What is the dimension of the state vector?

In summary: Psi \rangle \mapsto -\mathrm{i} \hbar \vec{\sigma}_{\sigma_j}.$$In other words, the commutator of position and spin is just the Lorentz force.Yes.
  • #1
Often ignored, but turned out to be a problem when trying to compute the commutator of position and spin. Pauli matrices clearly acting on two dimensional vectors while position on infinite dimensional vectors. But a system is described as a single state vector in Dirac notation. A system can of course have both position and spin. Then how many dimensions on Earth does a state vector have?
 
Physics news on Phys.org
  • #2
Its a countably infinite Hilbert space.

Thanks
Bill
 
  • #3
bhobba said:
Its a countably infinite Hilbert space.

Thanks
Bill
Then how can spin operators represented by 2X2 matrices act on the countably infinite dimensional vectors to have eigenvalue equations?
 
  • #4
Xiaomin Chu said:
Then how can spin operators represented by 2X2 matrices act on the countably infinite dimensional vectors to have eigenvalue equations?

Two countably infinite Hilbert spaces is also a countably infinite Hilbert space.

Thanks
Bill
 
  • #5
bhobba said:
Two countably infinite Hilbert spaces is also a countably infinite Hilbert space.

Thanks
Bill
You mean, tensor product?
 
  • #6
Xiaomin Chu said:
You mean, tensor product?

You can look at it that way.

But the usual way is via the observables forming a complete commuting set:
http://en.wikipedia.org/wiki/Complete_set_of_commuting_observables

The spin observable and position observable form a complete commuting set (ignoring the issue of the eigenfunctions of the position operator being continuous)

Thanks
Bill
 
  • #7
bhobba said:
You can look at it that way.

But the usual way is via the observables forming a complete commuting set:
http://en.wikipedia.org/wiki/Complete_set_of_commuting_observables

The spin observable and position observable form a complete commuting set (ignoring the issue of the eigenfunctions of the position operator being continuous)

Thanks
Bill
Thanks.
Is this correct:?
A state vector is a vector in a Hilbert space which is the tensor product of a CSCO's space. Each operator in CSCO operates on its own indices, so position and spin operators do not affect each other.
Then another problem, just the same as entanglement: tensor product will create "non-product state". What does it mean?
 
  • #8
Xiaomin Chu said:
A state vector is a vector in a Hilbert space which is the tensor product of a CSCO's space. Each operator in CSCO operates on its own indices, so position and spin operators do not affect each other.

Yes.

Xiaomin Chu said:
Then another problem, just the same as entanglement: tensor product will create "non-product state". What does it mean?

What does what mean?

I am however not up on things like simply separable states, Segre embeddings etc so may not be able to help.

Thanks
Bill
 
  • #9
As bhobba already said, in nonrelativistic QT (NOT in relativistic!) the spin commutes with the position and momentum variables. A complete set of free single-particle eigenstates are, e.g., the position-spin eigenstates ##|\vec{x},s,\sigma_z \rangle##, where the spin quantum-number ##s \in \{0,1/2,1,\ldots \}## is fixed. Both, position and spin, are represented by linear self-adjoint operators on an abstract separable Hilbert space. There are no matrices here!

