Quantum Mechanics without Hilbert Space

[Varon]

Von Neumann developed the concept of Hilbert Space in Quantum Mechanics. Supposed he didn't introduce it and we didn't use Hilbert Space now. What are its counterpart in pure Schroedinger Equation in one to one mapping comparison?

In details. I know that "the states of a quantum mechanical system are vectors in a certain Hilbert space, the observables are hermitian operators on that space, the symmetries of the system are unitary operators, and measurements are orthogonal projections" But this concept was developed by Von Neumann. Before he developed Hilbert Space. What are their counterpart in the pure Schroedinger equation up to Born interpretation of the amplitude square as the probability that electron can be found there?

Please answer more in words or conceptual and not with dense mathematical equations. Thanks.

 

[Varon]

Example. In wikipedia entry on the Schroedinger Equations. Not a single word about Hilbert Space is used.

http://en.wikipedia.org/wiki/Schrödinger_equation

In the book Deep Down Thing. Mathematical formulation of the Schroedinger equation is given in nice details but no Hilbert space is mentioned anywhere in the book.

This is also true for other books like Introducing Quantum Theory.

So it seems there is a disconnect between Hilbert Space and the original Schroedinger equations. As if SE can do away with Hilbert space entirely. Since Hilbert space is an add-on. What would be QM like without Hilbert space. (?)

 

[Chopin]

Example. In wikipedia entry on the Schroedinger Equations. Not a single word about Hilbert Space is used.

So it seems there is a disconnect between Hilbert Space and the original Schroedinger equations. As if SE can do away with Hilbert space entirely. Since Hilbert space is an add-on. What would be QM like without Hilbert space. (?)

QM without a Hilbert space would be very much like QM in its earliest days, before the Hilbert space formalism was developed. :)

To answer this question, one small correction needs to be made. The Schrodinger Equation describes the dynamics of a system (how things change over time), while the Hilbert space describes the state of a system (how it is at any given moment.) So the two aren't really equivalent. The real comparison to be made is between the Hilbert space and the wavefunction, which is a three-dimensional field of complex numbers that we denote by $$\Psi(x,y,z)$$.

The wavefunction can actually be considered shorthand for states out of a Hilbert space. When we talk about a wavefunction with $$\Psi(0,0,0) = 1$$, that's the same as talking about the Hilbert state where position = $$(0,0,0)$$. So the two are equivalent in a sense, but Hilbert spaces are more general than that. In addition to using a Hilbert state to describe a particle at a position, you can use it to describe a particle with a certain charge, or a certain momentum, or even something more abstract like which slit a particle goes through in the double-slit experiment. The wavefunction is just a convenient notation for labeling the Hilbert states that describe positions.

You can get through a lot of single-particle quantum mechanics without using Hilbert spaces. For instance, you can calculate the energy levels of the Hydrogen atom using just the Schrodinger Equation (that's how Schrodinger did it!) But if you want to describe particle interactions, or perturbation, or just talk in general about quantum superpositions, then wavefunctions don't cut it--you have to understand Hilbert spaces, and all of the linear algebra that goes with them.

 

[Hurkyl]

If you're doing calculus in a vector space with an inner product and assuming things are "well-behaved", then you're working in a Hilbert space whether or not you explicitly acknowledge that fact.

 

[Varon]

QM without a Hilbert space would be very much like QM in its earliest days, before the Hilbert space formalism was developed. :)

To answer this question, one small correction needs to be made. The Schrodinger Equation describes the dynamics of a system (how things change over time), while the Hilbert space describes the state of a system (how it is at any given moment.) So the two aren't really equivalent. The real comparison to be made is between the Hilbert space and the wavefunction, which is a three-dimensional field of complex numbers that we denote by $$\Psi(x,y,z)$$.

The wavefunction can actually be considered shorthand for states out of a Hilbert space. When we talk about a wavefunction with $$\Psi(0,0,0) = 1$$, that's the same as talking about the Hilbert state where position = $$(0,0,0)$$. So the two are equivalent in a sense, but Hilbert spaces are more general than that. In addition to using a Hilbert state to describe a particle at a position, you can use it to describe a particle with a certain charge, or a certain momentum, or even something more abstract like which slit a particle goes through in the double-slit experiment. The wavefunction is just a convenient notation for labeling the Hilbert states that describe positions.

You can get through a lot of single-particle quantum mechanics without using Hilbert spaces. For instance, you can calculate the energy levels of the Hydrogen atom using just the Schrodinger Equation (that's how Schrodinger did it!) But if you want to describe particle interactions, or perturbation, or just talk in general about quantum superpositions, then wavefunctions don't cut it--you have to understand Hilbert spaces, and all of the linear algebra that goes with them.

Thanks this is the clearest answer ever after days of agonizing about it.. :)

I was reading this book Introducing Quantum Theory by McEvoy. It says:

"Thus, the solution of Schroedinger's equation - the wave function for the system - was replaced by an infinite series - the wave functions of the individual states - which are natural harmonics of each other. That is to say, their frequencies are related in the ratio of whole numbers, or integers.

The method is shown by the graphs below. The bold curve indicates the initial function which is then replaced by the sum of the infinite series of the harmonic periodic waves.

Schroedinger's remarkable discovery was that the replacement waves described the individual states of the quantum system and their amplitudes gave the relative importance of that particular state to the whole system."

Question. Are the replacement waves (or Fourier components) describing the individual states of the quantum system equivalent to the basis vectors in Hilbert Space? If not. What is the equivalent of replacement waves (or Fourier components) in Hilbert Space?

 

[Chopin]

Are the replacement waves (or Fourier components) describing the individual states of the quantum system equivalent to the basis vectors in Hilbert Space?

Yes. Or more accurately, they are one possible basis for the Hilbert space. Just like any vector space, there are infinitely many sets of vectors which can serve as a basis for the space, each as good as any other. We may choose one which is especially convenient for us at the time, but we may later decide to span the space with a new set of vectors. This ability to look at things from multiple directions is the most important concept to understand about QM (in some ways, it's really the only thing to understand.) It is also exactly the same as any other vector space, so if you have experience with linear algebra, chances are you've already puzzled through this concept.

There's really nothing special about the SE that causes its solutions to work like this. It is a theorem that the solutions of any linear differential equation form a vector space--if you add two of them together or multiply by a scalar, the result is again a solution of the equation. The most natural set of basis vectors to use are those corresponding to the harmonics, in exactly the same way as the solutions to a classical standing wave equation are most easily described by a set of Fourier modes. The fact that a Hilbert space can be constructed out of a quantum mechanical system is just a consequence of the SE being linear (or alternatively, the other way around.)

