Writing Down Qubit States

In [1]:
from qiskit import *

In the previous chapter we saw that there are multiple ways to extract an output from a qubit. The two methods we've used so far are the z and x measurements.

In [2]:
# z measurement of qubit 0
measure_z = QuantumCircuit(1,1)
measure_z.measure(0,0);

# x measurement of qubit 0
measure_x = QuantumCircuit(1,1)
measure_x.h(0)
measure_x.measure(0,0);

Sometimes these measurements give results with certainty. Sometimes their outputs are random. This all depends on which of the infinitely many possible states our qubit is in. We therefore need a way to write down these states and figure out what outputs they'll give. For this we need some notation, and we need some math.

The z basis

If you do nothing in a circuit but a measurement, you are certain to get the outcome 0. This is because the qubits always start in a particular state, whose defining property is that it is certain to output a 0 for a z measurement.

We need a name for this state. Let's be unimaginative and call it $0$ . Similarly, there exists a qubit state that is certain to output a 1. We'll call this $1$.

These two states are completely mutually exclusive. Either the qubit definitely outputs a 0, or it definitely outputs a 1. There is no overlap.

One way to represent this with mathematics is to use two orthogonal vectors.

$$ |0\rangle = \begin{pmatrix} 1 \\\\\\ 0 \end{pmatrix} \, \, \, \, |1\rangle =\begin{pmatrix} 0 \\\\\\ 1 \end{pmatrix}. $$

This is a lot of notation to take in all at once. First let's unpack the weird $|$ and $\rangle$ . Their job is essentially just to remind us that we are talking about the vectors that represent qubit states labelled $0$ and $1$. This helps us distinguish them from things like the bit values 0 and 1 or the numbers 0 and 1. It is part of the bra-ket notation, introduced by Dirac.

If you are not familiar with vectors, you can essentially just think of them as lists of numbers which we manipulate using certain rules. If you are familiar with vectors from your high school physics classes, you'll know that these rules make vectors well-suited for describing quantities with a magnitude and a direction. For example, velocity of an object is described perfectly with a vector. However, the way we use vectors for quantum states is slightly different to this. So don't hold on too hard to your previous intuition. It's time to do something new!

In the example above, we wrote the vector as a vertical list of numbers. We call these column vectors. In Dirac notation, they are also called kets.

Horizontal lists are called row vectors. In Dirac notation they are bras. They are represented with a $\langle$ and a $|$.

$$ \langle 0| = \begin{pmatrix} 1 & 0\end{pmatrix} \, \, \, \, \langle 1| =\begin{pmatrix} 0 & 1 \end{pmatrix}. $$

The rules on how to manipulate vectors define what it means to add or multiply them. For example, to add two vectors we need them to be the same type (either both column vectors, or both row vectors) and the same length. Then we add each element in one list to the corresponding element in the other. For a couple of arbitrary vectors that we'll call $a$ and $b$, this works as follows.

$$ \begin{pmatrix} a_0 \\\\ a_1 \end{pmatrix} +\begin{pmatrix} b_0 \\\\ b_1 \end{pmatrix}=\begin{pmatrix} a_0+b_0 \\\\ a_1+b_1 \end{pmatrix}. $$

To multiple a vector by a number, we simply multiply every element in the list by that number:

$$ x \times\begin{pmatrix} a_0 \\\\ a_1 \end{pmatrix} = \begin{pmatrix} x \times a_0 \\\\ x \times a_1 \end{pmatrix} $$

Multiplying a vector with another vector is a bit more tricky, since there are multiple ways we can do it. One is called the 'inner product', and works as follows.

$$ \begin{pmatrix} a_0 & a_1 \end{pmatrix} \begin{pmatrix} b_0 \\\\ b_1 \end{pmatrix}= a_0~b_0 + a_1~b_1. $$

Note that the right hand side of this equation contains only normal numbers being multipled and added in a normal way. The inner product of two vectors therefore yields just a number. As we'll see, we can interpret this as a measure of how similar the vectors are.

