I Representation of Spin 1/2 quantum state

[cianfa72]

If you are looking at both degrees of freedom, then you have to look at operators on the tensor product Hilbert space. For example, you could look at the operator $I \otimes \sigma_z$, i.e., the identity on the configuration space degree of freedom, tensor product with the $\sigma_z$ spin operator. This operator's eigenvectors are vectors in the tensor product Hilbert space, in this case any configuration space vector tensor product with an eigenvector of $\sigma_z$.

So coming back to this topic, if the linear operators acting on each of the Hilbert spaces associated to the system's degrees of freedom (call such Hilbert spaces $U$ and $V$) are both hermitian/self-adjoint then, by spectral theorem, the collection of tensor products $\{u_i \otimes v_j\}$ (where $u_i$ and $v_j$ are eigenvectors of each hermitian operator respectively) will be a basis of eigenvectors for the tensor product space $U \otimes V$.

I believe the above is not true in general for "generic" linear operators since we cannot apply the spectral theorem to them.

 

[cianfa72]

Another question related to the topic. What is a simple vs multiple eigenvalue?

Does the difference boils down to the eigenvalue's geometric multeplicity (i.e. the dimension of the associated eigenspace) ?

 

[gentzen]

Another question related to the topic. What is a simple vs multiple eigenvalue?

Does the difference boils down to the eigenvalue's geometric multeplicity (i.e. the dimension of the associated eigenspace) ?

yes

 

[cianfa72]

Ok, so in case of entangled position/momentum plus spin the pure state of the entangled system cannot be given as the tensor product of a position/momentum pure state times a spin pure state (the entangled state is given as linear combination of tensor products of basis pure states from each Hilbert space, though).

What about the hermitian/self-adjoint operators tensor product ? Their eigenvalues should be related to the product of eigenvalues of each hermitian operator, right ?

 

[gentzen]

What about the hermitian/self-adjoint operators tensor product ? Their eigenvalues should be related to the product of eigenvalues of each operator, right ?

If $A$ has eigenvalues $\lambda_i$ with corresponding eigenvectors $v_i$ and $B$ has eigenvalues $\mu_j$ with corresponding eigenvectors $w_j$, then $A\otimes B$ has eigenvalues $\lambda_i\mu_j$ with corresponding eigenvectors $v_i\otimes w_j$:
$$(A\otimes B)\ v_i\otimes w_j = (Av_i)\otimes(Bw_j)=(\lambda_iv_i)\otimes(\mu_jw_j)=\lambda_i\mu_j\ v_i\otimes w_j$$

 

[cianfa72]

If $A$ has eigenvalues $\lambda_i$ with corresponding eigenvectors $v_i$ and $B$ has eigenvalues $\mu_j$ with corresponding eigenvectors $w_j$, then $A\otimes B$ has eigenvalues $\lambda_i\mu_j$ with corresponding eigenvectors $v_i\otimes w_j$

Of course, however for sure the set $\{v_i \otimes w_j\}$ will be a basis of eigenvectors for $A\otimes B$ when both $A$ and $B$ are hermitian/self-adjoint.

 

[PeterDonis]

if the linear operators acting on each of the Hilbert spaces associated to the system's degrees of freedom (call such Hilbert spaces $U$ and $V$) are both hermitian/self-adjoint then, by spectral theorem, the collection of tensor products $\{u_i \otimes v_j\}$ (where $u_i$ and $v_j$ are eigenvectors of each hermitian operator respectively) will be a basis of eigenvectors for the tensor product space $U \otimes V$.

Yes. This is because the operators acting on each Hilbert space are tensored with the identity on the other Hilbert space (so the spin-z operator $\sigma_z$ becomes $I \otimes \sigma_z$ when acting on the tensor product space), so they automatically commute with each other and hence can have a set of simultaneous eigenvectors.

 

[cianfa72]

so they automatically commute with each other and hence can have a set of simultaneous eigenvectors.

What does mean that those tensor product operators (e.g $I \otimes \sigma_z$ and $\vec{x} \otimes I$) commute with each other?

 

[PeterDonis]

What does mean that those tensor product operators (e.g $I \otimes \sigma_z$ and $\vec{x} \otimes I$) commute with each other?

Isn't it obvious? You know what "commute" means, right?

 

[cianfa72]

Isn't it obvious? You know what "commute" means, right?

It should mean that $$(I \otimes \sigma_z)(\vec{x} \otimes I) \ket{\psi} = (\vec{x} \otimes I) (I \otimes \sigma_z) \ket{\psi}, \forall \ket{\psi}$$ We can check that it is true using the definition of tensor product operators acting on a generic tensor product of type $\ket{\psi} = \ket{\alpha} \otimes \ket {\beta}$. Since both $I \otimes \sigma_z$ and $\vec{x} \otimes I$ are hermitian and commute each other, they have at least a common eigenbasis -- see also Commuting Operators Have the Same Eigenvectors.

However any eigenbasis of the first operator is an eigenbasis for the second (and the other way around) if and only if the commutating hermitian operators have both nondegerate eigenvalues -- see Common eigenfunctions of commuting operators.

 

[PeterDonis]

It should mean that $$(I \otimes \sigma_z)(\vec{x} \otimes I) \ket{\psi} = (\vec{x} \otimes I) (I \otimes \sigma_z) \ket{\psi}, \forall \ket{\psi}$$

Yes.

We can check that it is true using the definition of tensor product operators acting on a generic tensor product of type $\ket{\psi} = \ket{\alpha} \otimes \ket {\beta}$.

Yes.

Since both $I \otimes \sigma_z$ and $\vec{x} \otimes I$ are hermitian and commute each other, they have at least a common eigenbasis -- see also Commuting Operators Have the Same Eigenvectors.

Yes.

However any eigenbasis of the first operator is an eigenbasis for the second (and the other way around) if and only if the commutating hermitian operators have both nondegerate eigenvalues -- see Common eigenfunctions of commuting operators.

Yes.

 

[cianfa72]

Ok, so coming back to the case of entangled position/momentum plus spin state, such a state can be written as linear combination of tensor products basis $\{v_i \otimes w_j\}$ that will be an eigenbasis of the hermitian/self-adjoint tensor product operator $\vec{x} \otimes \sigma_z$. Note that $\{v_i\}$ and $\{w_j\}$ are eigenbasis of operator $\vec{x}$ and $\sigma_z$ respectively.