 

[Matterwave]

Some theories exist best in certain mathematical languages, but it doesn't mean the theory cannot take some other form. Another example of this from QM, is Einstein's Special Theory of Relativity. Einstein formulated it, and it was only later that Minkowski formulated the theory in his 4 dimensional language (with the Minkowski metric, and 4 vectors, etc).

 

[dextercioby]

In my mind, out of all topological vector spaces it's easiest to work in a Hilbert space. The scalar product offers the natural environment to set up a probabilistic interpretation. I cannot conceive QM without probability (densities), hence without a pre-Hilbert structure of the state space. Completion of this space with respect to the strong topology is a mathematical <must>, as some useful theorems require it.

 

[Varon]

Yes. Or more accurately, they are one possible basis for the Hilbert space. Just like any vector space, there are infinitely many sets of vectors which can serve as a basis for the space, each as good as any other. We may choose one which is especially convenient for us at the time, but we may later decide to span the space with a new set of vectors. This ability to look at things from multiple directions is the most important concept to understand about QM (in some ways, it's really the only thing to understand.) It is also exactly the same as any other vector space, so if you have experience with linear algebra, chances are you've already puzzled through this concept.

I was asking whether the replacement waves (or Fourier components) describing the individual states of the quantum system is equivalent to the basis vectors in Hilbert Space, and you said only one possible basis. Why not all basis since the Fourier has infinite series and it can map to the Hilbert space infinite dimensions. Both are infinite. Or are you saying that the Fourier components chosen is the basis chosen by the measurements? But before measurement, can't we map the infinite basis vectors to all the Fourier infinite components? They can fit since both are infinite.

There's really nothing special about the SE that causes its solutions to work like this. It is a theorem that the solutions of any linear differential equation form a vector space--if you add two of them together or multiply by a scalar, the result is again a solution of the equation. The most natural set of basis vectors to use are those corresponding to the harmonics, in exactly the same way as the solutions to a classical standing wave equation are most easily described by a set of Fourier modes. The fact that a Hilbert space can be constructed out of a quantum mechanical system is just a consequence of the SE being linear (or alternatively, the other way around.)

http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/psi.html

In the paper above is detailed review for those who finished quantum mechanics class. Nowhere is Hilbert Space mentioned. So HS formalism is only taken up by in more advanced class?

About Hilbert space. I'm quite familiar with this aspect having spent time reading on the Preferred Basis problem. I just want now to understand what's it connection to Fourier series.
So when the ray collapses to the basis vectors. Is the collapse counterpart in Fourier the choosing of one or a few of the component waves (akin to the choosing of a basis in the wave function collapse)? Also you said earlier that quantum superpositions need Hilbert space and Fourier series don't work. But isn't it that the main wave in Fourier is the sum of all component waves. So why can't we say the component waves is like the basis vectors and the ray is like the main Fourier wave (so quantum superposition should also be possible in Fourier waves)?

 

[Chopin]

I cannot conceive QM without probability (densities), hence without a pre-Hilbert structure of the state space.

Very true, although it's important to note that the wavefunction was not always viewed as a probability measure. When Schrodinger first developed the SE, he viewed $$\Psi$$ as a classical charge density, and you can get a ways into QM (like atomic energy levels) with that assumption. It's only when you get into things like scattering experiments that you have to start viewing it as a probability density.

I was asking whether the replacement waves (or Fourier components) describing the individual states of the quantum system is equivalent to the basis vectors in Hilbert Space, and you said only one possible basis. Why not all basis since the Fourier has infinite series and it can map to the Hilbert space infinite dimensions. Both are infinite. Or are you saying that the Fourier components chosen is the basis chosen by the measurements? But before measurement, can't we map the infinite basis vectors to all the Fourier infinite components? They can fit since both are infinite.

Any basis for the Hilbert space must have infinite cardinality, since it's infinite-dimensional. But you can always form a new basis for a space out of linear combinations of the old one. For instance, say we have vectors $$V_0, V_1, V_2, ...$$ that form the basis of the space. We can construct new vectors $$W_0 = V_0 + V_1, W_1 = V_1 + V_2, W_2 = V_2 + V_3, ...$$, and the result will also be a basis for the space. This works exactly like finite-dimensional vector spaces that are studied in linear algebra.

The reason that that specific basis is used when talking about the SE is that each of the components has a definite energy value (technically, because the basis is comprised of the eigenvectors of the Hamiltonian.) This means you can determine the energy level of any wave just by taking a weighted average of the energies of each component wave.

http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/psi.html

In the paper above is detailed review for those who finished quantum mechanics class. Nowhere is Hilbert Space mentioned. So HS formalism is only taken up by in more advanced class?

My understanding is that the Hilbert space concept was largely developed by Dirac, as a way of formalizing the work that had thus far gone on in the development of QM. As you can see by the paper you have linked to, you can do quite a bit without referring to it. But you're using it implicitly all along, by virtue of the fact that the solutions of the SE combine linearly. You're just not calling it out using vector space terminology. Dirac's innovation was developing a notation that made it easy to do so, and reinterpreting the complex magnitude of the wavefunction as an inner product between rays in the Hilbert space.

About Hilbert space. I'm quite familiar with this aspect having spent time reading on the Preferred Basis problem. I just want now to understand what's it connection to Fourier series.
So when the ray collapses to the basis vectors. Is the collapse counterpart in Fourier the choosing of one or a few of the component waves (akin to the choosing of a basis in the wave function collapse)? Also you said earlier that quantum superpositions need Hilbert space and Fourier series don't work. But isn't it that the main wave in Fourier is the sum of all component waves. So why can't we say the component waves is like the basis vectors and the ray is like the main Fourier wave (so quantum superposition should also be possible in Fourier waves)?

We can and do say that. The Fourier series is just forming a linear combination of basis states, which is exactly what the Hilbert space is. The two are isomorphic to each other--the Fourier decomposition of any wave (classical or quantum) can be viewed as expanding the wave into a basis defined by the harmonics. The Hilbert space formalism is just an extension of this concept that let's you do more advanced things with it.

 

[unusualname]

Even the complex numbers with norm |psi|^2 (ie psi.psi* or inner product psi1.psi2*) is a hilbert space.

So you cannot do QM without hilbert space, because nature has complex probability amplitudes, 'nuff said.