The inner product requires the first vector to be a bra and the second to be a ket. In fact, this is where their names come from. Dirac wanted to write the inner product as something like $\langle a | b \rangle$, which looks like the names of the vectors enclosed in brackets. Then he worked backwards to split the bracket into a bra and a ket.

If you try out the inner product on the vectors we already know, you'll find

$$ \langle 0 | 0\rangle = \langle 1 | 1\rangle = 1,\\\\ \langle 0 | 1\rangle = \langle 1 | 0\rangle = 0. $$

Here we are using a concise way of writing the inner products where, for example, $\langle 0 | 1 \rangle$ is the inner product of $\langle 0 |$ with $| 1 \rangle$. The top line shows us that the inner product of these states with themselves always gives a 1. When done with two orthogonal states, as on the bottom line, we get the outcome 0. These two properties will come in handy later on.

The x basis - part 1

So far we've looked at states for which the z measurement has a certain outcome. But there are also states for which the outcome of a z measurement is equally likely to be 0 or 1. What might these look like in the language of vectors?

A good place to start would be something like $|0\rangle + |1\rangle$ , since this includes both $|0\rangle$ and $|1\rangle$ with no particular bias towards either. But let's hedge our bets a little and multiply it by some number $x$ .

$$ x ~ (|0\rangle + |1\rangle) = \begin{pmatrix} x \\\\ x \end{pmatrix} $$

We can choose the value of $x$ to make sure that the state plays nicely in our calculations. For example, think about the inner product,

$$ \begin{pmatrix} x & x \end{pmatrix} \times \begin{pmatrix} x \\\\ x \end{pmatrix}= 2x^2. $$

We can get any value for the inner product that we want, just by choosing the appropriate value of $x$.

As mentioned earlier, we are going to use the inner product as a measure of how similar two vectors are. With this interpretation in mind, it is natural to require that the inner product of any state with itself gives the value $1$. This is already acheived for the inner products of $|0\rangle$ and $|1\rangle$ with themselves, so let's make it true for all other states too.

This condition is known as the normalization condition. In this case, it means that $x=\frac{1}{\sqrt{2}}$. Now we know what our new state is, so here's a few ways of writing it down.

$$ \begin{pmatrix} \frac{1}{\sqrt{2}} \\\\ \frac{1}{\sqrt{2}} \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\\\ 1 \end{pmatrix} = \frac{ |0\rangle + |1\rangle}{\sqrt{2}} $$

This state is essentially just $|0\rangle$ and $|1\rangle$ added together and then normalized, so we will give it a name to reflect that origin. We call it $|+\rangle$ .

The Born rule

Now we've got three states that we can write down as vectors. We can also calculate inner products for them. For example, the inner product of each with $\langle 0 |$ is

$$ \langle 0 | 0\rangle = 1 \\\\ \langle 0 | 1\rangle = 0 \\\\ \, \, \, \, \langle 0 | +\rangle = \frac{1}{\sqrt{2}}. $$

We also know the probabilities of getting various outcomes from a z measurement for these states. For example, let's use $p^z_0$ to denote the probability of the result 0 for a z measurement. The values this has for our three states are

$$ p_0^z( | 0\rangle) = 1,\\\\ p_0^z( | 1\rangle) = 0, \\\\ p_0^z( | +\rangle) = \frac{1}{2}. $$

As you might have noticed, there's a lot of similarlity between the numbers we get from the inner products and those we get for the probabilities. Specifically, the three probabilities can all be written as the square of the inner products:

$$ p_0^z(|a\rangle) = (~\langle0|a\rangle~)^2. $$

Here $|a\rangle$ represents any generic qubit state.

This property doesn't just hold for the 0 outcome. If we compare the inner products with $\langle 1 |$ with the probabilities of the 1 outcome, we find a similar relation.

$$ \\\\ p_1^z(|a\rangle) = (~\langle1|a\rangle~)^2. $$

The same also holds true for other types of measurement. All probabilities in quantum mechanics can be expressed in this way. It is known as the Born rule.

