Principles of Quantum Mechanics · Problem 10.1.2

Systems with N Degrees of Freedom — R. Shankar

Physics — Quantum Mechanics · direct-product-two-dimensional

Problem

Imagine a fictitious world in which the single-particle Hilbert space is two-dimensional. Let us denote the basis vectors by

|+\rangle

and

|-\rangle

. Let

\sigma_1^{(1)}=\begin{array}{c}+\\-\end{array}\left[\begin{array}{cc}a & b\\ c & d\end{array}\right] \quad \text{and} \quad \sigma_2^{(2)}=\begin{array}{c}+\\-\end{array}\left[\begin{array}{cc}e & f\\ g & h\end{array}\right]

be operators in

\mathbb{V}_1

and

\mathbb{V}_2

, respectively (the

\pm

signs label the basis vectors. Thus

b=\langle+|\sigma_1^{(1)}|-\rangle

etc.) The space

\mathbb{V}_1\otimes\mathbb{V}_2

is spanned by four vectors

|+\rangle\otimes|+\rangle

|+\rangle\otimes|-\rangle

|-\rangle\otimes|+\rangle

|-\rangle\otimes|-\rangle

. Show (using the method of images or otherwise) that (1)

\sigma_1^{(1)\otimes(2)}=\sigma_1^{(1)}\otimes I^{(2)}=\begin{array}{c}++\\+-\\-+\\--\end{array}\begin{bmatrix}a & 0 & b & 0\\ 0 & a & 0 & b\\ c & 0 & d & 0\\ 0 & c & 0 & d\end{bmatrix}

(2)

\sigma_2^{(1)\otimes(2)}=\begin{bmatrix}e & f & 0 & 0\\ g & h & 0 & 0\\ 0 & 0 & e & f\\ 0 & 0 & g & h\end{bmatrix}

(3)

(\sigma_1\sigma_2)^{(1)\otimes(2)}=\sigma_1^{(1)}\otimes\sigma_2^{(2)}=\begin{bmatrix}ae & af & be & bf\\ ag & ah & bg & bh\\ ce & cf & de & df\\ cg & ch & dg & dh\end{bmatrix}

Do part (3) in two ways, by taking the matrix product of

\sigma_1^{(1)\otimes(2)}

and

\sigma_2^{(1)\otimes(2)}

and by directly computing the matrix elements of

\sigma_1^{(1)}\otimes\sigma_2^{(2)}