Position representation of the state of the system

[fog37]

Commuting operators have the same set of eigenstates so they should surely span the same vector space.

Position eigenstates and momentum eigenstates form two bases of the same Hilbert space related by the unitary Fourier transform.

https://en.wikipedia.org/wiki/Position_and_momentum_space

Thanks Truecrimson.

I read the following :

The r and p operators are unitarily equivalent, with the unitary operator being given explicitly by the Fourier transform. Thus they have the same spectrum. In physical language, p acting on momentum space wave functions is the same as r acting on position space wave functions (under the image of the Fourier transform).

I see. I knew that the position operator and the momentum operator do not commute and the momentum space wave functions is the same as r acting on position space wave functions ( one the FT of the other). So they have the same Hilbert space. What about the energy operator and the angular momentum operators? Do they possibly have bases that form different Hilbert spaces? So there are indeed different Hilbert spaces for different operators...

 

[Orodruin]

Do they possibly have bases that form different Hilbert spaces? So there are indeed different Hilbert spaces for different operators...

No. A quantum state is an element of a Hilbert space. When you talk about quantum operators, those are operators on that Hilbert space. It therefore makes no sense to talk about different spaces for different operators unless you are describing two different systems.

 

[fog37]

Thank you.

I think my confusion is cleared up: the vector state ##|\Psi>## (wavefunction) of a physical system belongs to a Hilbert space. The various operators have their own complete bases that can be used to deconstruct the state ##|\Psi>## as a weighted superposition. The various bases (one for each operator) are all bases for the same Hilbert space.

I have read that the Hilbert space ##H_x## of one-dimensional wavefunctions like ##\Psi(x)## is different from the Hilbert space ##H_{(x,y,z)} ##of three-dimensional wavefunctions ##\Psi(x,y,z)##. The Hilbert space ##H_{(x,y,z)}## can be constructed from the Hilbert spaces ##H_x##, ##H_y##, and ##H_z## using the tensor product.

I guess spin observable has a different Hilbert space than the other external observables which all act on the same Hilbert space ##H_{(x,y,z)}## . The state space of a single particle with spin is given by the tensor product of two vector spaces: $$H_{(x,y,z)} \otimes H_s$$ since spin has its own vector space ##H_s##.

 

[bhobba]

What a nice thread - a joy to read .

Thanks
Bill

 

[Orodruin]

I have read that the Hilbert space ##H_x## of one-dimensional wavefunctions like ##\Psi(x)## is different from the Hilbert space ##H_{(x,y,z)} ##of three-dimensional wavefunctions ##\Psi(x,y,z)##. The Hilbert space ##H_{(x,y,z)}## can be constructed from the Hilbert spaces ##H_x##, ##H_y##, and ##H_z## using the tensor product.

Correct. Note that you can perfectly well use one basis (say position basis) to describe one of the factors in the tensor product and another (say momentum basis) for another. Also, some operators have degenerate eigenvalues. In those cases you should make sure you are working in an orthogonal basis before you start projecting.

I guess spin observable has a different Hilbert space than the other external observables which all act on the same Hilbert space ##H_{(x,y,z)}## . The state space of a single particle with spin is given by the tensor product of two vector spaces: $$H_{(x,y,z)} \otimes H_s$$ since spin has its own vector space ##H_s##.

Correct. The Hilbert space describing spin is a finite dimensional one.

On a TeXnical note, kets should be typeset using \lvert and \rangle. Compare ##\lvert \psi\rangle = \ldots## to ##|\psi>= \ldots##

 

[mike1000]

Hello Forum,
My understanding is that the state of the system is ##|\Psi>##. We can take the inner product between the state ##|\Psi>## and the eigenstates of the position operator ##\hat{x}##:
$$<x|\Psi>=\Psi(x)$$
The function ##\Psi(x)## is the wave function we are initially introduced to in beginner quantum mechanics. We call this the position representation of the state ##|\Psi>##. In position space, the eigenfunctions of the position operator ##\hat{x}## are delta functions ##\delta(x-x_0)## located at ##x_0##.
We could also take the inner product between the state ##|\Psi>## and the eigenstates of the momentum operator ##\hat{p_x}## to get a new function:
$$<p|\Psi>=\Psi(p)$$

What state is represented by the symbol Ψ?