Global and relative phases

Vectors are how we use math to represent the state of a qubit. With them we can calculate the probabilities of all the possible things that could ever be measured. These probabilities are essentially all that is physically relevant about a qubit. It is by measuring them that we can determine or verify what state our qubits are in. Any aspect of the state that doesn't affect the probabilities is therefore just a mathematical curiosity.

Let's find an example. Consider a state that looks like this:

$$ |\tilde 0\rangle = \begin{pmatrix} -1 \\\\ 0 \end{pmatrix} = -|0\rangle. $$

This is equivalent to multiplying the state $|0\rangle$ by $-1$. It means that every inner product we could calculate with $|\tilde0\rangle$ is the same as for $|0\rangle$, but multplied by $-1$.

$$ \langle a|\tilde 0\rangle = -\langle a| 0\rangle $$

As you probably know, any negative number squares to the same value as its positive counterpart: $(-x)^2 =x^2$.

Since we square inner products to get probabilities, this means that any probability we could ever calculate for $|\tilde0\rangle$ will give us the same value as for $|0\rangle$. If the probabilities of everything are the same, there is no observable difference between $|\tilde0\rangle$ and $|0\rangle$; they are just different ways of representing the same state.

This is known as the irrelevance of the global phase. Quite simply, this means that multplying the whole of a quantum state by $-1$ gives us a state that will look different mathematically, but which is actually completely equivalent physically.

The same is not true if the phase is relative rather than global. This would mean multiplying only part of the state by $-1$ , for example:

$$ \begin{pmatrix} a_0 \\\\ a_1 \end{pmatrix} \rightarrow \begin{pmatrix} a_0 \\\\ -a_1 \end{pmatrix}. $$

Doing this with the $|+\rangle$ state gives us a new state. We'll call it $|-\rangle$.

$$ |-\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\\\ -1 \end{pmatrix} = \frac{ |0\rangle - |1\rangle}{\sqrt{2}} $$

The values $p_0^z$ and $p_1^z$ for $|-\rangle$ are the same as for $|+\rangle$. These two states are thus indistinguishable when we make only z measurements. But there are other ways to distinguish them. To see how, consider the inner product of $|+\rangle$ and $|-\rangle$.

$$ \langle-|+\rangle = \langle+|-\rangle = 0 $$

The inner product is 0, just as it is for $|0\rangle$ and $|1\rangle$. This means that the $|+\rangle$ and $|-\rangle$ states are orthogonal: they represent a pair of mutually exclusive possible ways for a qubit to be a qubit.

The x basis - part 2

Whenever we find a pair of orthogonal qubit states, we can use it to define a new kind of measurement.

First, let's apply this to the case we know well: the z measurement. This asks a qubit whether it is $|0\rangle$ or $|1\rangle$. If it is $|0\rangle$, we get the result 0. For $|1\rangle$ we get 1. Anything else, such as $|+\rangle$, is treated as a superposition of the two.

$$ |+\rangle = \frac{|0\rangle+|1\rangle}{\sqrt{2}}. $$

For a superposition, the qubit needs to randomly choose between the two possibilities according to the Born rule.

We can similarly define a measurement based on $|+\rangle$ and $|-\rangle$. This asks a qubit whether it is $|+\rangle$ or $|-\rangle$. If it is $|+\rangle$, we get the result 0. For $|-\rangle$ we get 1. Anything else is treated as a superposition of the two. This includes the states $|0\rangle$ and $|1\rangle$, which we can write as

$$ |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}}, \, \, \, \, |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}}. $$

For these, and any other superpositions of $|+\rangle$ and $|-\rangle$, the qubit chooses its outcome randomly with probabilities

$$ p_0^x(|a\rangle) = (~\langle+|a\rangle~)^2,\\\\ p_1^x(|a\rangle) = (~\langle-|a\rangle~)^2. $$

This is the x measurement.

The conservation of certainty

Qubits in quantum circuits always start out in the state $|0\rangle$. By applying different operations, we can make them explore other states.

