# Peeter Joot's Blog.

• ## Archives

 ivor on Just Energy Canada nasty busin… A final pre-exam upd… on An updated compilation of note… Anon on About peeterjoot on About Anon on About
• ## People not reading this blog: 6,973,738,433 minus:

• 132,470 hits

# Posts Tagged ‘Hilbert space’

## PHY456H1F: Quantum Mechanics II. Lecture 13 (Taught by Prof J.E. Sipe). Spin and spinors (cont.)

Posted by peeterjoot on October 24, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Multiple wavefunction spaces.

Reading: See section 26.5 in the text [1].

We identified

\begin{aligned}\psi(\mathbf{r}) = \left\langle{{ \mathbf{r}}} \vert {{\psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.1)

with improper basis kets

\begin{aligned}{\left\lvert {\mathbf{r}} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.2)

Now introduce many function spaces

\begin{aligned}\begin{bmatrix}\psi_1(\mathbf{r}) \\ \psi_2(\mathbf{r}) \\ \dot{v}s \\ \psi_\gamma(\mathbf{r})\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)

with improper (unnormalizable) basis kets

\begin{aligned}{\left\lvert {\mathbf{r} \alpha} \right\rangle}, \qquad \alpha \in 1, 2, ... \gamma\end{aligned} \hspace{\stretch{1}}(2.4)

\begin{aligned}\psi_\alpha(\mathbf{r}) = \left\langle{{ \mathbf{r}\alpha}} \vert {{\psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.5)

for an abstract ket ${\left\lvert {\psi} \right\rangle}$

We will try taking this Hilbert space

\begin{aligned}H = H_o \otimes H_s\end{aligned} \hspace{\stretch{1}}(2.6)

Where $H_o$ is the Hilbert space of “scalar” QM, “o” orbital and translational motion, associated with kets ${\left\lvert {\mathbf{r}} \right\rangle}$ and $H_s$ is the Hilbert space associated with the $\gamma$ components ${\left\lvert {\alpha} \right\rangle}$. This latter space we will label the “spin” or “internal physics” (class suggestion: or perhaps intrinsic). This is “unconnected” with translational motion.

We build up the basis kets for $H$ by direct products

\begin{aligned}{\left\lvert {\mathbf{r} \alpha} \right\rangle} = {\left\lvert {\mathbf{r}} \right\rangle} \otimes {\left\lvert {\alpha} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.7)

Now, for a rotated ket we seek a general angular momentum operator $\mathbf{J}$ such that

\begin{aligned}{\left\lvert {\psi'} \right\rangle} = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.8)

where

\begin{aligned}\mathbf{J} = \mathbf{L} + \mathbf{S},\end{aligned} \hspace{\stretch{1}}(2.9)

where $\mathbf{L}$ acts over kets in $H_o$, “orbital angular momentum”, and $\mathbf{S}$ is the “spin angular momentum”, acting on kets in $H_s$.

Strictly speaking this would be written as direct products involving the respective identities

\begin{aligned}\mathbf{J} = \mathbf{L} \otimes I_s + I_o \otimes \mathbf{S}.\end{aligned} \hspace{\stretch{1}}(2.10)

We require

\begin{aligned}\left[{J_i},{J_j}\right] = i \hbar \sum \epsilon_{i j k} J_k\end{aligned} \hspace{\stretch{1}}(2.11)

Since $\mathbf{L}$ and $\mathbf{S}$ “act over separate Hilbert spaces”. Since these come from legacy operators

\begin{aligned}\left[{L_i},{S_j}\right] = 0\end{aligned} \hspace{\stretch{1}}(2.12)

We also know that

\begin{aligned}\left[{L_i},{L_j}\right] = i \hbar \sum \epsilon_{i j k} L_k\end{aligned} \hspace{\stretch{1}}(2.13)

so

\begin{aligned}\left[{S_i},{S_j}\right] = i \hbar \sum \epsilon_{i j k} S_k, \end{aligned} \hspace{\stretch{1}}(2.14)

as expected. We could, in principle, have more complicated operators, where this would not be true. This is a proposal of sorts. Given such a definition of operators, let’s see where we can go with it.

