4.1.1 Many quantum bits

The rules that we learned so far for one and two qubits generalize very naturally to quantum system of many qubits. For example, an arbitrary state of three qubits can be written as follows:

|ψ⟩\displaystyle\left|\psi\right\rangle =ψ000⁢|000⟩+ψ001⁢|001⟩+ψ010⁢|010⟩+ψ011⁢|011⟩\displaystyle=\psi_{000}\left|000\right\rangle+\psi_{001}\left|001\right% \rangle+\psi_{010}\left|010\right\rangle+\psi_{011}\left|011\right\rangle (4.1)
+ψ100⁢|100⟩+ψ101⁢|101⟩+ψ110⁢|110⟩+ψ111⁢|111⟩,\displaystyle+\,\psi_{100}\left|100\right\rangle+\psi_{101}\left|101\right% \rangle+\psi_{110}\left|110\right\rangle+\psi_{111}\left|111\right\rangle,

where ψi⁢j⁢k∈[−1,1]\psi_{ijk}\in[-1,1] and the squares of these amplitudes again sum to one, i.e.,

ψ0002+ψ0012+ψ0102+ψ0112+ψ1002+ψ1012+ψ1102+ψ1112=1.\displaystyle\psi_{000}^{2}+\psi_{001}^{2}+\psi_{010}^{2}+\psi_{011}^{2}+\psi_% {100}^{2}+\psi_{101}^{2}+\psi_{110}^{2}+\psi_{111}^{2}=1.

Note that, in total, there are 8=238=2^{3} amplitudes, one for each bitstring of three bits. We can also think of |ψ⟩\left|\psi\right\rangle as a vector with eight entries:

|ψ⟩=(ψ000ψ001ψ010ψ011ψ100ψ101ψ110ψ111).\displaystyle\left|\psi\right\rangle=\begin{pmatrix}\psi_{000}\\ \psi_{001}\\ \psi_{010}\\ \psi_{011}\\ \psi_{100}\\ \psi_{101}\\ \psi_{110}\\ \psi_{111}\end{pmatrix}.

More generally, a state of nn qubits can be specified by using 2n2^{n} amplitudes ψa1,…,an\psi_{a_{1},\dots,a_{n}}, one for each bitstring of nn bits:

|ψ⟩=ψ00⁢…⁢00⁢|00⁢…⁢00⟩+ψ00⁢…⁢01⁢|00⁢…⁢01⟩+…+ψ11⁢…⁢11⁢|11⁢…⁢11⟩\displaystyle\left|\psi\right\rangle=\psi_{00\dots 00}\left|00\dots 00\right% \rangle+\psi_{00\dots 01}\left|00\dots 01\right\rangle+\ldots+\psi_{11\dots 11% }\left|11\dots 11\right\rangle (4.2)

Again, each amplitude ψa1,…,an\psi_{a_{1},\dots,a_{n}} should be in [−1,1][-1,1] and their squares should sum to one:

ψ00⁢…⁢002+ψ00⁢…⁢012+…+ψ11⁢…⁢112=1\displaystyle\psi_{00\dots 00}^{2}+\psi_{00\dots 01}^{2}+\ldots+\psi_{11\dots 1% 1}^{2}=1 (4.3)

If some of the amplitudes ψa1,…,an\psi_{a_{1},\dots,a_{n}} are zero, we can simply leave them out. For example, the five-qubit quantum state

12⁢(|00000⟩+|11111⟩)\displaystyle\frac{1}{\sqrt{2}}\left(\left|00000\right\rangle+\left|11111% \right\rangle\right)

has 32 amplitudes, of which 30 are zero.

Since there are 2n2^{n} amplitudes, we can also think of |ψ⟩\left|\psi\right\rangle as a vector in a 2n2^{n}-dimensional space. What does Eq. 4.3 mean geometrically? For a single qubit, we saw in Section 2.1.2 that the states correspond to points on the unit circle, i.e., two-dimensional vectors of length one. By the Pythagorean theorem, it is true in any dimension that the sum of squares of a vector’s all entries is the square of its length. Thus, Eq. 4.3 means geometrically that |ψ⟩\left|\psi\right\rangle corresponds to a vector of length one or a unit vector in a 2n2^{n}-dimensional space.