 

[JesseM]

Varon, are you familiar with the notion that if you have a position wavefunction, you can find the amplitudes of different momentum eigenstates by expressing the position wavefunction as a Fourier series, where each term in the series is a momentum eigenstate? In terms of the position representation, each momentum eigenstate corresponds to a uniform probability of finding the particle anywhere in all of space, but with the phase of wavefunction varying like a sine wave (the "phase" corresponds to the direction in the complex plane that the complex amplitude is pointing, if you consider the amplitude of finding the particle at each possible position in space for a given momentum eigenstate). So, only in the momentum eigenstate does the particle have a single definite wavelength. When you think of it this way, you can show that something like the "uncertainty principle" would apply even to ordinary classical waves (the more localized a wavepacket is in space, the greater the spread in wavelengths in the Fourier series), although obviously classical waves aren't waves of probability so the interpretation of the quantum uncertainty principle is a bit different. Anyway, this gives you a possibly more intuitive way of understanding why position and momentum don't "commute" that just involves picturing waves in space, not matrices.

Here's a good basic intro with some pictures:

http://webs.morningside.edu/slaven/physics/uncertainty/uncertainty5.html

 

[Chopin]

Even the complex numbers with norm |psi|^2 (ie psi.psi* or inner product psi1.psi2*) is a hilbert space.

So you cannot do QM without hilbert space, because nature has complex probability amplitudes, 'nuff said.

Ha. Well said.

 

[Varon]

Varon, are you familiar with the notion that if you have a position wavefunction, you can find the amplitudes of different momentum eigenstates by expressing the position wavefunction as a Fourier series, where each term in the series is a momentum eigenstate? In terms of the position representation, each momentum eigenstate corresponds to a uniform probability of finding the particle anywhere in all of space, but with the phase of wavefunction varying like a sine wave (the "phase" corresponds to the direction in the complex plane that the complex amplitude is pointing, if you consider the amplitude of finding the particle at each possible position in space for a given momentum eigenstate). So, only in the momentum eigenstate does the particle have a single definite wavelength. When you think of it this way, you can show that something like the "uncertainty principle" would apply even to ordinary classical waves (the more localized a wavepacket is in space, the greater the spread in wavelengths in the Fourier series), although obviously classical waves aren't waves of probability so the interpretation of the quantum uncertainty principle is a bit different. Anyway, this gives you a possibly more intuitive way of understanding why position and momentum don't "commute" that just involves picturing waves in space, not matrices.

Here's a good basic intro with some pictures:

http://webs.morningside.edu/slaven/physics/uncertainty/uncertainty5.html

Yes I'm familiar with how HUP can be modeled by pure waves. In fact this is shown in details in the book Deep Down Things. But it never mention about Hilbert Space. so this position with many momentum component waves making it up. in Hilbert version, is each momentum wave equivalent to one axis component vector or axis in Hilbert space?

I was reading another book on the history of QM. Schroedinger actually thought the wave was physical. These days, we are told the wave function is not in 3D space because the Hilbert Space is multi dimensional, but if we model it by pure Fourier, each component even infinite is still in 3D space. So there is still possibility Schrodinger is right that the wave is in 3D? Why not?

 

[Varon]

Very true, although it's important to note that the wavefunction was not always viewed as a probability measure. When Schrodinger first developed the SE, he viewed $$\Psi$$ as a classical charge density, and you can get a ways into QM (like atomic energy levels) with that assumption. It's only when you get into things like scattering experiments that you have to start viewing it as a probability density.



Any basis for the Hilbert space must have infinite cardinality, since it's infinite-dimensional. But you can always form a new basis for a space out of linear combinations of the old one. For instance, say we have vectors $$V_0, V_1, V_2, ...$$ that form the basis of the space. We can construct new vectors $$W_0 = V_0 + V_1, W_1 = V_1 + V_2, W_2 = V_2 + V_3, ...$$, and the result will also be a basis for the space. This works exactly like finite-dimensional vector spaces that are studied in linear algebra.

The reason that that specific basis is used when talking about the SE is that each of the components has a definite energy value (technically, because the basis is comprised of the eigenvectors of the Hamiltonian.) This means you can determine the energy level of any wave just by taking a weighted average of the energies of each component wave.

Are you referring to Fourier when you mentioned that the component has a definite energy value? Or can you model this by Hilbert space too that each axis has a definite energy value? If so, it is possible to totally do away with Fourier and use pure Hilbert space even when dealing with eigenvectors of the Hamiltonian? Meaning you only focus on certain basis vectors and ignore the rest?

My understanding is that the Hilbert space concept was largely developed by Dirac, as a way of formalizing the work that had thus far gone on in the development of QM. As you can see by the paper you have linked to, you can do quite a bit without referring to it. But you're using it implicitly all along, by virtue of the fact that the solutions of the SE combine linearly. You're just not calling it out using vector space terminology. Dirac's innovation was developing a notation that made it easy to do so, and reinterpreting the complex magnitude of the wavefunction as an inner product between rays in the Hilbert space.



We can and do say that. The Fourier series is just forming a linear combination of basis states, which is exactly what the Hilbert space is. The two are isomorphic to each other--the Fourier decomposition of any wave (classical or quantum) can be viewed as expanding the wave into a basis defined by the harmonics. The Hilbert space formalism is just an extension of this concept that let's you do more advanced things with it.

what advanced things that Hilbert space can do that Fourier cant? Mathematically? Why can't you use pure Fourier to describe the double slit experiment?

 

[Chopin]

Are you referring to Fourier when you mentioned that the component has a definite energy value? Or can you model this by Hilbert space too that each axis has a definite energy value? If so, it is possible to totally do away with Fourier and use pure Hilbert space even when dealing with eigenvectors of the Hamiltonian? Meaning you only focus on certain basis vectors and ignore the rest?

You're always going to be taking the eigenvectors of the Hamiltonian and considering them the set of base states. In an infinite square well, those happen to form a Fourier series, but they aren't always. For instance, if you analyze something like a Coulomb potential, you'll again find a set of constant-energy solutions, which look sort of like sine waves, but their amplitudes and frequencies will vary as you move out from the center of the potential. These waves can't be considered a Fourier decomposition anymore, because they're not just sine waves, but you can still form all valid solutions to the equation by combining them linearly.

So it's the same idea as a Fourier series, where you have a set of base states that you combine together, but the math doesn't work out like a Fourier series anymore (you aren't adding sine waves, you're adding crazy Bessel functions or something.) This is what I mean when I say that a Hilbert space is a generalization of a Fourier series--the elements of a Fourier series are actually also a Hilbert space, even though you might not think about them in those terms.

what advanced things that Hilbert space can do that Fourier cant? Mathematically? Why can't you use pure Fourier to describe the double slit experiment?

