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Qubits

We're all familiar with classical computers where information is represented in bits. Bits can be either 00 or 11.

Having many of these 00s and 11s allows us to represent more complex things like numbers, text, images, and videos.

In quantum computing, we represent information using qubits (pronounced kew-bits). It's just a mashup of the word "quantum" and the word "bit".

Qubits also have the states 00 and 11 but they can exist in a superposition of the two.

A superposition is a fundamental concept from the world of quantum mechanics where a system can exist in a state such that, upon observation, it has some probability to be obvserved in any of the possible states. This is exactly the same concept talked about in the famous Schrödinger's cat thought experiment. Just as the cat exists in a superposition of being dead and alive, a qubit can exist in a superposition of being00 and 11.

We'll start to make use of Bra-ket notation by writing our qubit states 00 and 11 as ∣0⟩\ket{0} and ∣1⟩\ket{1}. So we are representing our qubits as vectors. In fact, these two vectors will form an orthonormal basis:

⟨0∣0⟩=1,  ⟨1∣1⟩=1,  ⟨0∣1⟩=0\braket{0|0} = 1,\ \ \braket{1|1} = 1,\ \ \braket{0|1} = 0

We refer to ∣0⟩\ket{0} and ∣1⟩\ket{1} as the computational basis states.

A superposition will be represented like this: a∣0⟩+b∣1⟩a\ket{0} + b\ket{1}where we are multiplying the vectors ∣0⟩\ket{0} and ∣1⟩\ket{1} by the complex numbers aa and bb respectively. And so our set of possible qubit states forms a vector space.

When we observe (or measure) a qubit in the computational basis, we will get either ∣0⟩\ket{0} or ∣1⟩\ket{1}. Not both, or none, we will always get one of the two. This phenonmenon is called collapse and represents how a quantum system breaks superposition when a system is observed.

We can calculate the exact probability of measuring each possible outcome using the complex coefficients measured earlier. The probability of measuring ∣0⟩\ket{0} is ∣a∣2|a|^2 and likewise the probability of measuring ∣1⟩\ket{1} is ∣b∣2|b|^2. We know from basic mathematics that probabilities must sum to 1, so we can see the constraint on the coefficients in the superposition:∣a∣2+∣b∣2=1|a|^2 + |b|^2 = 1

We can make this bit more mathematical by saying the probability of getting result ∣a⟩\ket{a} when measuring a qubit in state ∣ψ⟩\ket{\psi} is given by:

P(a∣ψ)=∣⟨a∣ψ⟩∣2P(a|\psi) = |\braket{a|\psi}|^2

This definition is a lot more powerful, because it allows us to measure the probability of measurement in any state, not just the computational basis states. What it means to measure in a different basis we'll get to later.

One last important result we'll state on these probabilities is that if we have ∣ψ⟩=a∣0⟩+b∣1⟩\ket{\psi} = a\ket{0} + b\ket{1}:

⟨ψ∣ψ⟩=(a∗⟨0∣+b∗⟨1∣)(a∣0⟩+b∣1⟩)=a∗a⟨0∣0⟩+a∗b⟨0∣1⟩+b∗a⟨1∣0⟩+b∗b⟨1∣1⟩=a∗a+b∗b=∣a∣2+∣b∣2=1\braket{\psi|\psi} = (a^*\bra{0} + b^*\bra{1})(a\ket{0} + b\ket{1}) \\[3ex] = a^*a\braket{0|0} + a^*b\braket{0|1} + b^*a\braket{1|0} + b^*b\braket{1|1} \\[3ex] = a^*a + b^*b \\[3ex] = |a|^2 + |b|^2 \\[3ex] = 1

This is the normalisation property and is the equivalent of saying measurement probabilities must sum to 1. If ⟨ψ∣ψ⟩≠1\braket{\psi|\psi} \neq 1 then we do not have a valid qubit state.

Visualising a qubit

We have a useful tool for visualising the qubit called the Bloch sphere (named after the physicist Felix Bloch).

The Bloch sphere places the state ∣0⟩\ket{0} at the north pole, and the state ∣1⟩\ket{1} at the south pole. Any superposition states lie somewhere in between and the latitude represents the probability of measuring either ∣0⟩\ket{0} or ∣1⟩\ket{1}. For example, a qubit that lies on the equator has an exact 50/50 chance.

