High School Understanding Quantum Theory: Wave vs Vector in Hilbert Space

[fanieh]

In the 1930s, John von Neumann consolidated ideas from Bohr, Heisenberg and Schrodinger and placed the new quantum theory in Hilbert space.

In Hilbert space, a vector represents the Schrodinger wave function.

I know they are equivalent..

But can we say it is more natural and intuitive to say a particle has wave function (waves of probability) and this is better when conveying to layman as it is easier to visualize waves than a vector?

Also can't we say the probability wave really exist in some higher space. When describing more than one particle.. the wave no longer occur in 3D space, but in higher dimensional mathematical space. Is it more logical to think this higher dimensional space really exist in actual. Notice that vector space is just for convenient arrangement of information. So we can't say our bank account database have objective existence in reality. But in the case of wave function. Perhaps we can say the waves have factual existence (like in MWI or BM)?

There is no version of MWI or BM of the vectors because they are obviously for arrangement of information only. But waves are different and more real, is this correct way to think of it?

I'm asking because when conveying to laymen. I plan to use pure wave function only or the language of waves without complicating it with state vectors and so on. But then I wonder if there is possibility waves are really more factual than vectors (which has no possibility of being factual or actual existing physically).

Lastly. How do you describe quantum field theory using the language of pure waves.
What is the equivalent of Fock space using the language of waves only? Can you do QFT without using any concept of vector but waves (or wave function) only?

 

[bhobba]

But can we say it is more natural and intuitive to say a particle has wave function (waves of probability) and this is better when conveying to layman as it is easier to visualize waves than a vector?

The issue is there is this pesky little thing called the truth.

States are not waves - but elements of a Hilbert space.

When expanded in terms of eigenfunctions of position the coefficients can sometimes look like waves (only sometimes) but they are not waves - they are elements of a Hilbert space.

But waves are different and more real, is this correct way to think of it?

No.

Lastly. How do you describe quantum field theory using the language of pure waves. What is the equivalent of Fock space using the language of waves only? Can you do QFT without using any concept of vector but waves (or wave function) only?

There is no equivalent, and you can't do it.

Explaining QFT to laymen is so difficult I would almost say its impossible, but at the lay level the following isn't too bad:
https://www.amazon.com/dp/0473179768/?tag=pfamazon01-20

If you want to explain QM to laymen I think something along the lines of the following is the best approach:
http://www.scottaaronson.com/democritus/lec9.html

Thanks
Bill

 

[fanieh]

The issue is there is this pesky little thing called the truth.

States are not waves - but elements of a Hilbert space.

When expanded in terms of eigenfunctions of position the coefficients can sometimes look like waves (only sometimes) but they are not waves - they are elements of a Hilbert space.

But wave functions and Hilbert space are mathematically equivalent.. why can't you model the latter with the former... any proof?
What others think?

No.
There is no equivalent, and you can't do it.

Explaining QFT to laymen is so difficult I would almost say its impossible, but at the lay level the following isn't too bad:
https://www.amazon.com/dp/0473179768/?tag=pfamazon01-20

If you want to explain QM to laymen I think something along the lines of the following is the best approach:
http://www.scottaaronson.com/democritus/lec9.html

Thanks
Bill

 

[dextercioby]

@fanieh You have chosen the "B" status for this discussion which is hint to me that your degree of mathematical preparation is too low to allow for a respondent to come up with an explanation for this part:

The issue is there is this pesky little thing called the truth.

States are not waves - but elements of a Hilbert space.

When expanded in terms of eigenfunctions of position the coefficients can sometimes look like waves (only sometimes) but they are not waves - they are elements of a Hilbert space.
But wave functions and Hilbert space are mathematically equivalent.. why can't you model the latter with the former... any proof?

 

[Zafa Pi]

But wave functions and Hilbert space are mathematically equivalent.. why can't you model the latter with the former... any proof?

Wave functions are vectors in Hilbert space. For example, position is a complex valued function on the reals in L², which is a Hilbert space.

But can we say it is more natural and intuitive to say a particle has wave function (waves of probability) and this is better when conveying to layman as it is easier to visualize waves than a vector?

In the case of planar polarized photons the state is a vector in the plane (also a Hilbert space), very easy to visualize.

 

[bhobba]

But wave functions and Hilbert space are mathematically equivalent.. why can't you model the latter with the former... any proof?

Knowing Dextercoiby's background the following is of course not meant for him, but for the OP that wrote it.