The double slit thing was probably a bad example. A better example of something that a Hilbert space can do that's beyond the Schrodinger wavefunction's capability is handling particles with spin. Say you have a single electron which can be either spin up or spin down. According to the laws of quantum mechanics, this means it can also be in a superposition of these two states--for instance, it could be 75% spin up and 25% spin down. We can describe this by naming two states, perhaps $$U$$ for Up and $$D$$ for Down. Then our new state would be $$X = .75U + .25D$$. You can see that we're making linear combinations of states here too, just like we would in a Fourier series if we said that a wavefunction was 75% fundamental and 25% 1st harmonic. It's just applying the same sort of concept to a different domain of states.

 

[JesseM]

Yes I'm familiar with how HUP can be modeled by pure waves. In fact this is shown in details in the book Deep Down Things. But it never mention about Hilbert Space. so this position with many momentum component waves making it up. in Hilbert version, is each momentum wave equivalent to one axis component vector or axis in Hilbert space?

Yes, but only in the momentum basis, a momentum wave (momentum eigenstate) would be a sum of an infinite number of different vectors in the position basis (where each basis vector corresponds to a wavefunction where the amplitude is perfectly localized to a single point, so you just have a spike at that point and zero amplitude everywhere else). Similarly each position basis vector (position eigenstate) is the sum of an infinite number of different momentum eigenstates. I talked a bit more about the idea of decomposing a state vector into a sum of basis vectors in [post=3250764]this post[/post].

I was reading another book on the history of QM. Schroedinger actually thought the wave was physical. These days, we are told the wave function is not in 3D space because the Hilbert Space is multi dimensional, but if we model it by pure Fourier, each component even infinite is still in 3D space. So there is still possibility Schrodinger is right that the wave is in 3D? Why not?

Schrödinger originally thought this way but although the idea might seem to work for single-particle wavefunctions, he soon realized it doesn't work for multiparticle wavefunctions. There's a good history of these ideas in a book called The Infamous Boundary by David Wick, in the chapter "Revolution, Part II: Schrödinger's Waves"...on p. 34 Wick writes:

"Schrödinger still hopes his wave might represent a spread-out electron. But there is a second difficulty: for two or more particles, more than three numbers are needed to describe their locations, so his wave exists in a fictitious space of more than three dimensions. Schrödinger has not forgotten this point; he even emphasizes it a few sentences later:

It has been stressed repeatedly that the [wave]-function itself cannot and must not in general be interpreted directly in terms of three-dimensional space however the one-electron problem leads towards this...

"Unfortunately for Schrödinger, this last fact is fatal to his view."

 

[Varon]

Yes, but only in the momentum basis, a momentum wave (momentum eigenstate) would be a sum of an infinite number of different vectors in the position basis (where each basis vector corresponds to a wavefunction where the amplitude is perfectly localized to a single point, so you just have a spike at that point and zero amplitude everywhere else). Similarly each position basis vector (position eigenstate) is the sum of an infinite number of different momentum eigenstates. I talked a bit more about the idea of decomposing a state vector into a sum of basis vectors in [post=3250764]this post[/post].

Schrödinger originally thought this way but although the idea might seem to work for single-particle wavefunctions, he soon realized it doesn't work for multiparticle wavefunctions. There's a good history of these ideas in a book called The Infamous Boundary by David Wick, in the chapter "Revolution, Part II: Schrödinger's Waves"...on p. 34 Wick writes:

"Schrödinger still hopes his wave might represent a spread-out electron. But there is a second difficulty: for two or more particles, more than three numbers are needed to describe their locations, so his wave exists in a fictitious space of more than three dimensions. Schrödinger has not forgotten this point; he even emphasizes it a few sentences later:

"Unfortunately for Schrödinger, this last fact is fatal to his view."

What exactly are the 3 numbers to describe the position? Axis? Hence if 2 particles then 6 axis. But this assumption is based on modeling it in Hilbert space. If you use plain Fourier, the 2 particles are still located in 3D space. You must include more waves and the waves are all in 3d!

Btw.. The integers called quantum numbers by Bohr, Sommerfield and Heisenberg which are said "to be related in a natural way to the numbers if nodes in a vibrating system"' what are their equivalent in Hilbert space via basis vectors?

 

[Varon]

You're always going to be taking the eigenvectors of the Hamiltonian and considering them the set of base states. In an infinite square well, those happen to form a Fourier series, but they aren't always. For instance, if you analyze something like a Coulomb potential, you'll again find a set of constant-energy solutions, which look sort of like sine waves, but their amplitudes and frequencies will vary as you move out from the center of the potential. These waves can't be considered a Fourier decomposition anymore, because they're not just sine waves, but you can still form all valid solutions to the equation by combining them linearly.

So it's the same idea as a Fourier series, where you have a set of base states that you combine together, but the math doesn't work out like a Fourier series anymore (you aren't adding sine waves, you're adding crazy Bessel functions or something.) This is what I mean when I say that a Hilbert space is a generalization of a Fourier series--the elements of a Fourier series are actually also a Hilbert space, even though you might not think about them in those terms.

Why did you mention about the coulomb potential? Are you giving it as example that it can only be modeled by Hilbert space and not by fourier?

A separate question, I know that Hilbert space is like phase space where a ray or point describes the whole quantum state. So you mean in the Schroedinger equation, not every basis is used or they only put the base states. This means it depends on your application what kind of data to put in the basis? For example. When calculating for kinetic energy of an electron, you supply a certain base state. When calculating for position, you put certain state, etc? I thought that one simply put the complete state. Why. What would happen if you put the complete state or use complete basis instead of just one basis?

The double slit thing was probably a bad example. A better example of something that a Hilbert space can do that's beyond the Schrodinger wavefunction's capability is handling particles with spin. Say you have a single electron which can be either spin up or spin down. According to the laws of quantum mechanics, this means it can also be in a superposition of these two states--for instance, it could be 75% spin up and 25% spin down. We can describe this by naming two states, perhaps $$U$$ for Up and $$D$$ for Down. Then our new state would be $$X = .75U + .25D$$. You can see that we're making linear combinations of states here too, just like we would in a Fourier series if we said that a wavefunction was 75% fundamental and 25% 1st harmonic. It's just applying the same sort of concept to a different domain of states.

 

[JesseM]

What exactly are the 3 numbers to describe the position? Axis? Hence if 2 particles then 6 axis. But this assumption is based on modeling it in Hilbert space. If you use plain Fourier, the 2 particles are still located in 3D space. You must include more waves and the waves are all in 3d!

