Tensor Product Exercises

These exercises practice finite-dimensional tensor products for distinguishable subsystems. They use the two-qubit basis order

$$\lvert00\rangle,\quad \lvert01\rangle,\quad \lvert10\rangle,\quad \lvert11\rangle,$$

with subsystem $A$ written first and subsystem $B$ second. Identity factors are written explicitly when an operator acts only on one subsystem.

For identical particles, do not use these exercises as the final word: after forming tensor products one must impose exchange symmetry. See the identical-particle and Fock-space pages for that layer.

All state vectors below are normalized. Helpful background. Product bases explains the coordinate order, and operators on composite systems explains identity factors.

Pauli operators and product order

For qubits, use

$$I= \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}, \qquad X= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix},$$

and

$$Y= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}, \qquad Z= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}.$$

Thus

$$Z\lvert0\rangle=\lvert0\rangle, \qquad Z\lvert1\rangle=-\lvert1\rangle.$$

Exercises

1. Let $\mathcal H_A$ have basis $\{\lvert0\rangle,\lvert1\rangle,\lvert2\rangle\}$ and let $\mathcal H_B$ have basis $\{\lvert g\rangle,\lvert e\rangle\}$. Construct a product basis of $\mathcal H_A\otimes\mathcal H_B$ and give the dimension.

Solution

Pair every basis vector of $A$ with every basis vector of $B$. With $A$ listed first, a natural order is

$$\lvert0g\rangle,\quad \lvert0e\rangle,\quad \lvert1g\rangle,\quad \lvert1e\rangle,\quad \lvert2g\rangle,\quad \lvert2e\rangle.$$

The dimension is

$$\dim(\mathcal H_A\otimes\mathcal H_B) = 3\cdot2 = 6.$$

2. Expand the product state

$$\bigl(\alpha\lvert0\rangle+\beta\lvert1\rangle\bigr) \otimes \bigl(\gamma\lvert0\rangle+\delta\lvert1\rangle\bigr)$$

in the standard two-qubit basis order.

Solution

Distribute the tensor product:

$$\lvert\psi\rangle = \alpha\gamma\lvert00\rangle +\alpha\delta\lvert01\rangle +\beta\gamma\lvert10\rangle +\beta\delta\lvert11\rangle.$$

In the basis order $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$, the coordinate column is

$$\begin{pmatrix} \alpha\gamma\\ \alpha\delta\\ \beta\gamma\\ \beta\delta \end{pmatrix}.$$

The coefficient array factors as $c_{ij}=a_i b_j$, so it has rank one when written as a $2\times2$ coefficient matrix.

3. Decide whether

$$\lvert\psi\rangle = \frac{1}{\sqrt{10}} \bigl( \lvert00\rangle +2\lvert01\rangle +\lvert10\rangle +2\lvert11\rangle \bigr)$$

is a product state. If it is, factor it.

Solution

The coefficient matrix in the $A,B$ order is

$$C = \frac{1}{\sqrt{10}} \begin{pmatrix} 1&2\\ 1&2 \end{pmatrix}.$$

The two rows are proportional, so the matrix has rank one. Therefore the state is a product state. One factorization is

$$\lvert\psi\rangle = \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2} \otimes \frac{\lvert0\rangle+2\lvert1\rangle}{\sqrt5}.$$

Multiplying the two normalized factors gives the four amplitudes $1/\sqrt{10}$, $2/\sqrt{10}$, $1/\sqrt{10}$, and $2/\sqrt{10}$.

4. Compute the matrices of $Z\otimes I$, $I\otimes X$, and $Z\otimes Z$ in the standard two-qubit basis order.

Solution

The first operator acts with $Z$ on the first qubit:

$$Z\otimes I = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix}.$$

The second operator flips the second qubit:

$$I\otimes X = \begin{pmatrix} 0&1&0&0\\ 1&0&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{pmatrix}.$$

The product operator multiplies the $Z$ eigenvalues:

$$Z\otimes Z = \begin{pmatrix} 1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&1 \end{pmatrix}.$$

5. Prove that local operators on different subsystems commute:

$$[A\otimes I_B,\ I_A\otimes B]=0.$$

Solution

Use the product rule for tensor-product operators:

$$(A\otimes I_B)(I_A\otimes B) = A\otimes B.$$

In the opposite order,

$$(I_A\otimes B)(A\otimes I_B) = A\otimes B.$$

The two products are the same, so their difference is zero:

$$[A\otimes I_B,\ I_A\otimes B]=0.$$

6. Compute the commutator

$$[X\otimes I,\ Z\otimes Z].$$

Solution

Because the second factor is the identity in the first operator,

$$[X\otimes I,\ Z\otimes Z] = [X,Z]\otimes Z.$$

The Pauli commutator is

$$[X,Z] = -2iY.$$

Therefore

$$[X\otimes I,\ Z\otimes Z] = -2i\,Y\otimes Z.$$

This shows a local operator need not commute with an interaction operator.

7. For real energy parameters $\varepsilon_A$, $\varepsilon_B$, and $J$, diagonalize the two-qubit Hamiltonian

$$H = \varepsilon_A Z\otimes I +\varepsilon_B I\otimes Z +J Z\otimes Z.$$

Solution

All three terms are diagonal in the standard product basis, so the product basis vectors are eigenvectors. Let $z_0=+1$ and $z_1=-1$. For $\lvert ab\rangle$,

$$E_{ab} = \varepsilon_A z_a +\varepsilon_B z_b +J z_a z_b.$$

Thus

$$\begin{aligned} E_{00}&=\varepsilon_A+\varepsilon_B+J,\\ E_{01}&=\varepsilon_A-\varepsilon_B-J,\\ E_{10}&=-\varepsilon_A+\varepsilon_B-J,\\ E_{11}&=-\varepsilon_A-\varepsilon_B+J. \end{aligned}$$

The coupling shifts energies according to whether the two $Z$ eigenvalues are aligned or anti-aligned.

8. The controlled-phase operator has matrix

$$U_{\mathrm{CZ}} = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&-1 \end{pmatrix}.$$

Show that it cannot be written as a single product $A\otimes B$ of any two one-qubit matrices.

Solution

A product operator takes a product vector to another product vector, or to zero. But controlled phase takes $|+\rangle\otimes|+\rangle$, with $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$, to a state with coefficient matrix

$$C=\frac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$

Its determinant is $-1/2$, so its rank is two: the output is entangled. This rules out every product $A\otimes B$, including nondiagonal factors.

9. A Hamiltonian can be diagonal in a product basis and still generate entanglement. Evolve $|+\rangle\otimes|+\rangle$ under $H=JZ\otimes Z$ for $J\ne0$. Find its concurrence and the earliest positive time at which it is maximally entangled.

Solution

Put $\theta=Jt/\hbar$. The even-parity basis states have energy $J$ and the odd-parity states have energy $-J$, so

$$|\psi(t)\rangle=\frac12\left[ e^{-i\theta}(|00\rangle+|11\rangle) +e^{i\theta}(|01\rangle+|10\rangle)\right].$$

The coefficient determinant is $-i\sin(2\theta)/2$. For a normalized pure two-qubit state, concurrence is twice the absolute value of this determinant. Thus

$$C(\psi(t))=|\sin(2Jt/\hbar)|, \qquad t_{\mathrm{first}}=\frac{\pi\hbar}{4|J|}.$$

The interaction changes relative phases between product-basis components; those phases generally prevent factorization. Having product eigenvectors does not imply that every superposition stays a product.

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.