The multi-qubit computational basis is the set of multi-qubit states containing all possible combinations of $|0\rangle$ and $|1\rangle.$ For example, for the 2-qubit case, we have $|00\rangle, |01\rangle, |10\rangle, |11\rangle.$ You may notice that these correspond to the numbers 0–3 in binary; this is not an accident! It is common to write out the integer value, especially for a large number of qubits, e.g., $|10\rangle \rightarrow |2\rangle,$ $|111\rangle \rightarrow |7\rangle.$ ***Exercise I.11.1.*** The two-qubit computational basis consists of 4 vectors: $\vert 0\rangle \otimes \vert 0\rangle,$ $\vert 0\rangle \otimes \vert 1\rangle,$ $\vert 1 \rangle \otimes \vert 0\rangle,$ and $\vert 1\rangle \otimes \vert 1\rangle.$ These vectors constitute every possible pairing of two qubits in the two possible single-qubit basis states. Apply the tensor product to evaluate the 4-dimensional basis vectors. What do you notice about these vectors?

Solution.

Let's work these out using the tensor product formula above: $$ \begin{align*} \vert 0\rangle \otimes \vert 0\rangle &= \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \\ \vert 0\rangle \otimes \vert 1\rangle &= \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \\ \vert 1\rangle \otimes \vert 0\rangle &= \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \\ \vert 1\rangle \otimes \vert 1\rangle &= \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}. \end{align*} $$ There are two things to note here: the first is that each vector contains only a single 1! This makes it clear that we have a basis, because we can create any $4 \times 1$ complex-valued vector by taking a linear combination of these basis states. The second point relates to where exactly the 1 sits in the vector. This is related to the binary expression of the product of basis states. For example, $\vert 0\rangle \otimes \vert 0\rangle$ has a 1 in the zero-th entry, which corresponds to `00` in binary. The vector $\vert 1\rangle \otimes \vert 0\rangle$ has a 1 in the second entry (indexing from 0) - `10` is 2 in binary. Thus, to construct an arbitrary basis vector, instead of taking a tensor product, you can just create a vector with a 1 at the element that corresponds to that basis state's binary representation! ▢

--- *Tip*. To avoid writing so many $\otimes$ symbols, multi-qubit states are often abbreviated by concatenating the bits and stuffing them inside a single ket. For example, the 4-qubit state $\vert 0\rangle \otimes \vert 1\rangle \otimes \vert 1\rangle \otimes \vert 0\rangle$ becomes $\vert 0110\rangle,$ which takes much less effort to write! --- ***Exercise I.11.2.*** Consider the number and size of the vectors in the 2-qubit computational basis; how many vectors do you expect to have in the 3-qubit computational basis, and what size will they be? Generalize this result to an $n$-qubit system (this should be a clear indication of why we need to actually *build* quantum computers, rather than just simulating them!).

Solution.

The 3-qubit computational basis will consist of 8 vectors of length 8. Every time another qubit is added, we multiply by 2, so an $n$-qubit system has $2^n$ basis vectors with $2^n$ elements. Consider that this also means that, to operate on such basis vectors, we'll need unitary matrices of size $2^n \times 2^n.$ This exponential dependence means that simulating quantum computers will become computationally intractable very quickly. ▢

--- ***Exercise I.11.3.*** Use the tensor product to compute the state vector of a two-qubit system where the first qubit is in state $\vert +\rangle,$ and the second is in state $\vert 1\rangle.$ Express that state as a linear combination of the two-qubit computational basis vectors, and verify that this state is still normalized. *Tip*. There are two approaches to working this problem. One is to express the states as two vectors and take the tensor product. The other is to work at the level of computational basis states, and leverage the fact that the tensor product is *distributive*, i.e. $(\vert a\rangle + \vert b\rangle)\otimes \vert c\rangle = \vert a\rangle \otimes \vert c\rangle + \vert b\rangle \otimes \vert c\rangle$ (and similarly for matrices).

Solution.

We would like to compute $\vert +\rangle \otimes \vert 1\rangle,$ where we recall that $$ \vert +\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad \vert 1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}. $$ Using the first method, we can compute $$ \vert +\rangle \otimes \vert 1\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}. $$ By inspecting the basis vectors above, we find that $\vert +\rangle \otimes \vert 1\rangle = \frac{1}{\sqrt{2}} \left( \vert 01\rangle + \vert 11\rangle \right).$ The modulus square of all the vector entries sums to 1, and so our state is still normalized. For the second method, we can use the distributive property of the tensor product and obtain the linear combination directly: $$ \vert +\rangle \otimes \vert 1\rangle = \frac{1}{\sqrt{2}} (\vert 0\rangle + \vert 1\rangle) \otimes \vert 1\rangle = \frac{1}{\sqrt{2}} (\vert 01\rangle + \vert 11\rangle). $$ We can obtain the state vector by rewriting the computational basis states in their vector form to obtain the same results as above. This method is much more compact, but does take some time to get used to. ▢

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Codercise I.11.1 — Preparing basis state

Codercise I.11.2 — Separable Operations

Codercise I.11.3 — Expectation value of two-qubit observable

Codercise I.11.4 — Double Trouble

The tensor product

Multi-qubit bases

Separable operations