Representation of Spin 1/2 quantum state

[cianfa72]

TL;DR Summary

About the spin representation of particle spin 1/2 quantum state in two-dimensional Hilbert space

Hi,

I'm aware of the wave function $\Psi$ of a quantum system represents basically the "continuous components" of a quantum state (a point/vector in the infinite-dimension Hilbert space) in a basis. If we take the $\delta(x - \bar x)$ eigenfunctions as basis on Hilbert space then the wave function $\Psi(x)$ is a function of the position $x$. On the other hand if we take the momentum eigenfunctions as basis in the same Hilbert space then the wave function $\Psi(p)$ is actually a function of momentum $p$.

What about the quantum spin 1/2 state ? I know it "lives" in a two-dimensional Hilbert space so a basis comprises just two vectors/kets ($| \uparrow \rangle$ and $| \downarrow \rangle$).

I'm confused how this two-dimensional Hilbert space is related to the former infinite-dimensional Hilbert space. Are the two actually different spaces ? Thanks.

 

[PeroK]

I'm confused how this two-dimensional Hilbert space is related to the former infinite-dimensional Hilbert space. Are the two actually different spaces ? Thanks.

Yes, but if you want to study the spatial dynamics and the spin of an electron, you need the tensor product of these two Hilbert spaces. In which case, a typical wavefunction is:$$\psi(x)\otimes \chi$$

 

[PeterDonis]

I'm aware of the wave function $\Psi$ of a quantum system represents basically the "continuous components" of a quantum state (a point/vector in the infinite-dimension Hilbert space) in a basis.

Then you are aware of something that is not the case as you state it.

What you are describing is the part of the wave function that represents the position/momentum of the system. But the system can have other degrees of freedom besides that, such as spin. In the case of a spin-1/2 particle, the particle's spin degree of freedom is represented in a two-dimensional Hilbert space, and the full Hilbert space that describes all of the particle's degrees of freedom is, as @PeroK says, the tensor product of the position/momentum Hilbert space with the spin Hilbert space.

 

[cianfa72]

Yes, but if you want to study the spatial dynamics and the spin of an electron, you need the tensor product of these two Hilbert spaces. In which case, a typical wavefunction is:$$\psi(x)\otimes \chi$$

Ah ok, but $\psi(x)$ are just the components of ket $| \Psi \rangle$ like $\chi$ the components of $| \chi \rangle$.

Hence $\psi(x) \otimes \chi$ is actually the tensor product of a vector in $\mathbb C^{\infty}$ Hilbert space times a vector in $\mathbb C^2$ Hilbert space.

Btw $\psi(x) \otimes \chi$ is a particular form of an element of the above tensor product space (i.e. it is not the general form of an element of the above tensor product).

 

[PeterDonis]

Ah ok, but $\psi(x)$ are just the components of ket $| \Psi \rangle$ like $\chi$ the components of $| \chi \rangle$.

Not if the particle has spin. If the particle has spin, $\psi(x)$ does not completely represent $\Psi$, so $\psi(x)$ are not "the components" of $\Psi$; $\psi(x)$ only represents the portion of $\Psi$ that is in the position/momentum Hilbert space. And if the position/momentum of the system is entangled with its spin (as it is, for example, after a quantum particle has passed through a Stern-Gerlach magnet), there is no way to separate out just the position/momentum part of the wave function, so you can't even write down a single $\psi(x)$ for the particle.

Hence $\psi(x) \otimes \chi$ is actually the tensor product of a vector in $\mathbb C^{\infty}$ Hilbert space times a vector in $\mathbb C^2$ Hilbert space.

If the particle's state is separable between those parts, then yes, the complete $\Psi$ can be represented as $\psi(x) \otimes \chi$. But, as above, if the particle's state is entangled between position/momentum and spin, then no, $\Psi$ will not be expressible as such a tensor product state; it will only be expressible as a normalized sum of multiple such tensor product states.