For matrix elements of $\mathbf{L}$ we have

\begin{aligned}{\left\langle {\mathbf{r}} \right\rvert} L_x {\left\lvert {\mathbf{r}'} \right\rangle} = -i \hbar \left( y \frac{\partial {}}{\partial {z}}-z \frac{\partial {}}{\partial {y}} \right) \delta(\mathbf{r}- \mathbf{r}')\end{aligned} \hspace{\stretch{1}}(2.15)

What are the matrix elements of ${\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle}$? From the commutation relationships we know

\begin{aligned}\sum_{\alpha'' = 1}^\gamma {\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha''} \right\rangle}{\left\langle {\alpha''} \right\rvert} S_j {\left\lvert {\alpha'} \right\rangle}-\sum_{\alpha'' = 1}^\gamma {\left\langle {\alpha} \right\rvert} S_j {\left\lvert {\alpha''} \right\rangle}{\left\langle {\alpha''} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle}=i \hbar \sum_k \epsilon_{ijk} {\left\langle {\alpha} \right\rvert} S_k {\left\lvert {\alpha''} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.16)

We see that our matrix element is tightly constrained by our choice of commutator relationships. We have $\gamma^2$ such matrix elements, and it turns out that it is possible to choose (or find) matrix elements that satisfy these constraints?

The ${\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle}$ matrix elements that satisfy these constraints are found by imposing the commutation relations

\begin{aligned}\left[{S_i},{S_j}\right] = i \hbar \sum \epsilon_{i j k} S_k, \end{aligned} \hspace{\stretch{1}}(2.17)

and with

\begin{aligned}S^2 = \sum_j S_j^2,\end{aligned} \hspace{\stretch{1}}(2.18)

(this is just a definition). We find

\begin{aligned}\left[{S^2},{S_i}\right] = 0\end{aligned} \hspace{\stretch{1}}(2.19)

and seeking eigenkets

\begin{aligned}S^2 {\left\lvert {s m_s} \right\rangle} &= s(s+1) \hbar^2 {\left\lvert {s m_s} \right\rangle} \\ S_z {\left\lvert {s m_s} \right\rangle} &= \hbar m_s {\left\lvert {s m_s} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.20)

Find solutions for $s = 1/2, 1, 3/2, 2, \cdots$, where $m_s \in \{-s, \cdots, s\}$. ie. $2 s + 1$ possible vectors ${\left\lvert {s m_s} \right\rangle}$ for a given $s$.

\begin{aligned}s = \frac{1}{{2}} &\implies \gamma = 2 \\ s = 1 &\implies \gamma = 3 \\ s = \frac{3}{2} &\implies \gamma = 4 \end{aligned}

We start with the algebra (mathematically the Lie algebra), and one can compute the Hilbert spaces that are consistent with these algebraic constraints.

We assume that for any type of given particle $S$ is fixed, where this has to do with the nature of the particle.

\begin{aligned}s = \frac{1}{{2}} &\qquad \text{A spinlatex 1/2particle} \\ s = 1 &\qquad \text{A spin $1$ particle} \\ s = \frac{3}{2} &\qquad \text{A spin $3/2$ particle}\end{aligned}

$S$ is fixed once we decide that we are talking about a specific type of particle.

A non-relativistic particle in this framework has two nondynamical quantities. One is the mass $m$ and we now introduce a new invariant, the spin $s$ of the particle.

This has been introduced as a kind of strategy. It is something that we are going to try, and it turns out that it does. This agrees well with experiment.

In 1939 Wigner asked, “what constraints do I get if I constrain the constraints of quantum mechanics with special relativity.” It turns out that in the non-relativistic limit, we get just this.

There’s a subtlety here, because we get into some logical trouble with the photon with a rest mass of zero ($m = 0$ is certainly allowed as a value of our invariant $m$ above). We can’t stop or slow down a photon, so orbital angular momentum is only a conceptual idea. Really, the orbital angular momentum and the spin angular momentum cannot be separated out for a photon, so talking of a spin $1$ particle really means spin as in $\mathbf{J}$, and not spin as in $\mathbf{L}$.

## Spin $1/2$ particles

Reading: See section 26.6 in the text [1].

Let’s start talking about the simplest case. This includes electrons, all leptons (integer spin particles like photons and the weakly interacting W and Z bosons), and quarks.

\begin{aligned}s &= \frac{1}{{2}} \\ m_s &= \pm \frac{1}{{2}}\end{aligned} \hspace{\stretch{1}}(2.22)

states

\begin{aligned}{\left\lvert {s m_s} \right\rangle} = {\left\lvert { \frac{1}{{2}}, \frac{1}{{2}} } \right\rangle},{\left\lvert { \frac{1}{{2}}, -\frac{1}{{2}} } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.24)

Note there is a convention

\begin{aligned}{\left\lvert { \frac{1}{{2}} \bar{\frac{1}{{2}}} } \right\rangle} &= {\left\lvert { \frac{1}{{2}}, -\frac{1}{{2}} } \right\rangle} \\ {\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle} &= {\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.25)

\begin{aligned}\begin{aligned}S^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} &= \frac{1}{{2}} \left( \frac{1}{{2}} + 1 \right) \hbar^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \\ &=\frac{3}{4} \hbar^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \\ \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.27)

\begin{aligned}S_z {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} = m_s \hbar {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \end{aligned} \hspace{\stretch{1}}(2.28)

For shorthand

\begin{aligned}{\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle} &= {\left\lvert { + } \right\rangle} \\ {\left\lvert { \frac{1}{{2}} \bar{\frac{1}{{2}}} } \right\rangle} &= {\left\lvert { - } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.29)

\begin{aligned}S^2 \rightarrow \frac{3}{4} \hbar^2 \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.31)

\begin{aligned}S_z \rightarrow \frac{\hbar}{2}\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.32)

One can easily work out from the commutation relationships that

\begin{aligned}S_x \rightarrow \frac{\hbar}{2}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.33)

\begin{aligned}S_y \rightarrow \frac{\hbar}{2}\begin{bmatrix}0 & -i \\ i & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.34)

We’ll start with adding $\mathbf{L}$ into the mix on Wednesday.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

## PHY456H1F: Quantum Mechanics II. Lecture 1 (Taught by Prof J.E. Sipe). Review: Composite systems

Posted by peeterjoot on September 15, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Peeter’s lecture notes from class. May not be entirely coherent.

# Composite systems.

This is apparently covered as a side effect in the text [1] in one of the advanced material sections. FIXME: what section?

Example, one spin one half particle and one spin one particle. We can describe either quantum mechanically, described by a pair of Hilbert spaces

\begin{aligned}H_1,\end{aligned} \hspace{\stretch{1}}(1.1)

of dimension $D_1$

\begin{aligned}H_2,\end{aligned} \hspace{\stretch{1}}(1.2)

of dimension $D_2$

Recall that a Hilbert space (finite or infinite dimensional) is the set of states that describe the system. There were some additional details (completeness, normalizable, $L2$ integrable, …) not really covered in the physics curriculum, but available in mathematical descriptions.

We form the composite (Hilbert) space

\begin{aligned}H = H_1 \otimes H_2\end{aligned} \hspace{\stretch{1}}(1.3)

\begin{aligned}H_1 : { {\lvert {\phi_1^{(i)}} \rangle} }\end{aligned} \hspace{\stretch{1}}(1.4)

for any ket in $H_1$

\begin{aligned}{\lvert {I} \rangle} = \sum_{i=1}^{D_1} c_i {\lvert {\phi_1^{(i)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.5)

where

\begin{aligned}{\langle { \phi_1^{(i)}} \rvert}{\lvert { \phi_1^{(j)}} \rangle} = \delta^{i j}\end{aligned} \hspace{\stretch{1}}(1.6)

Similarly

\begin{aligned}H_2 : { {\lvert {\phi_2^{(i)}} \rangle} }\end{aligned} \hspace{\stretch{1}}(1.7)

for any ket in $H_2$

\begin{aligned}{\lvert {II} \rangle} = \sum_{i=1}^{D_2} d_i {\lvert {\phi_2^{(i)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.8)

where

\begin{aligned}{\langle { \phi_2^{(i)}} \rvert}{\lvert { \phi_2^{(j)}} \rangle} = \delta^{i j}\end{aligned} \hspace{\stretch{1}}(1.9)

The composite Hilbert space has dimension $D_1 D_2$

basis kets:

\begin{aligned}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} = {\lvert { \phi^{(ij)}} \rangle},\end{aligned} \hspace{\stretch{1}}(1.10)

where

\begin{aligned}{\langle { \phi^{(ij)}} \rvert}{\lvert { \phi^{(kl)}} \rangle} = \delta^{ik} \delta^{jl}.\end{aligned} \hspace{\stretch{1}}(1.11)

Any ket in $H$ can be written

\begin{aligned}{\lvert {\psi} \rangle} &= \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \\ &= \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi^{(ij)}} \rangle}.\end{aligned}

Direct product of kets:

\begin{aligned}{\lvert {I} \rangle} \otimes {\lvert {II} \rangle} &\equiv\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}c_i d_j{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \\ &=\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}c_i d_j{\lvert { \phi^{(ij)}} \rangle} \end{aligned}

If ${\lvert {\psi} \rangle}$ in $H$ cannot be written as ${\lvert {I} \rangle} \otimes {\lvert {II} \rangle}$, then ${\lvert {\psi} \rangle}$ is said to be “entangled”.

FIXME: insert a concrete example of this, with some low dimension.

## Operators.

With operators $\mathcal{O}_1$ and $\mathcal{O}_2$ on the respective Hilbert spaces. We’d now like to build

\begin{aligned}\mathcal{O}_1 \otimes \mathcal{O}_2\end{aligned} \hspace{\stretch{1}}(1.12)

If one defines

\begin{aligned}\mathcal{O}_1 \otimes \mathcal{O}_2\equiv\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.13)

Q:Can every operator that can be defined on the composite space have a representation of this form?

No.

Special cases. The identity operators. Suppose that

\begin{aligned}{\lvert {\psi} \rangle}=\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.14)

then

\begin{aligned}(\mathcal{O}_1 \otimes \mathcal{I}_2) {\lvert {\psi} \rangle}=\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.15)

### Example commutator.

Can do other operations. Example:

\begin{aligned}\left[{ \mathcal{O}_1 \otimes \mathcal{I}_2 },{ \mathcal{I}_1 \otimes \mathcal{O}_2 }\right] = 0\end{aligned} \hspace{\stretch{1}}(1.16)

Let’s verify this one. Suppose that our state has the representation

\begin{aligned}{\lvert {\psi} \rangle} = \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.17)

so that the action on this ket from the composite operations are

\begin{aligned}(\mathcal{O}_1 \otimes \mathcal{I}_2){\lvert {\psi} \rangle} &= \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \\ (\mathcal{I}_1 \otimes \mathcal{O}_2){\lvert {\psi} \rangle} &= \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.18)

Our commutator is

\begin{aligned}\left[{(\mathcal{O}_1 \otimes \mathcal{I}_2)},{(\mathcal{I}_1 \otimes \mathcal{O}_2)}\right]{\lvert {\psi} \rangle} &=(\mathcal{O}_1 \otimes \mathcal{I}_2)(\mathcal{I}_1 \otimes \mathcal{O}_2) {\lvert {\psi} \rangle} -(\mathcal{I}_1 \otimes \mathcal{O}_2)(\mathcal{O}_1 \otimes \mathcal{I}_2){\lvert {\psi} \rangle} \\ &=(\mathcal{O}_1 \otimes \mathcal{I}_2)\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle}-(\mathcal{I}_1 \otimes \mathcal{O}_2)\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \\ &=\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle}-\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle} \\ &=0 \qquad \square\end{aligned}

### Generalizations.

Can generalize to

\begin{aligned}H_1 \otimes H_2 \otimes H_3 \otimes \cdots\end{aligned} \hspace{\stretch{1}}(1.20)

Can also start with $H$ and seek factor spaces. If $H$ is not prime there are, in general, many ways to find factor spaces

\begin{aligned}H = H_1 \otimes H_2 =H_1' \otimes H_2'\end{aligned} \hspace{\stretch{1}}(1.21)

A ket ${\lvert {\psi} \rangle}$, if unentangled in the first factor space, then it will be in general entangled in a second space. Thus ket entanglement is not a property of the ket itself, but instead is intrinsically related to the space in which it is represented.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.