It seems to me this question is putting the cart before the horse. How can we know the eigenstates of the operator before we know the wave function?

 

[PeterDonis]

How can we know the eigenstates of the operator before we know the wave function?

Um, because the eigenstates of the operator depend on the operator, not the wave function?

 

[mike1000]

Um, because the eigenstates of the operator depend on the operator, not the wave function?

What state is represented by ##|Ψ \rangle## and where did that come from in the equation $$<x|\Psi>=\Psi(x)$$

 

[hilbert2]

It seems to me this question is putting the cart before the horse. How can we know the eigenstates of the operator before we know the wave function?

That's a bit like asking "How can we know Newton's gravitation law before knowing the current velocity and position of Moon?". The set of eigenstates is a property of the system, and the wavefunction is the state of the system at some instant.

 

[mike1000]

That's a bit like asking "How can we know Newton's gravitation law before knowing the current velocity and position of Moon?". The set of eigenstates is a property of the system, and the wavefunction is the state of the system at some instant.

So where did |Ψ> come from in the OP's equation ##<x|\Psi>=\Psi(x)##

 

[hilbert2]

So where did |Ψ> come from in the OP's equation ##\langle x|Ψ \rangle##

It can be any possible state, just like we can say that the position vector ##\mathbf{x}## and momentum vector ##\mathbf{p}## tell about the mechanical state of the moon without having to specify what their actual values are.

 

[mike1000]

It can be any possible state, just like we can say that the position vector ##\mathbf{x}## and momentum vector ##\mathbf{p}## tell about the mechanical state of the moon without having to specify what their actual values are.

Are you saying that |Ψ> is a symbol which represents one of the eigenstates of the position operator ( or some linear combination of the eigenstates of the position operator?

 

[hilbert2]

It can be any linear combination of the eigenstates of any operator.

 

[mike1000]

It can be any linear combination of the eigenstates of any operator.

Ok, its symbolic.

Last question, how do you get the eigenvalues/states of an operator without letting the operator operate on something? For instance, the momentum operator takes the gradient of something. How do you get the eigenvalues of that operator without letting it operate on something?

 

[Nugatory]

Are you saying that |Ψ> is a symbol which represents one of the eigenstates of the position operator ( or some linear combination of the eigenstates of the position operator?

It is the state in which the system was originally prepared. In any particular problem we know what it is because the preparation procedure is part of the initial conditions in the problem statement.

No matter what the problem, we can say that ##\langle{x}|\Psi\rangle## is the amplitude for getting ##x## out of a position measurement. But we can't be more specific than that without knowing what specific physical situation we're working with, which is tantamount to knowing ##|\Psi\rangle##. (It may take a fair amount of work to get from the description of the preparation to ##|\Psi\rangle## written in the basis that we want, but in principle we have enough information to do it if the problem is properly specified)

 

[Nugatory]

Last question, how do you get the eigenvalues/states of an operator without letting the operator operate on something? For instance, the momentum operator takes the gradient of something. How do you get the eigenvalues of that operator without letting it operate on something?

That's just a differential equation: what constants ##\lambda## and functions ##\psi## are solutions of ##-i\hbar\nabla\psi=\lambda\psi##? Once we know that, we can choose to write any arbitrary state ##\Psi## (in the same Hilbert space) as a sum of these functions.

 

[fog37]

So where did |Ψ> come from in the OP's equation ##<x|\Psi>=\Psi(x)##

This is how I think about it: the physical system is in a certain quantum mechanical state which we indicate as ##|\Psi>##. This state, whatever it may be, is a vector that lives in a Hilbert vector space. By taking the inner product (bracket) of the state vector ##|\Psi>## with the eigenvectors (eigenstates) ##|x>## of the position operator ##\hat {x}## we obtain the function ##\Psi(x)## which associates a complex value to each ##x##. Each ##x## represents the eigenvalues of each position eigenvector ##|x>##...

