remedying that first, before trying to figure out any "intuition". You can't expect to have an intuitive grasp of something if you don't understand how it works or what predictions it makes or how those predictions are confirmed by experiments.

 

[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.