A couple of points.

Do the following look like waves to you?
https://www.quora.com/How-do-I-plot-the-hydrogen-atom-wave-functions

The Hilbert space is basis independent, like all vector spaces. A representation in terms of a particular basis is not the same as the vector itself. But yes - its isomorphic to it so in that that sense the statement is of course correct. So I will change my reply to thinking in terms of wave-functions enshrines a particular basis. That's OK, providing you know that's what you are doing ie its just something chosen purely for convenience - its not anything fundamental.

Thanks
Bill

 

[bhobba]

Wave functions are vectors in Hilbert space.

Let f(x) be the wave-function then its state is |u> = ∫ f(x) |x>. The f(x) of course also forms a vector space, but I think its wise to realize, while ispmprphic to the state space, is not actually the state.

My apologies to the OP. I didn't notice it was at the B level.

Its hard to answer your queries, with reasons, at that level. I can give you my answers, but I can't at the B level explain why. If you have studied linear algebra then the reasons I gave may be clearer.

Thanks
Bill

 

[fanieh]

First I was not asking how wave function is equivalent to Hilbert space.. but asking if you can formulate things in wave functions only without using the concept of Hilbert space (just for sake of discussions and understanding and not telling the world they should reject the latter). Let me emphasize this point. I know Hilbert space is more convenient to use.. but just asking if you will be given exercise to formulate everything in wave functions only, whether you can do that.

Also I understood the basic of Hilbert space which is akin to the phase space developed by Hamilton. So if you use 3 coordinates for position and 3 for momentum, then you have 6 dimensions to describe one particle.. and ten particles would required sixty dimension. In quantum theory, the analog of phase space is Hilbert space. And as in Hamiltonian's concept, a single point in Hilbert space represents the entire quantum state. I understand a vector is a mathematical entity that exhibits both direction and magnitude.. and unlike points, vectors can be added and multiplied, which is essential for quantum theory. Etc.

I'm just asking supposed we can suppress giving any concept of vectors and Hilbert space. Can we force everything in the language of pure wave functions and Fourier? This is because if Bohmian mechanics was right. It's all localized particles and wave functions. This is easier to visualize than vectors and Hilbert space. I know Hilbert space is more convenient and efficient.. but again.. I was asking if you can put all in the language of wave function and Fourier only? Again for sake of discussions only, Can't you understand my point.

About the momentum basis in Hilbert space, is it nothing but the observable in the language of wave functions, or the momentum observable?

 

[PeterDonis]

asking if you can formulate things in wave functions only without using the concept of Hilbert space

No, you can't, because the space of all wave functions is a Hilbert space.

You appear to be confused about the difference between the "wave function" picture and the "state vector" picture. They both use Hilbert spaces, and the Hilbert spaces they use are equivalent. The only difference is how the elements of the Hilbert space are represented; the wave function picture represents them as functions, and the state vector picture represents them as vectors. So using the wave function picture does not avoid the concept of Hilbert space; it just represents the elements of the Hilbert space differently.

 

[Zafa Pi]

Let f(x) be the wave-function then its state is |u> = ∫ f(x) |x>.

I'm confused. What is the state |x> supposed to be? Does ∫ f(x) |x> = ∫ f(x)dx•|x> or ∫ f(x) |x>dx ?

 

[Paul Colby]

I'm confused. What is the state |x> supposed to be? Does ∫ f(x) |x> = ∫ f(x)dx•|x> or ∫ f(x) |x>dx ?

I suggest reading Dirac's book. Not the easiest read but well worth it. Oh, $\vert x\rangle$ is an abstract vector space element labeled by an eigenvalue.

 

[Zafa Pi]

I suggest reading Dirac's book. Not the easiest read but well worth it. Oh, $\vert x\rangle$ is an abstract vector space element labeled by an eigenvalue.

Eigenvalue of what operator?
That doesn't answer my 2nd question.
What's wrong with https://en.wikipedia.org/wiki/Position_operator?

 

[fanieh]

No, you can't, because the space of all wave functions is a Hilbert space.

You appear to be confused about the difference between the "wave function" picture and the "state vector" picture. They both use Hilbert spaces, and the Hilbert spaces they use are equivalent. The only difference is how the elements of the Hilbert space are represented; the wave function picture represents them as functions, and the state vector picture represents them as vectors. So using the wave function picture does not avoid the concept of Hilbert space; it just represents the elements of the Hilbert space differently.