Try this out yourself using a single qubit, creating circuits using operations from the following list, and then doing the x and z measurements in the way described at the top of the page.

In [3]:
qc = QuantumCircuit(1)

qc.h(0) # the hadamard

qc.x(0) # x gate

qc.y(0) # y gate

qc.z(0) # z gate

# for the following, replace theta by any number
theta = 3.14159/4
qc.ry(theta,0); # y axis rotation

You'll find examples where the z measurement gives a certain result, but the x is completely random. You'll also find examples where the opposite is true. Furthermore, there are many examples where both are partially random. With enough experimentation, you might even uncover the rule that underlies this behavior:

$$ (p^z_0-p^z_1)^2 + (p^x_0-p^x_1)^2 = 1. $$

This is a version of Heisenberg's famous uncertainty principle. The $(p^z_0-p^z_1)^2$ term measures how certain the qubit is about the outcome of a z measurement. The $(p^x_0-p^x_1)^2$ term measures the same for the x measurement. Their sum is the total certainty of the two combined. Given that this total always takes the same value, we find that the amount of information a qubit can be certain about is a limited and conserved resource.

Here is a program to calculate this total certainty. As you should see, whatever gates from the above list you choose to put in qc, the total certainty comes out as $1$ (or as near as possible given statistical noise).

In [4]:
shots = 2**14 # number of samples used for statistics

uncertainty = 0
for measure_circuit in [measure_z, measure_x]:

    # run the circuit with a the selected measurement and get the number of samples that output each bit value
    counts = execute(qc+measure_circuit,Aer.get_backend('qasm_simulator'),shots=shots).result().get_counts()

    # calculate the probabilities for each bit value
    probs = {}
    for output in ['0','1']:
        if output in counts:
            probs[output] = counts[output]/shots
        else:
            probs[output] = 0
            
    uncertainty += ( probs['0'] -  probs['1'] )**2

# print the total uncertainty
print('The total uncertainty is',uncertainty )
The total uncertainty is 0.9984774887561798

Now we have found this rule, let's try to break it! Then we can hope to get a deeper understanding of what is going on. We can do this by simply implementing the operation below, and then recalculating the total uncertainty.

In [5]:
# for the following, replace theta by any number
theta = 3.14159/2
qc.rx(theta,0); # x axis rotation

For a circuit with a single rx with $\theta=\pi/2$, we will find that $(p^z_0-p^z_1)^2 + (p^x_0-p^x_1)^2=0$. This operation seems to have reduced our total certainty to zero.

All is not lost, though. We simply need to perform another identical rx gate to our circuit to go back to obeying $(p^z_0-p^z_1)^2 + (p^x_0-p^x_1)^2=1$. This shows that the operation does not destroy our certainty; it simply moves it somewhere else and then back again. So let's find that somewhere else.

The y basis - part 1

There are infinitely many ways to measure a qubit, but the z and x measurements have a special relationship with each other. We say that they are mutually unbiased. This simply means that certainty for one implies complete randomness for the other.

At the end of the last section, it seemed that we were missing a piece of the puzzle. We need another type of measurement to plug the gap in our total certainty, and it makes sense to look for one that is also mutually unbiased with x and z.

The first step is to find a state that seems random to both x and z measurements. Let's call it $|\circlearrowleft\rangle$, for no apparent reason.

$$ |\circlearrowleft\rangle = c_0 | 0 \rangle + c_1 | 1 \rangle $$

Now the job is to find the right values for $c_0$ and $c_1$. You could try to do this with standard positive and negative numbers, but you'll never be able to find a state that is completely random for both x and z measurements. To achieve this, we need to use complex numbers.

Complex numbers

Hopefully you've come across complex numbers before, but here is a quick reminder.

Normal numbers, such as the ones we use for counting bananas, are known as real numbers. We cannot solve all possible equations using only real numbers. For example, there is no real number that serves as the square root of $-1$. To deal with this issue, we need more numbers, which we call complex numbers.