 

[fog37]

It is the state in which the system was originally prepared. In any particular problem we know what it is because the preparation procedure is part of the initial conditions in the problem statement.

Hi Nugatory, are we always able to prepare the system in the state we want to? What if we wanted to prepare it in a specific eigenstate of a certain observable?

 

[Nugatory]

Hi Nugatory, are we always able to prepare the system in the state we want to?

We can in a thought experiment but not in general.

What if we wanted to prepare it in a specific eigenstate of a certain observable?

The easiest way is to measure that observable and use only the systems that came out with the right value. For example, if we wanted a beam of particles prepared in the state "spin up in the vertical direction" we'd pass the beam through a vertically oriented Stern-Gerlach device; that will produce two beams, one of particles prepared in the spin-up state and the other of particles in the spin-down state.

 

[stevendaryl]

There is a confusion throughout physics and mathematics about the notation: [itex]f(x)[/itex]. Sometimes it means a function, and the [itex]x[/itex] is there just to indicate the functional dependency. Sometimes it means the value of the function at some unspecified (or only partly specified) point [itex]x[/itex].

This confusion comes into play when people write: [itex]\langle x |\Psi \rangle = \Psi(x)[/itex]. Some people say that the item on the right is the wave function, a function. But that's not in keeping with the idea that [itex]\langle x |\Psi \rangle[/itex] is the product of a bra, [itex]\langle x|[/itex] and a ket, [itex]|\Psi\rangle[/itex]. The product of a bra and a ket always produces a number.

It's possible to make this stuff more clear using the [itex]\lambda[/itex] notation, but it's a lot of effort, and it's easy enough to just explain what you really mean in words.

Using the lambda notation (from Church's lambda calculus), you would say that [itex]f(x)[/itex] means the value of [itex]f[/itex] at point [itex]x[/itex]. If you mean the function, you make that clear by "lambda-abstraction": [itex]\lambda x . f(x)[/itex] is the function that takes an arbitrary [itex]x[/itex] and returns [itex]f(x)[/itex].

With the lambda notation, you would not say that the wave function is [itex]\langle x|\Psi\rangle[/itex], you would say that the wave function is [itex]\lambda x. \langle x|\Psi\rangle[/itex].

 

[stevendaryl]

There is a confusion throughout physics and mathematics about the notation: [itex]f(x)[/itex].

I should say "ambiguity" rather than "confusion", because the mathematicians and physicists are not confused by it, although students often are.

 

[vanhees71]

Mathematically the right notation is to give a function with its domain and co-domain, i.e., for a wave function for a particle moving in one dimension you have a function ##\psi:\mathbb{R} \rightarrow \mathbb{C}##, which should be square integrable, i.e., normalizable to 1 in the sense
$$\int_{-\infty}^{\infty} \mathrm{d} x |\psi(x)|^2=1.$$
The value of ##\psi## at the argument ##x## is called ##\psi(x)##. You can also write ##x \mapsto \psi(x)## to emphasize that ##\psi## is a function.

The relation between the basis-free Dirac notation to the wave function is that you use the generalized position eigenbasis ##|x \rangle##, which are no states but belong to a larger space than the Hilbert space of states (it's pretty technical; if you are interested check the key word "rigged Hilbert space" or Ballentine's textbook for an introduction):
$$\psi(x)=\langle x|\psi \rangle,$$
where ##\psi(x)## is the value of ##\psi## at the argument ##x##.

On the other hand, physicists are sloppy and just say "##\psi(x)## is the wave function". Of course, what they mean with that is in fact what the more rigorous definitions of the mathematicians say.

