I Position representation of the state of the system

[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: f(x). Sometimes it means a function, and the x is there just to indicate the functional dependency. Sometimes it means the value of the function at some unspecified (or only partly specified) point x.

This confusion comes into play when people write: \langle x |\Psi \rangle = \Psi(x). 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 \langle x |\Psi \rangle is the product of a bra, \langle x| and a ket, |\Psi\rangle. The product of a bra and a ket always produces a number.

It's possible to make this stuff more clear using the \lambda 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 f(x) means the value of f at point x. If you mean the function, you make that clear by "lambda-abstraction": \lambda x . f(x) is the function that takes an arbitrary x and returns f(x).

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

 

[stevendaryl]

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

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.

 

[fog37]

Ok, thanks that is progress.

What about two independent fermions having a total antisymmetric wavefunction $$\Psi(x_1, x_2)=\psi_{a}(x_1) \psi_{b}(x_2)-\psi_a(x_2) \psi_b(x_1)$$?
This is the sum of two products not a single product but it does not mean entanglement, correct? Why not?