But the wavefunction itself assigns a single amplitude to each combination of positions for both particles, it doesn't give a separate set of amplitudes for each particle individually to be found at various positions.

Btw.. The integers called quantum numbers by Bohr, Sommerfield and Heisenberg which are said "to be related in a natural way to the numbers if nodes in a vibrating system"' what are their equivalent in Hilbert space via basis vectors?

Are you talking about the quantum numbers for electrons in an atom? If so I think the idea is that the numbers characterize different stationary states for the electron, each of which is an eigenstate for energy and one where the probabilities of finding the electron in different positions don't change with time (though the complex amplitude can change, meaning the phase of the amplitude vector can change while the magnitude doesn't, the wiki article has some graphic illustrations and some more mathematical details are http://itl.chem.ufl.edu/3417_s98/lectures/super_1.html ). Some details on what the numbers represent individually here. And for each set of quantum numbers you can plot the distinctive shape of the probability density function in space (often they just plot a surface of constant probability density), see here or here.

 

[Chopin]

Why did you mention about the coulomb potential? Are you giving it as example that it can only be modeled by Hilbert space and not by fourier?

I think part of your confusion may come from the fact that you're using "Fourier" to talk about two different things. First, you're using it to describe a series of sinusoidal waves, with frequencies that are integral multiples of each other. Second, you're using it to describe the act of modeling a system with a wavefunction $$\Psi$$, solving the Schrodinger Equation to find various solutions $$\Psi_0, \Psi_1, \Psi_2, ...$$, and then finally observing that you can build any solution out of a linear combination of these solutions.

For the infinite square well, those solutions happen to be a bunch of sine waves, with frequencies that are integral multiples of each other. Hooray, it's a Fourier series! That means that you can model any solution to the equation as a combination of these waves, which is basically just the spectral decomposition of the function.

What I was trying to illustrate with my Coulomb potential, though, is that the solutions to the equation need not be a bunch of sinusoids. For the Coulomb potential, you can still find a series of solutions $$\Psi_0, \Psi_1, \Psi_2, ...$$, but they're not going to be sinusoids, they're going to be some god-awful set of Bessel functions or something (math gurus, is that right? If it isn't, it's some god-awful set of some other kind of function, at least, so my point is the same.) So the set of solutions no longer forms a Fourier series, so you can't just take the spectral decomposition of the solution anymore. Therefore, in Sense 1 from above, we can no longer talk about this being a Fourier series.

However, your Sense 2 from above still applies--we can form any solution to the equation by linearly combining these basis solutions. This comes from the fact that the SE is linear, so those solutions $$\Psi_0, \Psi_1, \Psi_2, ...$$ form a vector space, which comprises all of the solutions to the equation. The neat thing about this vector space is that you can pick any basis you want, and it will be possible to expand out a function in terms of it. If you're interested in finding energy levels, it will be convenient to find a basis for which each vector has a definite energy. If you're interested in finding position, it will be convenient to find one for which each vector has a definite position, etc. This vector space is the Hilbert space--there's nothing scary or mystical about it, it's just a way of enumerating all of the different solutions to the equation, and makes explicit the fact that any linear combination of them is again a solution.

 

[Varon]

But the wavefunction itself assigns a single amplitude to each combination of positions for both particles, it doesn't give a separate set of amplitudes for each particle individually to be found at various positions.

Why can't you use Fourier which is only located in 3d in modelling the amplitude of positions for both particles? Unless you mean Fourier can be in more than 3d too?

Schroedinger is a genius. He could have easily thought that two electrons would produce more than 3 space. But maybe he thinks Fourier is enough to describe two particles,

Are you talking about the quantum numbers for electrons in an atom? If so I think the idea is that the numbers characterize different stationary states for the electron, each of which is an eigenstate for energy and one where the probabilities of finding the electron in different positions don't change with time (though the complex amplitude can change, meaning the phase of the amplitude vector can change while the magnitude doesn't, the wiki article has some graphic illustrations and some more mathematical details are http://itl.chem.ufl.edu/3417_s98/lectures/super_1.html ). Some details on what the numbers represent individually here. And for each set of quantum numbers you can plot the distinctive shape of the probability density function in space (often they just plot a surface of constant probability density), see here or here.

Im taiking of n as the size of the orbital, k as the shape of the orbit and m as the direction in which the orbit is point. My book Introducing Quantum Theory mentions them and the book never talk about Hilbert Space so wonder how they are arranged in Hilbert space.

 

[Chopin]

Why can't you use Fourier which is only located in 3d in modelling the amplitude of positions for both particles? Unless you mean Fourier can be in more than 3d too?

Because it doesn't work that way. If you want to write a wavefunction for two particles, you write it as $$\Psi(x_0,y_0,z_0,x_1,y_1,z_1)$$, which means "the probability of finding particle 0 at $$(x_0,y_0,z_0)$$ AND particle 1 at $$(x_1,y_1,z_1)$$." Therefore, you're now performing calculations in a 6-dimensional space. This is a concept that took me forever to wrap my head around, and led to a lot of confusion until I did.

Im taiking of n as the size of the orbital, k as the shape of the orbit and m as the direction in which the orbit is point. My book Introducing Quantum Theory mentions them and the book never talk about Hilbert Space so wonder how they are arranged in Hilbert space.

I'm going to go ahead and bust out the bra-ket notation here, to give some real examples of how this works. If we want to describe the orbitals as vectors in a Hilbert space, we might label them as $$|\psi_{nkm}\rangle$$, so the vector $$|\psi_{032}\rangle$$ would represent the orbital with n=0, k=3, m=2.

Using this notation, we can now describe a particle which is in a superposition of these orbitals, by saying something like $$|\phi\rangle = .75 |\psi_{012}\rangle + .25 |\psi_{321}\rangle$$.

So far, this is just a slightly different way of writing our states, and doesn't really add anything useful. Where it really gets powerful, though, is if you want to test an arbitrary vector to see what states it's in. Suppose I give you a vector $$|\phi\rangle$$, and you want to know what the probability is that it's in state $$|\psi_{021}\rangle$$. You can take the inner product of $$|\phi\rangle$$ and $$|\psi_{021}\rangle$$, which we denote as $$\langle\psi_{021}|\phi\rangle$$, and that value squared will give you the probability you're looking for.

You can do the same thing with wavefunctions, but it's much more cumbersome. The equivalent calculation using a wave function is $$\int{\psi_{021}(x)^*\phi(x) dx}$$, which is a lot harder to carry out. The Hilbert space just gives you a way to talk about these things without worrying about what their shape in space looks like. That makes it much easier to write out the calculations.