Note that you apparently are trying to use the notation $\Psi$ in two inconsistent ways: as the complete state of the particle, and as just the position/momentum part of the state. In what I posted above, I am using $\Psi$ in the first way, to represent the complete state.

 

[PeroK]

Ah ok, but $\psi(x)$ are just the components of ket $| \Psi \rangle$ like $\chi$ the components of $| \chi \rangle$.

Hence $\psi(x) \otimes \chi$ is actually the tensor product of a vector in $\mathbb C^{\infty}$ Hilbert space times a vector in $\mathbb C^2$ Hilbert space.

Btw $\psi(x) \otimes \chi$ is a particular form of an element of the above tensor product space (i.e. it is not the general form a an element of the above tensor product).

I just used the simplest form that a full wavefunction would take. Using ket notation doesn't change anything.

 

[cianfa72]

Note that you apparently are trying to use the notation $\Psi$ in two inconsistent ways: as the complete state of the particle, and as just the position/momentum part of the state. In what I posted above, I am using $\Psi$ in the first way, to represent the complete state.

Your $\Psi$ is actually the ket $|\Psi \rangle$ and it is not its "components" in a basis of the tensor product of the position/momentum Hilbert space times the spin two-dimensional Hilbert space ?

 

[cianfa72]

Not if the particle has spin. If the particle has spin, $\psi(x)$ does not completely represent $\Psi$, so $\psi(x)$ are not "the components" of $\Psi$; $\psi(x)$ only represents the portion of $\Psi$ that is in the position/momentum Hilbert space.

So for a given full particle state $| \Psi \rangle$, $\psi(x)$ makes sense only in case the full particle state is separable (i.e. it is the "components" of the state's part that "lives" in the position/momentum Hilbert space).

 

[PeterDonis]

Your $\Psi$ is actually the ket $|\Psi \rangle$ and it is not its "components" in a basis of the tensor product of the position/momentum Hilbert space times the spin two-dimensional Hilbert space ?

It could be either; I purposely used very general notation.

So for a given full particle state $| \Psi \rangle$, $\psi(x)$ makes sense only in case the full particle state is separable (i.e. it is the "components" of the state's part that "lives" in the position/momentum Hilbert space).

Yes. More precisely, the state of the particle must be separable between the position/momentum and spin degrees of freedom.

 

[cianfa72]

More precisely, the state of the particle must be separable between the position/momentum and spin degrees of freedom.

For position/momentum Hilbert space there are at least two basis type: position and momentum.

Does the same apply to spin-1/2 two-dimensional Hilbert space? In other words does exist different basis types for it ?

 

[PeterDonis]

For position/momentum Hilbert space there are at least two basis type: position and momentum.

There are an infinite number of possible bases. Position and momentum are the two that are usually used because they have obvious physical interpretations.

I don't know what you mean by basis "types". A basis is a basis.

Does the same apply to spin-1/2 two-dimensional Hilbert space? In other words does exist different basis types for it ?

Again, I don't know what you mean by basis "types". There are an infinite number of possible sets of basis vectors for the spin-1/2 Hilbert space.

 

[cianfa72]

There are an infinite number of possible sets of basis vectors for the spin-1/2 Hilbert space.

Yes, of course. I wanted just know if even for spin-1/2 Hilbert space, there were "special" basis with given names (or let me say type) as the position basis or momentum basis in the position/momentum infinite-dimension Hilbert space.

 

[anuttarasammyak]

For an example Pauli matrices as basis of spinor may be one you seek.

 

[cianfa72]

For an example Pauli matrices as basis of spinor may be one you seek.

Sorry, how can Pauli 2x2 matrices be a basis of a two-components complex vector (as spin-1/2 vector is) ?

 

[PeterDonis]

I wanted just know if even for spin-1/2 Hilbert space, there were "special" basis with given names (or let me say type) as the position basis or momentum basis in the position/momentum infinite-dimension Hilbert space.