You don't get my point. What I was asking was supposed China hated Hilbert so much that China wanted all textbook of quantum mechanics never to mention Hilbert. Can you present the concepts using wave mechanics only?

 

[Nugatory]

You don't get my point. What I was asking was supposed China hated Hilbert so much that China wanted all textbook of quantum mechanics never to mention Hilbert. Can you present the concepts using wave mechanics only?

No. You might be able to do a superficial treatment of single-particle systems in the position basis, but that is explicitly not presenting some of the key concepts.

 

[fanieh]

No. You might be able to do a superficial treatment of single-particle systems in the position basis, but that is explicitly not presenting some of the key concepts.

In the 1930s, John von Neumann consolidated ideas from Bohr, Heisenberg, and Schroedinger and placed the new quantum theory in Hilbert Space. I'd like to understand the pre-Hilbert era. Remember it was around 1927 when quantum field theory was developed.. without the concept of Hilbert Space.

Can you give an example where something modeled using Hilbert space can *never* be modeled using wave mechanics and fourier?

 

[PeterDonis]

In the 1930s, John von Neumann consolidated ideas from Bohr, Heisenberg, and Schroedinger and placed the new quantum theory in Hilbert Space. I'd like to understand the pre-Hilbert era. Remember it was around 1927 when quantum field theory was developed.. without the concept of Hilbert Space.

You're misunderstanding what Von Neumann did. He didn't change the formulation of QM from one that didn't use a Hilbert space to one that did. All he did was formalize the concept of a Hilbert space as applied to QM, by recognizing certain features of the formulations of QM (both Schrodinger and Heisenberg) that were already being used. In other words, pre-Von Neumann QM was already using Hilbert spaces implicitly, without realizing it. For example, the space of all possible wave functions of a single particle moving in one spatial dimension is a Hilbert space, so Schrodinger was implicitly using that Hilbert space without realizing it when he developed wave mechanics.

 

[fanieh]

You're misunderstanding what Von Neumann did. He didn't change the formulation of QM from one that didn't use a Hilbert space to one that did. All he did was formalize the concept of a Hilbert space as applied to QM, by recognizing certain features of the formulations of QM (both Schrodinger and Heisenberg) that were already being used. In other words, pre-Von Neumann QM was already using Hilbert spaces implicitly, without realizing it. For example, the space of all possible wave functions of a single particle moving in one spatial dimension is a Hilbert space, so Schrodinger was implicitly using that Hilbert space without realizing it when he developed wave mechanics.

I know pre-Von Neumann QM was already using Hilbert spaces implicitly, without realizing it. I understand Hilbert space contains infinite dimensions, but these are not geometric. Rather, each dimension represents a state of possible existence for a quantum system. All possible states coexist and add up to the wave function before a measurement is made or a given possibility is selected. I know the electron unmeasured is a complicated pattern in an infinite-dimensional Hilbert space, and it is not in our three-dimensional space and its attributes are not clearly defined. And in the world of the wave function, the electron does not have inherent properties but has incompletely defined potentialities.

But if you want to be a Bohmian, and you want to share Bohmian to friends. And telling them the local particles have trajectories. And the wave function in Bohmian really exist in higher dimension. It's better to describe wave functions because you can tell them somehow spacetime becomes infinite dimensional and this is where the wave function literally live. It's difficult to do this with vectors because it doesn't make sense the dimensions of the Hilbert space really exist in higher dimensions.

My point it. The wave functions of Schrodinger have more possibility of being actually there physically than the vectors of Hilbert space.. is it not? (what others think about this since Peterdonis is a math or shut up and calculate guy and doesn't want to even explore this aspect). Although the wave functions of Schrodinger needs higher dimensional space. If there is really higher dimensional space.. it is more logical to imagine they are composed of waves than vectors. Is it not?

 

[PeterDonis]

I understand Hilbert space contains infinite dimensions

Some Hilbert spaces do, but not all. For example, the Hilbert space of the possible states of a spin-1/2 particle with no other degrees of freedom has just two dimensions.

The wave functions of Schrodinger have more possibility of being actually there physically than the vectors of Hilbert space.. is it not?

I don't see how you can assign any useful meaning to "how possible" a particular mathematical structure is of "actually being there physically". But even if we leave that aside and assume there is some way of giving this a useful meaning, it would seem that two mathematical structures that are mathematically equivalent should have the same "possibility of being actually there physically". And wave functions and state vectors are mathematically equivalent.

the wave functions of Schrodinger needs higher dimensional space

So do state vectors. The two are mathematically equivalent.