 

[fog37]

Hello, since we are on the topic of functions and wavefunctions, in the case of the hydrogen atom (single electron), the total wavefunction includes both spatial and a spin parts: $$\Psi(x,y,z,s)= \psi_{n,\ell,m_{\ell}}(x,y,z) \chi(s)$$
The spatial part is actually a function ##\psi_{n,\ell,m_{\ell}}(x,y,z)## but the spin part ##\chi(s)## is not really a function since it does not have a position representation. I would call the "function" ##\chi(s)## a symbol that we place beside the spatial wavefunction... Correct?

 

[Orodruin]

It is a complex vector with two entries (for the case of spin 1/2).

 

[fog37]

Ok, thanks.
But the spatial wavefunction ##\psi_{n,\ell,m_{\ell}}## is not really multiplied by the spin complex-valued two dimensional vector, but simply paired with it :
$$\psi_{n,\ell,m_{\ell}} \begin{bmatrix} s_1 \\ s_2&\end{bmatrix}$$
where ##\begin{bmatrix} s_1 \\ s_2&\end{bmatrix}## is equal to ##\chi(s)##

 

[Orodruin]

Ok, thanks.
But the spatial wavefunction ##\psi_{n,\ell,m_{\ell}}## is not really multiplied by the spin complex-valued two dimensional vector, but simply paired with it :

Why do you think there is a difference?

 

[vanhees71]

In general the state is not simply a product of spatial and spin degrees of freedom. The general pure state in position representation is a spinor-valued wave function. An important example is the state of a particle with spin and an associated magnetic moment that run through a Stern-Gerlach apparatus (i.e., an inhomogeneous magnetic field), where you get entanglement between position and spin-##z## component (with the ##z## direction determined by the direction of the magnetic field).

 

[fog37]

Why do you think there is a difference?

Ok. So I can/could carry out the multiplication ##\psi_{n,\ell,m_{\ell}} \begin{bmatrix} s_1 \\ s_2&\end{bmatrix}## to obtain $$ \begin{bmatrix} \psi_{n,\ell,m_{\ell}} s_1 \\ \psi_{n,\ell,m_{\ell}} s_2&\end{bmatrix}$$

 

[fog37]

In general the state is not simply a product of spatial and spin degrees of freedom. The general pure state in position representation is a spinor-valued wave function. An important example is the state of a particle with spin and an associated magnetic moment that run through a Stern-Gerlach apparatus (i.e., an inhomogeneous magnetic field), where you get entanglement between position and spin-##z## component (with the ##z## direction determined by the direction of the magnetic field).

Hivanhees71, I just learned about the SG experiment but I did not know that it was an example of position and z-component of spin entanglement. I understand that the total wavefunction, even for a single particle like the 47th electron in silver is expressed as a product of the spatial part and the spin part. Where is the entaglement in that case? From what I know, entanglement must involve two (or more particles) and the total wavefunction cannot be just a product...I am clearly confused about this.
In the case of two identical electrons, the wavefunction is antisymmetric and is not just a product but a sum of two products. That does not imply entanglement though...

 

[Orodruin]

Ok. So I can/could carry out the multiplication ##\psi_{n,\ell,m_{\ell}} \begin{bmatrix} s_1 \\ s_2&\end{bmatrix}## to obtain $$ \begin{bmatrix} \psi_{n,\ell,m_{\ell}} s_1 \\ \psi_{n,\ell,m_{\ell}} s_2&\end{bmatrix}$$

Yes, but see vanhees' comment about entanglement between the spin degree of freedom and position.

 

[vanhees71]

Hivanhees71, I just learned about the SG experiment but I did not know that it was an example of position and z-component of spin entanglement. I understand that the total wavefunction, even for a single particle like the 47th electron in silver is expressed as a product of the spatial part and the spin part. Where is the entaglement in that case? From what I know, entanglement must involve two (or more particles) and the total wavefunction cannot be just a product...I am clearly confused about this.
In the case of two identical electrons, the wavefunction is antisymmetric and is not just a product but a sum of two products. That does not imply entanglement though...