You come to a matrix-differential-operator algebra on the Hilbert space ##L^2(\mathbb{R}^3,\mathbb{C}^{2s+1}##, i.e., the Hilbert space of square Lebesgue-integrable ##\mathbb{C}^{2s+1}##-valued functions when choosing the position representation. So let ##|\Psi \rangle## be a normalizable (true) Hilbert-space vector, then the mapping from the abstract Hilbert space ##\mathcal{H}## to this function-Hilbert space is given by
$$|\Psi \rangle \mapsto \Psi_{\sigma}(\vec{x})=\langle \vec{x},\sigma|\Psi \rangle, \quad \vec{x} \in \mathbb{R}^3,\sigma \in \{-s,-s+1,\ldots,s-1,2 \}.$$
In this representation operators are represented by "matrices". The matrix elements for position and spin-##z## components are very simple to calculate, because we've chosen the common (generalized) eigenvectors as a (generalized) basis for our position-spin wave-mechanics representation:
$$\hat{\vec{x}} \mapsto \langle \vec{x}_1,\sigma_1|\hat{\vec{x}} \vec{x}_2,\sigma_2 \rangle=\vec{x}_2 \langle \vec{x}_1,\sigma_1|\vec{x}_2,\sigma_2 \rangle =\vec{x}_2 \delta^{(3)} (\vec{x}_1-\vec{x}_2) \delta_{\sigma_1 \sigma_2},$$
$$\hat{\sigma}_z \mapsto \langle \vec{x}_1,\sigma_1|\hat{\sigma}_z \vec{x}_2,\sigma_2 \rangle=\sigma_2 \langle \vec{x}_1,\sigma_1|\vec{x}_2,\sigma_2 \rangle=\sigma_2 \delta^{(3)} (\vec{x}_1-\vec{x}_2) \delta_{\sigma_1 \sigma_2}.$$
Thus in the position-spin representation one has
$$\hat{\vec{x}},\sigma |\Psi \rangle \mapsto \sum_{\sigma'=-s}^s \int_{\mathbb{R}^3} \langle \vec{x},\sigma|\hat{\vec{x}} \vec{x}',\sigma' \rangle \Psi_{\sigma'}(\vec{x}') = \vec{x} \Psi_{\sigma}(\vec{x}),$$
$$\hat{\sigma}_z,\sigma |\Psi \rangle \mapsto \sum_{\sigma'=-s}^s \int_{\mathbb{R}^3} \langle \vec{x},\sigma|\hat{\sigma_z} \vec{x}',\sigma' \rangle \Psi_{\sigma'}(\vec{x}') = \sigma \Psi_{\sigma}(\vec{x}).$$
For other operators one has to use the commutator algebra (which follows from the corresponding representation of the Galilei group, which is characterized by the mass as a central charge of this group and the spin-quantum number ##s## of the particle, which is a Casimir operator of the group). E.g., from these considerations you come to
$$\hat{\vec{p}} |\Psi \rangle \mapsto -\mathrm{i} \hbar \vec{\nabla} \Psi_{\sigma}(\vec{x}).$$
 
  • #10
Xiaomin Chu said:
Then another problem, just the same as entanglement: tensor product will create "non-product state". What does it mean?
If you wanted to say that the tensor product allows non-product states, i.e. entangled states, you are right. Position and spin degrees of freedom can be entangled.

An example of this is the Stern-Gerlach experiment. Initially, you have a product state and the position and spin degrees of freedom evolve independently because the Hamiltonian acts independently on the position space and the spin space. The magnetic field gradient then introduces a term into the Hamiltonian which acts on both spaces. After the interaction, you have a superposition state where the position and spin degrees of freedom are entangled. Spin up is correlated with moving along a certain upper trajectory, spin down with moving along a certain lower trajectory.
 
  • Like
Likes vanhees71
  • #11
vanhees71 said:
As bhobba already said, in nonrelativistic QT (NOT in relativistic!) the spin commutes with the position and momentum variables. A complete set of free single-particle eigenstates are, e.g., the position-spin eigenstates ##|\vec{x},s,\sigma_z \rangle##, where the spin quantum-number ##s \in \{0,1/2,1,\ldots \}## is fixed. Both, position and spin, are represented by linear self-adjoint operators on an abstract separable Hilbert space. There are no matrices here!

