I Representation of Spin 1/2 quantum state

[cianfa72]

No, if we look at just one degree of freedom then we are just looking at the Hilbert space associated with that degree of freedom by itself, i.e., $\mathbb{H}_1$. There is no tensor product.

Sorry, the linear operator itself is an element of the tensor product space $\mathbb H_1 \otimes \mathbb H_1^*$. Indeed any linear operator acting on a vector space can be written in this way (basically it acts by contraction on the vector it acts on).

 

[PeterDonis]

Sorry, the linear operator itself is an element of the tensor product space $\mathbb H_1 \otimes \mathbb H_1^*$.

Why would you think that? An operator maps vectors to vectors. It operates on the vector space in which the vectors it is operating on are elements; in this case, $\mathbb{H}_1$.

Indeed any linear operator acting on a vector space can be written in this way (basically it acts by contraction on the vector it acts on).

Where are you getting this from?

 

[PeterDonis]

it acts by contraction on the vector it acts on

No, that's a covector. A covector is a linear map from vectors to numbers: "contraction" is the operation of applying a covector to a vector to get the corresponding number. The linear operators we are talking about here map vectors to vectors. That's not the same thing.

Also, a covector is an element of the conjugate vector space, which in this case I think is what you mean by the notation $\mathbb{H}_1^*$. It is not an operator in the tensor product space you wrote down.

 

[gentzen]

Why would you think that? An operator maps vectors to vectors. It operates on the vector space in which the vectors it is operating on are elements; in this case, $\mathbb{H}_1$.


Where are you getting this from?

No idea where he got this from. It is correct in the finite dimensional case, but in the general case, you only get the Hilbert-Schmidt operators in this way:

The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on $H$. They also form a Hilbert space, denoted by $B_{HS}(H)$ or $B_2(H)$, which can be shown to be naturally isometrically isomorphic to the tensor product of Hilbert spaces
$$H^* \otimes H,$$
where $H^*$ is the dual space of $H$.

 

[PeterDonis]

It is correct in the finite dimensional case

Why? Please see my post #33.

 

[gentzen]

Why? Please see my post #33.

Because I am a mathematician, and being "naturally isometrically isomorphic" is sufficiently equal for me. On the other hand, the infinite dimensional case is important to me, and there the statement is just plain and simple wrong.

 

[cianfa72]

The linear operators we are talking about here map vectors to vectors. That's not the same thing.

Why not ? A linear map $A$ on a vector space $V$ is defined by its action on a basis $\{v_i\}$. If we take the dual-vectors $\{v_i^*\}$ associated to the basis vector then we can write $A$ as the sum of tensor product of type $u_i \otimes v_i^*$, where $u_i$ are the images of basis vectors under $A$.

 

[PeterDonis]

being "naturally isometrically isomorphic" is sufficiently equal for me

What is naturally isomorphic to what in the finite dimensional case?

 

[PeterDonis]

A linear map $A$ on a vector space $V$ is defined by its action on a basis $\{v_i\}$.

Yes. But there are different possible actions. One possible action would be to map each basis vector to a particular number: that is the action of a covector, which, as I said, is called "contraction"--and "contraction" is the term you used in a previous post. Another possible action would be to map each basis vector to some vector. That is the action of the kind of operator we are talking about here. Those are two different actions. A number is not the same thing as a vector, and mapping a vector to another vector is not the same thing as contracting a vector with a covector to get a number.

 

[PeterDonis]

If we take the dual-vectors $\{v_i^*\}$ associated to the basis vector then we can write $A$ as the sum of tensor product of type $u_i \otimes v_i^*$, where $u_i$ are the images of basis vectors under $A$.

Ok, let's try a specific example: the operator $\sigma_x$ operating on a qubit, i.e., a normalized vector in the 2-dimensional Hilbert space $\mathbb{C}^2$. Its action on the eigenvectors of $\sigma_z$, which form a basis of $\mathbb{C}^2$, is as follows:

$$
\ket{z+} \to \ket{z-}
$$

$$
\ket{z-} \to \ket{z+}
$$

This defines an action on the vector space $\mathbb{C}^2$, i.e., it maps vectors in $\mathbb{C}^2$ to other vectors in $\mathbb{C}^2$. And I just defined it without doing any of the rigmarole you described in the above quote. No dual vectors, no tensor products.

How does what you said in the quote above work for this case?

 

[cianfa72]

A number is not the same thing as a vector, and mapping a vector to another vector is not the same thing as contracting a vector with a covector to get a number.

What do you get if you contract the second "slot" of the tensor product $u \otimes v^*$ with a vector $z$ ? Note that $v^*$ is a covector/dual-vector.

You get a vector, that is you have a linear map.

 

[PeterDonis]

What do you get if you contract the second "slot" of the tensor product $u \otimes v^*$ with a vector $z$ ? Note that $v^*$ is a covector/dual-vector.