 

[fanieh]

Some Hilbert spaces do, but not all. For example, the Hilbert space of the possible states of a spin-1/2 particle with no other degrees of freedom has just two dimensions.
I don't see how you can assign any useful meaning to "how possible" a particular mathematical structure is of "actually being there physically". But even if we leave that aside and assume there is some way of giving this a useful meaning, it would seem that two mathematical structures that are mathematically equivalent should have the same "possibility of being actually there physically". And wave functions and state vectors are mathematically equivalent.
So do state vectors. The two are mathematically equivalent.

If you are a programmer.. and asked to write a program where the software needs to model the games internal environment using quantum mechanics. Would you use wave functions or Hilbert spaces? Can you just use wave functions in the program? Meaning you deal with waves only and not use any coordinate system for the basis in Hilbert space.

I'm reading a book about wave mechanics and Heisenberg matrix mechanics. They didn't even mention about Hilbert but uses algebra in pure wave function language only.

 

[Paul Colby]

if you are a programmer..

Hypothetical question or do you need a programming approach?

 

[fanieh]

No, you can't, because the space of all wave functions is a Hilbert space.

You appear to be confused about the difference between the "wave function" picture and the "state vector" picture. They both use Hilbert spaces, and the Hilbert spaces they use are equivalent. The only difference is how the elements of the Hilbert space are represented; the wave function picture represents them as functions, and the state vector picture represents them as vectors. So using the wave function picture does not avoid the concept of Hilbert space; it just represents the elements of the Hilbert space differently.

Do you have example of how they can model the Fock space where the wave function picture represents them as functions? I want to compare it mathematically to one where the state vector picture represents them as vectors.

 

[PeterDonis]

Do you have example of how they can model the Fock space where the wave function picture represents them as functions?

https://en.wikipedia.org/wiki/Fock_space#Wave_Function_Interpretation

 

[fanieh]

The issue is there is this pesky little thing called the truth.

States are not waves - but elements of a Hilbert space.

When expanded in terms of eigenfunctions of position the coefficients can sometimes look like waves (only sometimes) but they are not waves - they are elements of a Hilbert space.
No.
There is no equivalent, and you can't do it.

Explaining QFT to laymen is so difficult I would almost say its impossible, but at the lay level the following isn't too bad:
https://www.amazon.com/dp/0473179768/?tag=pfamazon01-20

If you want to explain QM to laymen I think something along the lines of the following is the best approach:
http://www.scottaaronson.com/democritus/lec9.html

Thanks
Bill

I asked: "Lastly. How do you describe quantum field theory using the language of pure waves. What is the equivalent of Fock space using the language of waves only? Can you do QFT without using any concept of vector but waves (or wave function) only?"

You replied "no". But Peterdonis said above it was possible.. see https://en.wikipedia.org/wiki/Fock_space#Wave_Function_Interpretation
So "no" was a mistake and it should be "yes"? If still no. Please explain. Would appreciate it a lot.

 

[fanieh]

https://en.wikipedia.org/wiki/Fock_space#Wave_Function_Interpretation

Ok. Anyway. About the momentum basis in Hilbert space, is it nothing but the momentum observable in the language of Schoedinger wave functions? Like Schroedinger referred to it as momentum observable.. and later Von Neumann referred to it as momentum basis?

If not. What is the equivalent of basis in Schrodeinger pure wave function language?

 

[PeterDonis]

But Peterdonis said above it was possible

Note that that Wikipedia article just gives one example; it does not say that it is possible in general, for any Fock space. (And even the example given has some limitations, as the article notes.)

@bhobba made a good general point earlier, that your questions really go beyond what can be answered at the "B" level. At the "B" level the best advice we can give is to learn more.

What is the equivalent of basis in Schrodeinger pure wave function language?

You are confusing a bunch of different things. You really should take the time to learn QM in detail before asking these questions, because, as mentioned just now, this subject is not a "B" level subject.

Briefly, though:

A basis of a Hilbert space is a set of elements of the space which has two properties: (a) the elements are all linearly independent; and (b) any element of the space can be written as a linear combination of one or more of the basis elements. In state vector language, this is just the usual set of basis vectors that you're used to from vector analysis. In wave function language, elements are functions, and a linear combination of elements is just a weighted sum or integral (an integral is just a sum over a continuous domain) over the basis functions.