You come to a matrix-differential-operator algebra on the Hilbert space ##L^2(\mathbb{R}^3,\mathbb{C}^{2s+1}##, i.e., the Hilbert space of square Lebesgue-integrable ##\mathbb{C}^{2s+1}##-valued functions when choosing the position representation. So let ##|\Psi \rangle## be a normalizable (true) Hilbert-space vector, then the mapping from the abstract Hilbert space ##\mathcal{H}## to this function-Hilbert space is given by
$$|\Psi \rangle \mapsto \Psi_{\sigma}(\vec{x})=\langle \vec{x},\sigma|\Psi \rangle, \quad \vec{x} \in \mathbb{R}^3,\sigma \in \{-s,-s+1,\ldots,s-1,2 \}.$$
In this representation operators are represented by "matrices". The matrix elements for position and spin-##z## components are very simple to calculate, because we've chosen the common (generalized) eigenvectors as a (generalized) basis for our position-spin wave-mechanics representation:
$$\hat{\vec{x}} \mapsto \langle \vec{x}_1,\sigma_1|\hat{\vec{x}} \vec{x}_2,\sigma_2 \rangle=\vec{x}_2 \langle \vec{x}_1,\sigma_1|\vec{x}_2,\sigma_2 \rangle =\vec{x}_2 \delta^{(3)} (\vec{x}_1-\vec{x}_2) \delta_{\sigma_1 \sigma_2},$$
$$\hat{\sigma}_z \mapsto \langle \vec{x}_1,\sigma_1|\hat{\sigma}_z \vec{x}_2,\sigma_2 \rangle=\sigma_2 \langle \vec{x}_1,\sigma_1|\vec{x}_2,\sigma_2 \rangle=\sigma_2 \delta^{(3)} (\vec{x}_1-\vec{x}_2) \delta_{\sigma_1 \sigma_2}.$$
Thus in the position-spin representation one has
$$\hat{\vec{x}},\sigma |\Psi \rangle \mapsto \sum_{\sigma'=-s}^s \int_{\mathbb{R}^3} \langle \vec{x},\sigma|\hat{\vec{x}} \vec{x}',\sigma' \rangle \Psi_{\sigma'}(\vec{x}') = \vec{x} \Psi_{\sigma}(\vec{x}),$$
$$\hat{\sigma}_z,\sigma |\Psi \rangle \mapsto \sum_{\sigma'=-s}^s \int_{\mathbb{R}^3} \langle \vec{x},\sigma|\hat{\sigma_z} \vec{x}',\sigma' \rangle \Psi_{\sigma'}(\vec{x}') = \sigma \Psi_{\sigma}(\vec{x}).$$
For other operators one has to use the commutator algebra (which follows from the corresponding representation of the Galilei group, which is characterized by the mass as a central charge of this group and the spin-quantum number ##s## of the particle, which is a Casimir operator of the group). E.g., from these considerations you come to
$$\hat{\vec{p}} |\Psi \rangle \mapsto -\mathrm{i} \hbar \vec{\nabla} \Psi_{\sigma}(\vec{x}).$$
Thanks
 
  • #12
kith said:
If you wanted to say that the tensor product allows non-product states, i.e. entangled states, you are right. Position and spin degrees of freedom can be entangled.

An example of this is the Stern-Gerlach experiment. Initially, you have a product state and the position and spin degrees of freedom evolve independently because the Hamiltonian acts independently on the position space and the spin space. The magnetic field gradient then introduces a term into the Hamiltonian which acts on both spaces. After the interaction, you have a superposition state where the position and spin degrees of freedom are entangled. Spin up is correlated with moving along a certain upper trajectory, spin down with moving along a certain lower trajectory.
Thanks
 

What is the dimension of the state vector?

The dimension of the state vector refers to the number of independent variables needed to fully describe the state of a physical system. It is often denoted as "n" and is dependent on the complexity of the system being studied.

How is the dimension of the state vector determined?

The dimension of the state vector is determined by the number of coordinates or variables needed to describe the state of a system. For example, a simple pendulum can be described by two variables - the angle and angular velocity, thus its state vector has a dimension of 2.

Why is the dimension of the state vector important?

The dimension of the state vector is important because it helps scientists and engineers to understand and model complex systems. By knowing the dimension of the state vector, they can determine the number of equations needed to fully describe the system and make accurate predictions.

Can the dimension of the state vector change?

Yes, the dimension of the state vector can change depending on the physical system being studied. For example, a simple pendulum swinging in two dimensions has a state vector of 2, but if it is allowed to swing in three dimensions, its state vector would have a dimension of 3.

How does the dimension of the state vector relate to other concepts in physics?

The dimension of the state vector is closely related to other concepts in physics such as degrees of freedom and phase space. The number of degrees of freedom in a system is equal to the dimension of the state vector, and phase space is a mathematical space that represents all possible states of a system, with each point representing a different state vector.

Suggested for: What is the dimension of the state vector?

Replies
3
Views
175
Replies
1
Views
757
Replies
7
Views
384
Replies
6
Views
478
Replies
18
Views
1K
Replies
6
Views
836
Replies
8
Views
616
Replies
0
Views
392
Back
Top