You get a vector, that is you have a linear map.

How would this work for the specific case I gave in post #40?

 

[gentzen]

What is naturally isomorphic to what in the finite dimensional case?

The space of Hilbert-Schmidt operators on a Hilbert space $H$ is always naturally isometrically isomorphic to the tensor product of the Hilbert space $H$ with its dual $H^*$. And in the finite dimensional case, all linear operators acting on the Hilbert space $H$ are Hilbert-Schmidt operators. Which is no longer true in the infinite dimensional case. Hence the second sentence of cianfa72's post is plain and simple wrong for that case:

Sorry, the linear operator itself is an element of the tensor product space $\mathbb H_1 \otimes \mathbb H_1^*$. Indeed any linear operator acting on a vector space can be written in this way (basically it acts by contraction on the vector it acts on).

 

[PeterDonis]

in the finite dimensional case, all linear operators acting on the Hilbert space $H$ are Hilbert-Schmidt operators.

Ah, that was the piece I was missing.

I think, though, that you mean all linear operators from $H$ to $H$, correct? As I have been saying, a covector, i.e., an element of $H^*$, is also a linear operator on $H$, but it doesn't map $H$ to $H$, it maps $H$ to its underlying scalar field ($\mathbb{C}$ in the case we are discussing). Or maybe I am using looser terminology, which is typical for physicists as opposed to mathematicians.

 

[cianfa72]

The space of Hilbert-Schmidt operators on a Hilbert space $H$ is always naturally isometrically isomorphic to the tensor product of the Hilbert space $H$ with its dual $H^*$.

Yes, that's was basically my point. As you pointed out this is not longer true in infinite dimensional case.

 

[gentzen]

Ah, that was the piece I was missing.

I think, though, that you mean all linear operators from $H$ to $H$, correct?

Oh yes, indeed. Now I see your point.

As I have been saying, a covector, i.e., an element of $H^*$, is also a linear operator on $H$, but it doesn't map $H$ to $H$, it maps $H$ to its underlying scalar field ($\mathbb{C}$ in the case we are discussing). Or maybe I am using looser terminology, which is typical for physicists as opposed to mathematicians.

No, you are right. I was the one using looser terminology here, and didn't even notice.

 

[PeterDonis]

Yes, that's was basically my point.

But while this isomorphism exists, it is not necessary to use it in order to define the action of an operator from $H$ to $H$. I gave an example of such a definition without using this isomorphism in post #40.

As you pointed out this is not longer true in infinite dimensional case.

The isomorphism between the space of Hilbert-Schmidt operators from $H$ to $H$ and the tensor product of $H$ with its dual still exists in the infinite dimensional case. What is no longer true in the infinite dimensional case is that all linear operators from $H$ to $H$ are Hilbert-Schmidt operators. For example, the position and momentum operators are not Hilbert-Schmidt operators.

 

[cianfa72]

But while this isomorphism exists, it is not necessary to use it in order to define the action of an operator from $H$ to $H$.

Yes definitely, it is not necessary (it is just a way to look at it).

What is no longer true in the infinite dimensional case is that all linear operators from $H$ to $H$ are Hilbert-Schmidt operators. For example, the position and momentum operators are not Hilbert-Schmidt operators.

Ah ok, got it. So in this case there is not a natural isomorphism between a linear map on $\mathbb H$ and an element of $\mathbb H^* \otimes \mathbb H$.

 

[PeterDonis]

Ah ok, got it.

No, you don't. See below. You need to read more carefully.

So in this case there is not a natural isomorphism between a linear map on $\mathbb H$ and an element of $\mathbb H^* \otimes \mathbb H$

Wrong. There is an isomorphism between the space of Hilbert-Schmidt operators from $H$ to $H$, and the space $H \otimes H^*$. That is true regardless of whether $H$ is finite or infinite dimensional. Isomorphisms are between spaces--sets with particular structure that the isomorphism preserves--not between individual elements.

If ##H# is finite dimensional, then all linear operators from $H$ to $H$ are Hilbert-Schmidt operators. So the isomorphism described above includes all linear operators from $H$ to $H$.

If $H$ is infinite dimensional, then there are some linear operators from $H$ to $H$ that are not Hilbert-Schmidt operators. So the isomorphism described above, which still exists in this case, does not include all linear operators from $H$ to $H$. It only includes the ones that are Hilbert-Schmidt operators.

 

[cianfa72]

If $H$ is infinite dimensional, then there are some linear operators from $H$ to $H$ that are not Hilbert-Schmidt operators. So the isomorphism described above, which still exists in this case, does not include all linear operators from $H$ to $H$. It only includes the ones that are Hilbert-Schmidt operators.

Ah ok, the isomorphism includes just some of the linear operators from $H$ to $H$. However it doesn't include all of them.

 

[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.