Some example states are shown below:

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∣0⟩\ket{0}
bloch-sphere-1
∣1⟩\ket{1}
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12∣0⟩+12∣1⟩\frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1}

As you can see, the state 12∣0⟩+12∣1⟩\frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1} lies on the equator as there is a probability (12)2=1/2(\frac{1}{\sqrt{2}})^2 = 1/2 to measure ∣0⟩\ket{0} and likewise to measure ∣1⟩\ket{1}.

A qubit can exist in any state on the Bloch sphere and you can see there are a few other states labelled on the diagram. These labellings are common practice and are mathematically defined as:

∣+⟩=12∣0⟩+12∣1⟩∣−⟩=12∣0⟩−12∣1⟩∣R⟩=12∣0⟩+i2∣1⟩∣L⟩=12∣0⟩−i2∣1⟩\ket{+} = \frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1} \\[3ex] \ket{-} = \frac{1}{\sqrt{2}}\ket{0} - \frac{1}{\sqrt{2}}\ket{1} \\[3ex] \ket{R} = \frac{1}{\sqrt{2}}\ket{0} + \frac{i}{\sqrt{2}}\ket{1} \\[3ex] \ket{L} = \frac{1}{\sqrt{2}}\ket{0} - \frac{i}{\sqrt{2}}\ket{1}

These states seem fairly arbitrary for now, but we will see how they might arise later in the course.

A key point to note about the bloch sphere is that it can't uniquely represent every possible choice of co-efficients of ∣0⟩\ket{0} and ∣1⟩\ket{1}. To see this we'll first have to show how to calculate a qubit's point on the bloch sphere.

Just like polar coordinates on earth where we use latitude and longitude, any position on a sphere can be represented with just two angles. The angle θ\theta represents the angle down from the "north pole" (∣0⟩\ket{0}). e.g for state ∣1⟩\ket{1} we have θ=π\theta = \pi. The second angle ϕ\phi represents the rotation around the "equator". e.g. for state ∣+⟩\ket{+} we have ϕ=0\phi = 0 and for state ∣−⟩\ket{-} we have ϕ=π\phi = \pi. For convenience, we constrain 0≤θ≤π0 \leq \theta \leq \pi and 0≤ϕ<2π0 \leq \phi < 2\pi .

The actual position of the qubit on the bloch sphere will be called the bloch vector and with a little trigonometry, we find it is given by:

r=(cos⁡ϕsin⁡θ,sin⁡ϕsin⁡θ,cos⁡θ)\mathbf{r} = (\cos{\phi}\sin{\theta},\sin{\phi}\sin{\theta},\cos{\theta})

where ∣0⟩\ket{0} is at (0,0,1)(0, 0, 1), ∣+⟩\ket{+} is at (1,0,0)(1,0,0) etc.

Given these constraints we can see the following equation shows us how we can take a point on the bloch sphere back to the qubit state:

∣ψ⟩=cos⁡θ2∣0⟩+eiϕsin⁡θ2∣1⟩\ket{\psi} = \cos{\frac{\theta}{2}}\ket{0} + e^{i\phi}\sin{\frac{\theta}{2}}\ket{1}

Try plugging in a few values of θ\theta and ϕ\phi to convince yourself this works.

But hang on...

The coefficient of ∣0⟩\ket{0} is cos⁡θ2\cos{\frac{\theta}{2}} which, since our angle θ\theta is always real, is a real number. But we said coefficients could be complex numbers here, so how do we plot a qubit with a complex coefficient of ∣0⟩\ket{0} on our bloch sphere?

To anwer this question we'll talk about something called global phase.

Global Phase

We already saw earlier that we have a nice way to say mathematically the probability of measuring a state ∣ψ⟩\ket{\psi} in some basis ∣a⟩\ket{a} as ∣⟨a∣ψ⟩∣2|\braket{a|\psi}|^2. We want to see what happens to this probability when we multiply through by a "global phase".

A global phase, just means multiplying the state by a complex number of magnitude 1. Thanks to the polar representation of complex numbers , we have a very easy way to represent such a number: eiθe^{i\theta}.

First we need to check whether we're allowed to do this. If we multply some state ∣ψ⟩\ket{\psi} by eiθe^{i\theta} to get a new state ∣ψ′⟩\ket{\psi^\prime}, is this new state normalisable? Let's check:

⟨ψ′∣ψ′⟩=(e−iθ⟨ψ∣)(eiθ∣ψ⟩)=e−iθeiθ⟨ψ∣ψ⟩=⟨ψ∣ψ⟩\braket{\psi^\prime|\psi^\prime} = (e^{-i\theta}\bra{\psi})(e^{i\theta}\ket{\psi}) = e^{-i\theta}e^{i\theta}\braket{\psi|\psi} = \braket{\psi|\psi}

So as long as our original state ∣ψ⟩\ket{\psi} is normalisable we can multiply by a global phase.

