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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)


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


\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)


\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)


\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.


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?


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)


\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}


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.


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


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