 

[Varon]

I think part of your confusion may come from the fact that you're using "Fourier" to talk about two different things. First, you're using it to describe a series of sinusoidal waves, with frequencies that are integral multiples of each other. Second, you're using it to describe the act of modeling a system with a wavefunction $$\Psi$$, solving the Schrodinger Equation to find various solutions $$\Psi_0, \Psi_1, \Psi_2, ...$$, and then finally observing that you can build any solution out of a linear combination of these solutions.

For the infinite square well, those solutions happen to be a bunch of sine waves, with frequencies that are integral multiples of each other. Hooray, it's a Fourier series! That means that you can model any solution to the equation as a combination of these waves, which is basically just the spectral decomposition of the function.

What I was trying to illustrate with my Coulomb potential, though, is that the solutions to the equation need not be a bunch of sinusoids. For the Coulomb potential, you can still find a series of solutions $$\Psi_0, \Psi_1, \Psi_2, ...$$, but they're not going to be sinusoids, they're going to be some god-awful set of Bessel functions or something (math gurus, is that right? If it isn't, it's some god-awful set of some other kind of function, at least, so my point is the same.) So the set of solutions no longer forms a Fourier series, so you can't just take the spectral decomposition of the solution anymore. Therefore, in Sense 1 from above, we can no longer talk about this being a Fourier series.

However, your Sense 2 from above still applies--we can form any solution to the equation by linearly combining these basis solutions. This comes from the fact that the SE is linear, so those solutions $$\Psi_0, \Psi_1, \Psi_2, ...$$ form a vector space, which comprises all of the solutions to the equation. The neat thing about this vector space is that you can pick any basis you want, and it will be possible to expand out a function in terms of it. If you're interested in finding energy levels, it will be convenient to find a basis for which each vector has a definite energy. If you're interested in finding position, it will be convenient to find one for which each vector has a definite position, etc. This vector space is the Hilbert space--there's nothing scary or mystical about it, it's just a way of enumerating all of the different solutions to the equation, and makes explicit the fact that any linear combination of them is again a solution.

Thanks. I'm only thinking of Sense1 my background of Fourier came from music. So you are saying there is a Sense2 in which you add the solutions even if they don't involve sine waves? What is the particular concept called? I thought Fourier only involves sine waves and superpositions of them.

About Hilbert space being mystical. Lol. Hilbert space is the home of Schrodinger cat in a ghostly superposition of being dead or alive. Before measurement, we don't even know what occurs...whether*branches are splitted into Many worlds or Bohmian mechanics with instantaneous wave function or observers having the power to collapse the wave function, etc. So yes Virginia, Hilbert space is a mystical place. :)

 

[Varon]

Because it doesn't work that way. If you want to write a wavefunction for two particles, you write it as $$\Psi(x_0,y_0,z_0,x_1,y_1,z_1)$$, which means "the probability of finding particle 0 at $$(x_0,y_0,z_0)$$ AND particle 1 at $$(x_1,y_1,z_1)$$." Therefore, you're now performing calculations in a 6-dimensional space. This is a concept that took me forever to wrap my head around, and led to a lot of confusion until I did.



I'm going to go ahead and bust out the bra-ket notation here, to give some real examples of how this works. If we want to describe the orbitals as vectors in a Hilbert space, we might label them as $$|\psi_{nkm}\rangle$$, so the vector $$|\psi_{032}\rangle$$ would represent the orbital with n=0, k=3, m=2.

Using this notation, we can now describe a particle which is in a superposition of these orbitals, by saying something like $$|\phi\rangle = .75 |\psi_{012}\rangle + .25 |\psi_{321}\rangle$$.

So far, this is just a slightly different way of writing our states, and doesn't really add anything useful. Where it really gets powerful, though, is if you want to test an arbitrary vector to see what states it's in. Suppose I give you a vector $$|\phi\rangle$$, and you want to know what the probability is that it's in state $$|\psi_{021}\rangle$$. You can take the inner product of $$|\phi\rangle$$ and $$|\psi_{021}\rangle$$, which we denote as $$\langle\psi_{021}|\phi\rangle$$, and that value squared will give you the probability you're looking for.

You can do the same thing with wavefunctions, but it's much more cumbersome. The equivalent calculation using a wave function is $$\int{\psi_{021}(x)^*\phi(x) dx}$$, which is a lot harder to carry out. The Hilbert space just gives you a way to talk about these things without worrying about what their shape in space looks like. That makes it much easier to write out the calculations.

Thanks for the info. Note that wiki said:

"The modern usage of the term wave function refers to a complex vector or function, i.e. an element in a complex Hilbert space"

Therefore I think you must mention "Hilbertless wave function" to refer to this cumbersome method. Or maybe there is a more correct term for it?

 

[JesseM]

Why can't you use Fourier which is only located in 3d in modelling the amplitude of positions for both particles? Unless you mean Fourier can be in more than 3d too?

Yes, you can use Fourier analysis in any number of dimensions.

Im taiking of n as the size of the orbital, k as the shape of the orbit and m as the direction in which the orbit is point. My book Introducing Quantum Theory mentions them and the book never talk about Hilbert Space so wonder how they are arranged in Hilbert space.

I don't really understand what you mean by "arranged in Hilbert space". Each quantum state vector in Hilbert space can be broken down into a sum of eigenstates of observables like energy and position (you need a complete set of commuting observables to uniquely break down any quantum state vector into such a sum), the feature of a stationary state is that while the state vector itself may change over time, at all times the same single energy eigenstate is the only energy eigenstate in the sum (the complex amplitude on all the other energy eigenstates would be zero), and while the complex amplitude on position eigenstates in the sum may change, the amplitude-squared on each position eigenstate (representing the probability of finding the electron at that position) remains constant.

 

[Chopin]

Thanks. I'm only thinking of Sense1 my background of Fourier came from music. So you are saying there is a Sense2 in which you add the solutions even if they don't involve sine waves? What is the particular concept called? I thought Fourier only involves sine waves and superpositions of them.

Sense 1 is the only correct meaning of the term "Fourier". What I meant was that I think you may have been mistakenly using that term to describe Sense 2, which doesn't have anything to do with Fourier series at all. That's just adding solutions together, which is a property of any linear space. However, adding Fourier components happens to be a special case of this phenomenon where the components are sine waves, so it's understandable where the confusion came from.