So, for example, in Schrodinger language we can express any wave function as an integral over the position basis functions, which are just delta functions. In fact, the wave function $\psi(x)$ in the position representation is exactly that: a weighted integral of delta functions, with the weight of each delta function equal to the value of the wave function at that particular value of $x$. We can even write this out:

$$
\psi(x) = \int dx' \psi(x') \delta(x' - x)
$$

Notice that this is actually just an identity satisfied by the delta function.

 

[fanieh]

Note that that Wikipedia article just gives one example; it does not say that it is possible in general, for any Fock space. (And even the example given has some limitations, as the article notes.)

@bhobba made a good general point earlier, that your questions really go beyond what can be answered at the "B" level. At the "B" level the best advice we can give is to learn more.
You are confusing a bunch of different things. You really should take the time to learn QM in detail before asking these questions, because, as mentioned just now, this subject is not a "B" level subject.

Briefly, though:

A basis of a Hilbert space is a set of elements of the space which has two properties: (a) the elements are all linearly independent; and (b) any element of the space can be written as a linear combination of one or more of the basis elements. In state vector language, this is just the usual set of basis vectors that you're used to from vector analysis. In wave function language, elements are functions, and a linear combination of elements is just a weighted sum or integral (an integral is just a sum over a continuous domain) over the basis functions.

So, for example, in Schrodinger language we can express any wave function as an integral over the position basis functions, which are just delta functions. In fact, the wave function $\psi(x)$ in the position representation is exactly that: a weighted integral of delta functions, with the weight of each delta function equal to the value of the wave function at that particular value of $x$. We can even write this out:

$$
\psi(x) = \int dx' \psi(x') \delta(x' - x)
$$

Notice that this is actually just an identity satisfied by the delta function.

In short, the basis in Hilbert space is the coordinate axis?

I know that as the vector representing the state moves and changes direction, the magnitude of all the possibilities also change. And the projection of the state vector on each axis is a measure of the possibility of an electron being at a particular position.

So the axis is the basis?

When the axis has value or multiplying a vector by its complex conjugate produces a real number. Then the value exists. I thought this was akin to the eigenvalues of the observable in QM.

 

[PeterDonis]

In short, the basis in Hilbert space is the coordinate axis?

Coordinate basis vectors are one example of a basis, yes. (Note that it's the basis vectors--the vectors that point along the coordinate axes--that are the basis, not the axes themselves.)

Again, I strongly suggest that you take some time to learn QM from a textbook. We are getting to the point where answering your questions is amounting to giving you a course in QM, and that's beyond the scope of PF.

 

[fanieh]

Coordinate basis vectors are one example of a basis, yes. (Note that it's the basis vectors--the vectors that point along the coordinate axes--that are the basis, not the axes themselves.)

Again, I strongly suggest that you take some time to learn QM from a textbook. We are getting to the point where answering your questions is amounting to giving you a course in QM, and that's beyond the scope of PF.

Yes I know so let's iimit this thread to the coordinate basis vectors and it's connection to Schroedinger pure wave mechanics. Notice that when the vector is in the coordinate basis. It has value. In QM. This is similar to collapse or when it is in eigenvalue of an observable.. this is why I said when the vector is in coordinate basis.. it has value of let's say momentum. In QM. When it has value of momentum. it has value of the observable momentum. Hence I said the coordinate basis vector is akin to the observable in QM with values because both is in collapse (or with single outcome) thing. Is this wrong??

 

[PeterDonis]

when the vector is in the coordinate basis. It has value.

What do you mean by this? A vector is a vector. It's not a value.

Is this wrong?

It doesn't make sense to me so I can't tell. This is why I keep advising you to go learn QM from a textbook. You don't seem to be familiar enough with the basic terminology in the field to formulate your questions in a way that makes them understandable.

 

[fanieh]

What do you mean by this? A vector is a vector. It's not a value.
It doesn't make sense to me so I can't tell. This is why I keep advising you to go learn QM from a textbook. You don't seem to be familiar enough with the basic terminology in the field to formulate your questions in a way that makes them understandable.

I read from a textbook. It's about single outcome. When the vector is in the coordinate axis. It has single outcome, correct?

In QM, when it is in the eigenvalue of an observable. It has single outcome too correct? ( I use the word single outcome instead of collapse).

Hence aren't they related since they have both single outcome?