Once you have picked a set of Cartesian coordinate axes $x$, $y$, and $z$, the two spin components with respect to one of those axes (traditionally $z$) can be viewed as a "special" basis with a given name (such as spin-z up and spin-z down).

 

[PeterDonis]

Pauli matrices as basis of spinor

I think you mean spinors that are eigenvectors of one of the Pauli matrices (traditionally $\sigma_z$). The matrices themselves are operators, not vectors.

 

[PeterDonis]

Sorry, how can Pauli 2x2 matrices be a basis of a two-components complex vector (as spin-1/2 vector is) ?

See my post #16 for clarification of what I think was meant.

 

[cianfa72]

Once you have picked a set of Cartesian coordinate axes $x$, $y$, and $z$, the two spin components with respect to one of those axes (traditionally $z$) can be viewed as a "special" basis with a given name (such as spin-z up and spin-z down).

Are you talking of the Block sphere's representation of the projective space $\mathbf P(\mathbb H_2)$ ? The $\mathbb H_2$ ray represented by $| \uparrow \rangle$ is mapped on the North pole whereas the ray represented by $| \downarrow \rangle$ on the South pole.

Apart from this, is there a "connection/relation" between the Block sphere representation and the cartesian axes $x$, $y$ and $z$ ?

 

[PeterDonis]

Are you talking of the Block sphere's representation of the projective space $\mathbf P(\mathbb H_2)$ ? The $\mathbb H_2$ ray represented by $| \uparrow \rangle$ is mapped on the North pole whereas the ray represented by $| \downarrow \rangle$ on the South pole.

The Bloch sphere, yes.

Apart from this, is there a "connection/relation" between the Block sphere representation and the cartesian axes $x$, $y$ and $z$ ?

To use the Bloch sphere as a model, you need to pick which directions on the Bloch sphere correspond to which orientations of the actual, physical spin measurement device in your experiment. The usual way of doing that is to pick Cartesian coordinate axes in space and describe the orientation using them; then you can describe directions on the Bloch sphere corresponding to orientations of your experimental device the same way.

For example, you say the ket $\ket{\uparrow}$ maps to the North Pole and $\ket{\downarrow}$ maps to the South Pole. But those kets only name specific spin-1/2 states once you have picked an orientation. The usual convention is to pick spin-z (spin along the $z$ axis) as the orientation, so $\ket{\uparrow}$ means spin-z up and $\ket{\downarrow}$ means spin-z down. Then the North and South poles on the Bloch sphere correspond to the $z$ axis.

 

[Delta Kilo]

Apart from this, is there a "connection/relation" between the Bloch sphere representation and the cartesian axes $x$, $y$ and $z$ ?

Yes, they both transform under rotation of coordinate system in a consistent way, but according to different rules, SO(3) for vectors and SU(2) for spinors. The angle between vector in 3D space and spin direction remains the same, although spinor acquires a factor ot -1 when rotated by 360 degrees.

 

[cianfa72]

To use the Bloch sphere as a model, you need to pick which directions on the Bloch sphere correspond to which orientations of the actual, physical spin measurement device in your experiment. The usual way of doing that is to pick Cartesian coordinate axes in space and describe the orientation using them; then you can describe directions on the Bloch sphere corresponding to orientations of your experimental device the same way.

Ok, so an orientation of an experimental physical device in space corresponds to a direction in the Bloch sphere model. However I'm confused since two antipodal points on the Bloch sphere correspond actually two different rays in the state space $\mathbb H_2$ (each ray represents physically equivalent quantum states).

 

[PeterDonis]

so an orientation of an experimental physical device in space corresponds to a direction in the Bloch sphere model.

More precisely, choosing coordinate axes to describe the orientation of a physical device corresponds to choosing coordinates on the Bloch sphere so that the North and South Poles correspond to the axis of the spin measurement.

two antipodal points on the Bloch sphere correspond actually two different rays in the state space $\mathbb H_2$ (each ray represents physically equivalent quantum states).