About Hilbert space being mystical. Lol. Hilbert space is the home of Schrodinger cat in a ghostly superposition of being dead or alive. Before measurement, we don't even know what occurs...whether*branches are splitted into Many worlds or Bohmian mechanics with instantaneous wave function or observers having the power to collapse the wave function, etc. So yes Virginia, Hilbert space is a mystical place. :)

All of the mystical stuff comes from the interpretation of the Hilbert space, not from the actual math of the space itself. The math of the Hilbert space is really just pretty banal manipulations of vectors and inner products--exactly like good old vectors and matricies from high school math. For instance, the famous cat is just the sum of two state vectors $$|\psi_{cat}\rangle = \frac{1}{2}|\psi_{alive}\rangle + \frac{1}{2}|\psi_{dead}\rangle$$ (perhaps we should call this Schrodinger's Ket?) This is just the sum of two vectors, like any other two vectors.

The mysterious stuff comes when we try to interpret what that means--it happens that we've just added together two states that we normally think of as being mutually exclusive. That's when all of the probability and wavefunction collapse voodoo starts to come out. But that's all at the end, when we're interpreting results. All of the math we did before that, inside of the Hilbert space, is plain old arithmetic on vectors--completely exact, no probabilities or mystical behavior at all.

Thanks for the info. Note that wiki said:

"The modern usage of the term wave function refers to a complex vector or function, i.e. an element in a complex Hilbert space"

Therefore I think you must mention "Hilbertless wave function" to refer to this cumbersome method. Or maybe there is a more correct term for it?

The wavefunction is just a nifty way of describing the state in the position basis--it assigns a complex number to every position in space. This is a perfectly good vector in the Hilbert space, it just wasn't thought of that way originally. It was only when Dirac came along and formalized the theory that we started thinking of it that way. The integral I gave above is just a dot product across an infinite number of basis vectors, which is exactly what the bra-ket notation says, it's just a much more compact way of writing it, that doesn't make explicit reference to the position basis.

 

[Varon]

Sense 1 is the only correct meaning of the term "Fourier". What I meant was that I think you may have been mistakenly using that term to describe Sense 2, which doesn't have anything to do with Fourier series at all. That's just adding solutions together, which is a property of any linear space. However, adding Fourier components happens to be a special case of this phenomenon where the components are sine waves, so it's understandable where the confusion came from.

When I asked this question "Why did you mention about the coulomb potential? Are you giving it as example that it can only be modeled by Hilbert space and not by fourier?" I was only thinking in terms of Sense1. But you didn't reply yes or no. So I thought you were trying to say Fourier is still somehow used in coulomb potential in some way. But now knowing the context that I only meant Sense1. Then your answer is "yes" to the question "Are you giving it as example that it can only be modeled by Hilbert space and not by fourier?", right? Look. English is not my native language, and since I'm a novice in QM. I need a yes or no answer to be definite of my question. So just answer "right" if you are saying that the coulomb potential can only be modeled by Hilbert space and not by Fourier which I only understand use Sense1 which involve sine waves. This is to be 100% sure of your answer before I leave this thread.

All of the mystical stuff comes from the interpretation of the Hilbert space, not from the actual math of the space itself. The math of the Hilbert space is really just pretty banal manipulations of vectors and inner products--exactly like good old vectors and matricies from high school math. For instance, the famous cat is just the sum of two state vectors $$|\psi_{cat}\rangle = \frac{1}{2}|\psi_{alive}\rangle + \frac{1}{2}|\psi_{dead}\rangle$$ (perhaps we should call this Schrodinger's Ket?) This is just the sum of two vectors, like any other two vectors.

The mysterious stuff comes when we try to interpret what that means--it happens that we've just added together two states that we normally think of as being mutually exclusive. That's when all of the probability and wavefunction collapse voodoo starts to come out. But that's all at the end, when we're interpreting results. All of the math we did before that, inside of the Hilbert space, is plain old arithmetic on vectors--completely exact, no probabilities or mystical behavior at all.



The wavefunction is just a nifty way of describing the state in the position basis--it assigns a complex number to every position in space. This is a perfectly good vector in the Hilbert space, it just wasn't thought of that way originally. It was only when Dirac came along and formalized the theory that we started thinking of it that way. The integral I gave above is just a dot product across an infinite number of basis vectors, which is exactly what the bra-ket notation says, it's just a much more compact way of writing it, that doesn't make explicit reference to the position basis.

Hmm... so the words "wave function" always involve the position basis. Can you please give example of QM application that doesn't use the position basis and hence Hilbert Space is used. Maybe coulomb potential is one? What else to get the idea. Now if you no longer use wave function but Hilbert space. What is the proper term for that. For example. What is the replacement for the term "Solve for the wave function"... maybe "solve for the Hilbert space"?, this doesn't seem right. What is the right term to use? Thanks.

 

[Truecrimson]

so the words "wave function" always involve the position basis.

This is not quite true. In fact, you already saw the definition in the Wikipedia page that a wave function refers to an element of a complex Hilbert space. For example, there is also a momentum space wave function, which is related to the position space wave function by the Fourier transform.

http://en.wikipedia.org/wiki/Momentum_space

You should really think of a wave function as an element in a Hilbert space (a vector space). How you actually write it down when, say, you want to integrate it, ultimately depends on your choice of basis. This is exactly the same as writing a vector as a column of numbers. If the basis is not known, there is no way to know what those number mean. And, of course, you have the freedom to choose a basis that simplifies your problem. For example, the functional form of the Hamiltonian operator depends on the basis. In particular, the momentum operator in the momentum space is just a multiplication by momentum. So for a case of a free particle, changing to the momentum basis reduces the Schroedinger equation from a second order PDE to a first order ODE.

 

[qsa]

you can check your PM to see a system that generates QM without hilbert space using a fundamental entity. It is similar to Walsh function which is a variant of Fourier.

http://en.wikipedia.org/wiki/Walsh_function

 

[Chopin]

When I asked this question "Why did you mention about the coulomb potential? Are you giving it as example that it can only be modeled by Hilbert space and not by fourier?" I was only thinking in terms of Sense1. But you didn't reply yes or no. So I thought you were trying to say Fourier is still somehow used in coulomb potential in some way. But now knowing the context that I only meant Sense1. Then your answer is "yes" to the question "Are you giving it as example that it can only be modeled by Hilbert space and not by fourier?", right? Look. English is not my native language, and since I'm a novice in QM. I need a yes or no answer to be definite of my question. So just answer "right" if you are saying that the coulomb potential can only be modeled by Hilbert space and not by Fourier which I only understand use Sense1 which involve sine waves. This is to be 100% sure of your answer before I leave this thread.

No, the solutions to the Coulomb potential cannot be modeled by a Fourier series, since the component solutions aren't sine waves.