Note that the number of amplitudes grows very rapidly with the number of qubits. This explains why it quickly becomes impossible to directly store quantum states on a classical computer. For example, to represent a quantum state of n=300n=300 qubits, one would need more amplitudes than there are atoms in the observable universe! Because of this, you cannot have more than 10 qubits in Quirky because we don’t want your web browser to run out of memory!

As in Eq. 3.50, we can use the tensor product “⊗\otimes” to combine quantum states on any number of qubits. If we have two basis states, we define their tensor product simply by concatenating the bitstrings. Generalizing the two-qubit case from Eq. 3.49,

|a1,…,an⟩⊗|b1,…,bm⟩=|a1,…,an,b1,…,bm⟩.\displaystyle\left|a_{1},\dots,a_{n}\right\rangle\otimes\left|b_{1},\dots,b_{m% }\right\rangle=\left|a_{1},\dots,a_{n},b_{1},\dots,b_{m}\right\rangle. (4.4)

For example,

|101⟩⊗|01⟩=|10101⟩.\displaystyle\left|101\right\rangle\otimes\left|01\right\rangle=\left|10101% \right\rangle.

In general, if |α⟩\left|\alpha\right\rangle is an arbitrary quantum state of nn qubits and |β⟩\left|\beta\right\rangle is an arbitrary quantum state of mm qubits, then their tensor product or combined state is a state of n+mn+m qubits. To compute this state we simply “multiply out” by using the distributivity law and then apply Eq. 4.4 for each term. For example, the tensor product of two maximally entangled states is the following:

|Φ+⟩⊗|Φ+⟩\displaystyle\quad\left|\Phi^{+}\right\rangle\otimes\left|\Phi^{+}\right\rangle
=(12⁢|00⟩+12⁢|11⟩)⊗(12⁢|00⟩+12⁢|11⟩)\displaystyle=\left(\frac{1}{\sqrt{2}}\left|00\right\rangle+\frac{1}{\sqrt{2}}% \left|11\right\rangle\right)\otimes\left(\frac{1}{\sqrt{2}}\left|00\right% \rangle+\frac{1}{\sqrt{2}}\left|11\right\rangle\right)
=12⁢12⁢|00⟩⊗|00⟩+12⁢12⁢|00⟩⊗|11⟩+12⁢12⁢|11⟩⊗|00⟩+12⁢12⁢|11⟩⊗|11⟩\displaystyle=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\left|00\right\rangle\otimes% \left|00\right\rangle+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\left|00\right% \rangle\otimes\left|11\right\rangle+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\left|% 11\right\rangle\otimes\left|00\right\rangle+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2% }}\left|11\right\rangle\otimes\left|11\right\rangle
=12⁢|0000⟩+12⁢|0011⟩+12⁢|1100⟩+12⁢|1111⟩.\displaystyle=\frac{1}{2}\left|0000\right\rangle+\frac{1}{2}\left|0011\right% \rangle+\frac{1}{2}\left|1100\right\rangle+\frac{1}{2}\left|1111\right\rangle.
Exercise 4.1 (Tensoring Bell states).

Compute the tensor product |Φ−⟩⊗|Ψ−⟩\left|\Phi^{-}\right\rangle\otimes\left|\Psi^{-}\right\rangle of the two Bell states in Eqs. 3.70 and 3.72.

Solution.
|Φ−⟩⊗|Ψ−⟩\displaystyle\quad\left|\Phi^{-}\right\rangle\otimes\left|\Psi^{-}\right\rangle
=(12⁢|00⟩−12⁢|11⟩)⊗(12⁢|01⟩−12⁢|10⟩)\displaystyle=\left(\frac{1}{\sqrt{2}}\left|00\right\rangle-\frac{1}{\sqrt{2}}% \left|11\right\rangle\right)\otimes\left(\frac{1}{\sqrt{2}}\left|01\right% \rangle-\frac{1}{\sqrt{2}}\left|10\right\rangle\right)
=12⁢|0001⟩−12⁢|0010⟩−12⁢|1101⟩+12⁢|1110⟩.\displaystyle=\frac{1}{2}\left|0001\right\rangle-\frac{1}{2}\left|0010\right% \rangle-\frac{1}{2}\left|1101\right\rangle+\frac{1}{2}\left|1110\right\rangle.