Now let's see how this global phase affects measurement of some basis ∣a⟩\ket{a}:

P(a∣ψ′)=∣⟨a∣ψ′⟩∣2=∣⟨a∣eiθ∣ψ⟩∣2=∣eiθ⟨a∣ψ⟩∣2=(∣eiθ∣∣⟨a∣ψ⟩∣)2=∣⟨a∣ψ⟩∣2=P(a∣ψ)P(a|\psi^\prime) = |\braket{a|\psi^\prime}|^2 = |\bra{a}e^{i\theta}\ket{\psi}|^2 = |e^{i\theta}\braket{a|\psi}|^2 = (|e^{i\theta}||\braket{a|\psi}|)^2 = |\braket{a|\psi}|^2 = P(a|\psi)

These two results are quite interesting. They show us that multiplying a qubit by a global phase doesn't change the probability of measuring it in any basis. This is a very important property of qubits, and it means that the global phase of a qubit is unobservable. We would say that the two states ∣ψ⟩\ket{\psi} and ∣ψ′⟩\ket{\psi^\prime} are physically equivalent.

Ok, so now we're ready to answer our original question. How do we plot our qubit with a complex coefficient of ∣0⟩\ket{0} on the bloch sphere? Let's specify our coefficients in polar form:

∣ψ⟩=r0eiθ0∣0⟩+r1eiθ1∣1⟩\ket{\psi} = r_{0}e^{i\theta_0}\ket{0} + r_{1}e^{i\theta_1}\ket{1}

Now that we know we can multiply by global phase, let's multiply through by e−iθ0e^{-i\theta_0}:

∣ψ′⟩=e−iθ0∣ψ⟩=r0∣0⟩+r1ei(θ1−θ0)∣1⟩\ket{\psi^\prime} = e^{-i\theta_0}\ket{\psi} = r_{0}\ket{0} + r_{1}e^{i(\theta_1 - \theta_0)}\ket{1}

So we've managed to make the coefficient of ∣0⟩\ket{0} real without physically changing the qubit.

We can then match the parameters to the bloch sphere equation to get the angles:

∣ψ′⟩=r0∣0⟩+r1ei(θ1−θ0)∣1⟩cos⁡θ2+eiϕsin⁡θ2∴ϕ=θ1−θ0cos⁡θ2=r0sin⁡θ2=r1∴tan⁡θ2=r1r0θ=2tan⁡−1r1r0\ket{\psi^\prime} = r_{0}\ket{0} + r_{1}e^{i(\theta_1 - \theta_0)}\ket{1} \\[3ex] \cos{\frac{\theta}{2}} + e^{i\phi}\sin{\frac{\theta}{2}} \\[3ex] \therefore \phi = \theta_1 - \theta_0 \\[3ex] \cos{\frac{\theta}{2}} = r_{0} \\[3ex] \sin{\frac{\theta}{2}} = r_{1} \\[3ex] \therefore \tan{\frac{\theta}{2}} = \frac{r_{1}}{r_{0}} \\[3ex] \theta = 2\tan^{-1}{\frac{r_{1}}{r_{0}}}

So we can see that we can plot any qubit state on the bloch sphere, even if the coefficient of ∣0⟩\ket{0} is complex.

The process we've gone through here is a very important one and extends well beyond the bloch sphere. There will be many applications in the future when it will be useful for us to multiply by some global phase in order to make a coefficient real. Since multiplying by a global phase represents a rotation on the complex plane, it is always possible to make any given coefficient real through this process.

Exercises

Exercise 1

Calculate the probability of measuring ∣R⟩\ket{R} when in the state ∣+⟩\ket{+}.

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Exercise 2

Without further calculation, by observation of the bloch sphere calculate some of the other observation probabilities for different states. e.g. the probability of measuring ∣0⟩\ket{0} when in the state ∣L⟩\ket{L}, the probability of measuring ∣0⟩\ket{0} when in the state ∣−⟩\ket{-} etc.

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Exercise 3

Show that the state∣ψ⟩=i∣0⟩+2∣1⟩\ket{\psi} = i\ket{0} + 2\ket{1} is not normalised.

Then normalise it. That means, find a factor that we can multiply by to meet to the normalisation requirement.

Find the position of this normalised state on the bloch sphere (the angles θ\theta and ϕ\phi).

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