Antipodal points on the Bloch sphere correspond to orthogonal "up" and "down" spin states along a given axis. Choosing an axis includes choosing an orientation for the axis: for example, if we choose the $z$ axis, we also have to choose which direction on the $z$ axis corresponds to increasing $z$, and that direction will correspond to the point on the Bloch sphere that we designate as the "North" pole (the spin "up" state).

 

[cianfa72]

Antipodal points on the Bloch sphere correspond to orthogonal "up" and "down" spin states along a given axis. Choosing an axis includes choosing an orientation for the axis

So basically pick an axis as the orientation in space of an experimental physical device correspond to pick an orthogonal basis in $\mathbb H_2$ (from the rays associated at each of them).

 

[PeterDonis]

So basically pick an axis as the orientation in space of an experimental physical device correspond to pick an orthogonal basis in $\mathbb H_2$ (from the rays associated at each of them).

Yes.

 

[vanhees71]

The great appeal of the Dirac formulation is that this confusion is avoided from the very beginning. There are vectors in Hilbert space, which are completely independent of any choice of the basis. The wave function is then a component of these vectors with respect to a (generalized) basis of the Hilbert space.

The normalized kets $|\psi \rangle$ of the Hilbert space represent (modulo the usual cautions concerning rays vs. vectors) pure states of the system. Then the wave function, e.g., in the position representation is
$$\psi(\vec{x})=\langle \vec{x}|\psi \rangle$$
is a "probability amplitude", i.e.,
$$P(x)=|\psi(\vec{x})|^2$$
is the probability density for the position of the particle.

The probability amplitude for the momentum of the particle is given by the corresponding momentum-wave function, which is
$$\tilde{\psi}(\vec{p}) = \langle \vec{p}|\psi \rangle.$$
So far we describe particles without spin (e.g., pions).

If you have spin, neither $\vec{x}$ nor $\vec{p}$ are complete sets of compatible observables but in addition you have one spin component of the particle as an additional compatible degree of freedom. Usually one chooses $\sigma_z$ as this additional degree of freedom. So now the probability amplitudes have also to depend on the corresponding common eigenvectors of $\vec{x}$ or $\vec{p}$ and $\sigma_z$, i.e., the corresponding wave functions now are given by
$$\psi(\vec{x},\sigma_z)=\langle \vec{x},\sigma_z|\psi \rangle$$
and
$$\tilde{\psi}(\vec{p},\sigma_z) = \langle \vec{p},\sigma_z|\psi \rangle.$$
The $\sigma_z$ have discrete eigenvalues ($\sigma_z=\in \{-s \hbar,-(s-1) \hbar,\ldots,(s-1) \hbar, s \hbar \}$, where $s$ is the spin of the particle $s \in \{0,1/2,1,\ldots \}$. It refers to the eigenvalue $s(s+1) \hbar^2$ of $\vec{\sigma}^2$, which is also an observable compatible with $\vec{x}$ and $\sigma_z$ (or $\vec{p}$ and $\sigma_z$).

Now the meaning of the wave functions is that
$$\mathrm{d} P=|\psi(\vec{x}),\sigma_z)|^2 \mathrm{d}^3 x$$
is the probability to find a particle at a volume (of infinitesimal size) $\mathrm{d}^3 x$ around the position $\vec{x}$ and having a spin-$z$ component with value $\sigma_z$ and anlogously for $\tilde{\psi}(\vec{p},\sigma_z)$.

Last but not least it must be emphasized that all I said above about spin is only valid in non-relativistic QM (not in relativistic QFT).