In the infinite square well, the solutions are a set of sine waves, so combining them together forms a Fourier series. For the Coulomb potential, the solutions are complicated Bessel functions. You can still add them together, but since they aren't sine waves, it isn't really a Fourier series anymore.

More importantly, though, "Hilbert space" and "Fourier" aren't the same kind of thing, so you can't even really compare them. "Hilbert space" and "wavefunction" are the same kind of thing. Specifically, the wavefunction is just a shorthand for talking about the position basis of the Hilbert space.

Hmm... so the words "wave function" always involve the position basis. Can you please give example of QM application that doesn't use the position basis and hence Hilbert Space is used. Maybe coulomb potential is one? What else to get the idea. Now if you no longer use wave function but Hilbert space. What is the proper term for that. For example. What is the replacement for the term "Solve for the wave function"... maybe "solve for the Hilbert space"?, this doesn't seem right. What is the right term to use? Thanks.

The best example of problems that don't involve the position basis are operators with discrete eigenvectors. An easy example is photon polarization. A beam of photons can either be polarized horizontally or vertically. So to describe a photon's polarization, you need a two-dimensional Hilbert space. We can denote a photon that's polarized vertically as $$|V\rangle$$, and one that's polarized horizontally as $$|H\rangle$$. But a photon can also be in a superposition of horizontal and vertical. We denote this as $$|V\rangle + |H\rangle$$.

All of these states are vectors in a two-dimensional Hilbert space that describes the range of polarizations that a photon can have. This sort of problem is impossible to describe using a wavefunction, because there's no way to describe the problem in the position basis. A good text that describes these sorts of problems is the Feynman Lectures--he spends several chapters talking about discrete-valued problems like polarization and spin, and only then moves on to continuous operators like position and momentum.

 

[Varon]

This is not quite true. In fact, you already saw the definition in the Wikipedia page that a wave function refers to an element of a complex Hilbert space. For example, there is also a momentum space wave function, which is related to the position space wave function by the Fourier transform.

Chopin, hope you can comment on the above that it is more correct to refer to the wave function as an element of a complex Hilbert Space that is not necessarily position. You said: " "Hilbert space" and "wavefunction" are the same kind of thing. Specifically, the wavefunction is just a shorthand for talking about the position basis of the Hilbert space." But Truecrimson disagreed and said the wave function simply refers to an element of a complex Hilbert Space that is not necessarily position". Who is correct?

http://en.wikipedia.org/wiki/Momentum_space

You should really think of a wave function as an element in a Hilbert space (a vector space). How you actually write it down when, say, you want to integrate it, ultimately depends on your choice of basis. This is exactly the same as writing a vector as a column of numbers. If the basis is not known, there is no way to know what those number mean. And, of course, you have the freedom to choose a basis that simplifies your problem. For example, the functional form of the Hamiltonian operator depends on the basis. In particular, the momentum operator in the momentum space is just a multiplication by momentum. So for a case of a free particle, changing to the momentum basis reduces the Schroedinger equation from a second order PDE to a first order ODE.

 

[Chopin]

Chopin, hope you can comment on the above that it is more correct to refer to the wave function as an element of a complex Hilbert Space that is not necessarily position. You said: " "Hilbert space" and "wavefunction" are the same kind of thing. Specifically, the wavefunction is just a shorthand for talking about the position basis of the Hilbert space." But Truecrimson disagreed and said the wave function simply refers to an element of a complex Hilbert Space that is not necessarily position". Who is correct?

The wavefunction is almost always used as a way of describing the state in position basis. You could transform it into a wavefunction over momentum basis if you wanted to, though, because those two observables are both continuous. They're just two different bases for the Hilbert space, so the two different wavefunctions are just different representations of the same state. It's like if you have a vector, and then view it from a rotated coordinate system--the vector hasn't changed at all, but the numbers that describe it do, because the coordinate system is different.

What you can't do, though, is use a wavefunction to describe a particle with polarization or spin, because that observable is discrete.

 

[Varon]

The wavefunction is almost always used as a way of describing the state in position basis. You could transform it into a wavefunction over momentum basis if you wanted to, though, because those two observables are both continuous. They're just two different bases for the Hilbert space, so the two different wavefunctions are just different representations of the same state. It's like if you have a vector, and then view it from a rotated coordinate system--the vector hasn't changed at all, but the numbers that describe it do, because the coordinate system is different.

What you can't do, though, is use a wavefunction to describe a particle with polarization or spin, because that observable is discrete.

Ok. Thanks for all help. Are you a professor or student of physics?
Anyway. In the cosmology forum, I posted a question how much dynamics can occur in a superposition and a guy called Chalnoth answered: (what oscillations is he talking about?):

Chalnoth wrote:

"A large, complex object like a rock can't really be in a coherent superposition, let alone a galaxy.

Basically, the way we know about objects in coherent superpositions is through oscillation: we can observe the results of an object oscillating through, for instance, interference effects. But complex wavefunctions have oscillation times that tend to be very long, often much longer than the age of the universe.

And when your oscillation time is that long, there just isn't any way for the different components of the same wavefunction to obtain any information about one another. In fact, the different components of the wavefunction, when they are complex enough, interact so weakly with one another that they might as well be in different universes.

So anything as large as a galaxy in a superposition of states will behave exactly as if there was no superposition at all."

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

How true? I wanted to understand how much dynamics can exist in a superposition like whether a universe can be in superposition (in one branch) and evolve planets without having to use the branching in Many Worlds. Or maybe we can treat superposition as Potentia where there is potential to evolve and a realm of possibilities. But I can't understand how possibilities can suddenly give rise to planets... so maybe macroscopic superposition can only happen in Many Worlds and impossible as actually happening in real time in an object in one branch, isn't it. Supposed it could happen, how much dynamics can occur in a macroscopic superposition in one world I wonder. None according to Chalnoth, do you fully agree? And what is the oscillation thing he was describing?

 

[Chopin]

The oscillation thing refers to the fact that any definite-energy solution to the Schrodinger Equation looks like this:

$$\Psi(\textbf{x},t) = e^{iEt/\hbar}F(\textbf{x})$$

That is, it has a spatially stationary solution, but the entire thing rotates through the complex plane at a frequency defined by the energy E. If the system is in a superposition of these states, the various base states rotate at different frequencies, and the changing interference effects that come out of this leads to a spatially-changing waveform (think about beat frequencies in classical wave theory.)

I don't exactly know what all of the superposition stuff in that post is about. You'd probably be best off making a new post for that, since it's not really related to the original topic of this post.