It's best you think in (Gaussian) wave packets, i.e., you shoot a particle (described as a Gaussian wave packet) into the magnetic field of the SG apparatus. Now there's a force acting in ##z## direction if ##\sigma_z=+1/2## and in ##-z## direction if ##\sigma_z=-1/2##. Thus you get a state (after the SG apparatus), which is roughly of the form
$$\psi(\vec{x})=\psi_+(\vec{x}) \chi_{+} + \psi_{-}(\vec{x}) \chi_{-},$$
where ##\psi_{\pm}## are Gaussian wave packets with centers deflected up or down, respectively. If you choose the whole setup right, you have practically two separate beams (separate by location), where these partial beams have clearly defined values ##\sigma_z=\pm 1/2##. That's the most simple example for a preparation procedure for (nearly) pure ##\sigma_z=\pm 1/2## states.

 

[ftr]

It's best you think in (Gaussian) wave packets, i.e., you shoot a particle (described as a Gaussian wave packet) into the magnetic field of the SG apparatus. Now there's a force acting in ##z## direction if ##\sigma_z=+1/2## and in ##-z## direction if ##\sigma_z=-1/2##. Thus you get a state (after the SG apparatus), which is roughly of the form
$$\psi(\vec{x})=\psi_+(\vec{x}) \chi_{+} + \psi_{-}(\vec{x}) \chi_{-},$$
where ##\psi_{\pm}## are Gaussian wave packets with centers deflected up or down, respectively. If you choose the whole setup right, you have practically two separate beams (separate by location), where these partial beams have clearly defined values ##\sigma_z=\pm 1/2##. That's the most simple example for a preparation procedure for (nearly) pure ##\sigma_z=\pm 1/2## states.

This sounds really interesting. Can you show the calculation of how much the packet is shifted and its energy in both cases i.e. up or down.
If it is too complicated can you link to a reference. Thanks, highly appreciated.

 

[vanhees71]

A full quantum treatment of the SG experiment (amazingly that's not treated in usual textbooks, although it's not very complicated) can be found in

https://arxiv.org/abs/quant-ph/0409206

 

[fog37]

It's best you think in (Gaussian) wave packets, i.e., you shoot a particle (described as a Gaussian wave packet) into the magnetic field of the SG apparatus. Now there's a force acting in ##z## direction if ##\sigma_z=+1/2## and in ##-z## direction if ##\sigma_z=-1/2##. Thus you get a state (after the SG apparatus), which is roughly of the form
$$\psi(\vec{x})=\psi_+(\vec{x}) \chi_{+} + \psi_{-}(\vec{x}) \chi_{-},$$
where ##\psi_{\pm}## are Gaussian wave packets with centers deflected up or down, respectively. If you choose the whole setup right, you have practically two separate beams (separate by location), where these partial beams have clearly defined values ##\sigma_z=\pm 1/2##. That's the most simple example for a preparation procedure for (nearly) pure ##\sigma_z=\pm 1/2## states.

Hi vanhees71, thank you for the infos.

I guess I am still missing a fundamental concept: in the case of a single particle, the total wavefunction is (always?) the product of the spatial part and the spin part. When is it not a product?

For instance, let's remain in 1D and assume the particles are independent from each other. If the system was composed of only 2 particles, the total system wavefunction would be ##\Psi(x_1,x_2,s_1, s_2)##. Because of the required antisimmetry, this wavefunction is the sum of two products.
If there are 3 or more particles, we can use the Slater determinant and express the total wavefunction of the multi-particle system as a summation of various products.
But doesn't a summation of products mean entanglement? I naively believe that if the total wavefunction cannot be expressed as a separable function then entanglement must be involved...

 

[vanhees71]

For a non-relativistic spin-1/2 particle you can think of the the wave function as being a two-component spinor,
$$\psi(\vec{x})=\begin{pmatrix} \psi_{1/2}(\vec{x}) \\ \psi_{-1/2}(\vec{x}) \end{pmatrix} \in \mathbb{C}^2.$$
It's not necessarily a funktion which is a product of a complex function of position with a two-component spinor.