To define complex numbers we start by accepting the fact that $-1$ has a square root, and that its name is $i$. Any complex number can then be written

$$ x = x_r + i~x_i . $$

Here $x_r$ and $x_i$ are both normal numbers (positive or negative), where $x_r$ is known as the real part and $x_i$ as the imaginary part.

For every complex number $x$ there is a corresponding complex conjugate $x^*$

$$ x^* = x_r - i~x_i . $$

Multiplying $x$ by $x^*$ gives us a real number. It's most useful to write this as

$$ |x| = \sqrt{x~x^*}. $$

Here $|x|$ is known as the magnitude of $x$ (or, equivalently, of $x^*$ ).

If we are going to allow the numbers in our quantum states to be complex, we'll need to upgrade some of our equations.

First, we need to ensure that the inner product of a state with itself is always 1. To do this, the bra and ket versions of the same state must be defined as follows:

$$ |a\rangle = \begin{pmatrix} a_0 \\\\ a_1 \end{pmatrix}, ~~~ \langle a| = \begin{pmatrix} a_0^* & a_1^* \end{pmatrix}. $$

Then we just need a small change to the Born rule, where we square the magnitudes of inner products, rather than just the inner products themselves.

$$ p_0^z(|a\rangle) = |~\langle0|a\rangle~|^2,\\\\ p_1^z(|a\rangle) = |~\langle1|a\rangle~|^2,\\\\ p_0^x(|a\rangle) = |~\langle+|a\rangle~|^2,\\\\ p_1^x(|a\rangle) = |~\langle-|a\rangle~|^2. $$

The irrelevance of the global phase also needs an upgrade. Previously, we only talked about multiplying by -1. In fact, we can multiply a state by any complex number whose magnitude is 1. This will give us a state that will look different, but which is actually completely equivalent. This includes multiplying by $i$, $-i$ or infinitely many other possibilities.

The y basis - part 2

Now that we have complex numbers, we can define the following pair of states.

$$ |\circlearrowright\rangle = \frac{ | 0 \rangle + i | 1 \rangle}{\sqrt{2}}, ~~~~ |\circlearrowleft\rangle = \frac{ | 0 \rangle -i | 1 \rangle}{\sqrt{2}} $$

You can verify yourself that they both give random outputs for x and z measurements. They are also orthogonal to each other. They therefore define a new measurement, and that basis is mutally unbiased with x and z. This is the third and final fundamental measurement for a single qubit. We call it the y measurement, and can implement it with

In [6]:
# y measurement of qubit 0
measure_y = QuantumCircuit(1,1)
measure_y.sdg(0)
measure_y.h(0)
measure_y.measure(0,0);

With the x, y and z measurements, we now have everything covered. Whatever operations we apply, a single isolated qubit will always obey

$$ (p^z_0-p^z_1)^2 + (p^y_0-p^y_1)^2 + (p^x_0-p^x_1)^2 = 1. $$

To see this, we can incorporate the y measurement into our measure of total certainty.

In [7]:
shots = 2**14 # number of samples used for statistics

uncertainty = 0
for measure_circuit in [measure_z, measure_x, measure_y]:

    # run the circuit with a the selected measurement and get the number of samples that output each bit value
    counts = execute(qc+measure_circuit,Aer.get_backend('qasm_simulator'),shots=shots).result().get_counts()

    # calculate the probabilities for each bit value
    probs = {}
    for output in ['0','1']:
        if output in counts:
            probs[output] = counts[output]/shots
        else:
            probs[output] = 0
            
    uncertainty += ( probs['0'] -  probs['1'] )**2

# print the total uncertainty
print('The total uncertainty is',uncertainty )
The total uncertainty is 0.986548513174057

For more than one qubit, this relation will need another upgrade. This is because the qubits can spend their limited certainty on creating correlations that can only be detected when multiple qubits are measured. The fact that certainty is conserved remains true, but it can only be seen when looking at all the qubits together.

Before we move on to entanglement, there is more to explore about just a single qubit. As we'll see in the next section, the conservation of certainty leads to a particularly useful way of visualizing single-qubit states and gates.

In [ ]: