# Peeter Joot's (OLD) Blog.

• ## Archives

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 320,607

## 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.