What is the dimension of the state vector?

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
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Its a countably infinite Hilbert space.

Thanks
Bill
 
  • #3
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
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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
Two countably infinite Hilbert spaces is also a countably infinite Hilbert space.

Thanks
Bill
You mean, tensor product?
 
  • #6
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  • #7
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
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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.

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
vanhees71
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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
kith
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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.
 
  • #11
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
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
 

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