Now often the spinor notation is used. Then you put the wave function in a column vector (in this case you say spinor rather than vector to emphasize that it refers to the spin of the particle) and write
$$\Psi(\vec{x})=\begin{pmatrix} \psi(\vec{x},s \hbar) \\ \psi(\vec{x},(s-1) \hbar) \\ \vdots \\ \psi(\vec{x},-s \hbar \end{pmatrix}.$$
This is a column spinor with $(2s+1)$ components. For $s=1/2$ you have two-component spinors, also called "Weyl spinors".

In this representation the spin components are represented with help of the Pauli matrices,
$$\vec{s}=\frac{\hbar}{2} \vec{\sigma}.$$

 

[cianfa72]

If you have spin, neither $\vec{x}$ nor $\vec{p}$ are complete sets of compatible observables but in addition you have one spin component of the particle as an additional compatible degree of freedom. Usually one chooses $\sigma_z$ as this additional degree of freedom. So now the probability amplitudes have also to depend on the corresponding common eigenvectors of $\vec{x}$ or $\vec{p}$ and $\sigma_z$, i.e., the corresponding wave functions now are given by
$$\psi(\vec{x},\sigma_z)=\langle \vec{x},\sigma_z|\psi \rangle$$ and
$$\tilde{\psi}(\vec{p},\sigma_z) = \langle \vec{p},\sigma_z|\psi \rangle.$$

Sorry, if we have spin then there is also a spin additional component. Now the probability amplitudes will depend on the corresponding common eigenvectors (e.g. of $\vec{x}$ and $\sigma_z$). That means both operators act on the same underlying system's Hilbert space.

So actually there is not a separation of the system state in the tensor product of position/momentum Hilbert space times Hilbert spin space.

 

[cianfa72]

So actually there is not a separation of the system state in the tensor product of position/momentum Hilbert space times Hilbert spin space.

I am confused about the following point: do the hermitian/self-adjoint operators $\vec{x}$ and $\sigma_z$ have eigenvectors in the same Hilbert space (which is a tensor product Hilbert space), or each one has eigenvectors only in its associated Hilbert space (i.e. the infinite-dimensional Hilbert space for $\vec{x}$ and the 2-dimensional Hilbert space for $\sigma_z$) ?

 

[PeterDonis]

I am confused about the following point: do the hermitian/self-adjoint operators $\vec{x}$ and $\sigma_z$ have eigenvectors in the same Hilbert space (which is a tensor product Hilbert space), or each one has eigenvectors only in its associated Hilbert space (i.e. the infinite-dimensional Hilbert space for $\vec{x}$ and the 2-dimensional Hilbert space for $\sigma_z$) ?

If you are just looking at one degree of freedom, say spin, then you just look at the operator on that degree of freedom, so $\sigma_z$ for spin.

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 $\sigma_z$.

 

[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 $\sigma_z$.

Ok, so if we look at one degree of freedom a linear operator is just an element of the tensor product space $\mathbb H_1 \otimes \mathbb H_1^*$ where $\mathbb H_1$ is the n-dimensional separable Hilbert space associated to that degree of freedom.

Then if we look at both degrees of freedom, the "composite" linear operator acting on $\mathbb H_2 \otimes \mathbb H_1$ is an element of the tensor product Hilbert space $$\mathbb H_2 \otimes \mathbb H_2^* \otimes \mathbb H_1 \otimes \mathbb H_1^*$$ where $\mathbb H_2$ is the m-dimensional separable Hilbert space associated to the latter degree of freedom.

 

[PeterDonis]

Ok, so if we look at one degree of freedom a linear operator is just an element of the tensor product space $\mathbb H_1 \otimes \mathbb H_1^*$ where $\mathbb H_1$ is the n-dimensional separable Hilbert space associated to that degree of freedom.

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.

Then if we look at both degrees of freedom, the "composite" linear operator acting on the cartesian product $\mathbb H_2 \times \mathbb H_1$ is an element of the tensor product Hilbert space

Of the tensor product of the Hilbert spaces for both degrees of freedom. Which would just be $\mathbb{H}_1 \otimes \mathbb{H}_2$.

 

[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.