Position representation of the state of the system

[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?