# Dead or Alive: Can you be both?

Difficulty

by Jérémy Ribeiro

At the heart of Quantum Mechanics lies quantum superposition. This strange phenomenon is often described as the capacity of a quantum system to be in multiple incompatible states at the same time. The most famous example of this is Schrödinger’s cat, which would be both dead and alive at the same time. But how can this be? How can we humanly make sense of that apparent contradiction? Well… I think we cannot! More precisely, I think there is a problem of language in here. Exactly what a quantum scientist means by being “in superposition”, I think, is quite far from what the layman has in mind.

## A simple analogy

To start explaining what a quantum scientist has in mind when he/she says that a state is in superposition I will use a simple analogy: Shapes.

What? How is that related to the topic?

You’ll see! How would you describe or draw a shape that is both a disk and a rectangle?

That does not make any sense! Maybe something like this:

Yeah you see, it does not make sense to you, and you struggle to draw anything because I said something that does not make sense. This is exactly what happens when someone says that Schrödinger’s cat is both dead and alive! It is not clear what he/she means, and stated like that it is non-sense. When a quantum scientist says that a physical system is in superposition of two states (dead and alive), he/she means that it is in a state that is neither the first (dead) nor the second (alive) but it is in another state that possesses some of the characteristics of both (dead and alive).

Hmm…This is quite hard to visualize for me. Don’t you have an example?

Yes! For the example of the shape it could look like this:

Oh I see!

This is a relatively good analogy. This shape is neither a rectangle nor a disk, however it has some of the properties of both. Moreover I like this analogy because in quantum mechanics you cannot “see” the quantum state the physical system is in. In other words, if someone gives you a system in a certain unknown state, you cannot learn the state. If you try to measure it, you will only see a “projection” of it…

A what?

A projection is somehow a shadow, like in the picture above. It is as if we could not see the object itself but only the shadows. And you see the shape of the shadows?

Oh! They are a rectangle and a disk!

Note that we are not obliged to project the light on the object in this manner, we could use another angle to project the light, and we would get other shadows:

It is the same in quantum mechanics: It would correspond to changing the measurement you perform on the system.

However there is a very important point that this analogy does not capture. In quantum mechanics, when we measure the system we disturb it. Therefore, in general, the state after measurement is not the same as the one before. On the contrary, the shape in the pictures does not change after we shine a light it on. Here is the limit of the analogy.

To go beyond this simple and limited analogy, we will have to learn the quantum language, which we will do…

Wait! That sounds terribly complicated!

Well, do you know what an arrow is?

Like in Robin Hood?

Yeah, well no. More like an arrow you can draw!

Like this?

Yes! You see, you already speak quantum!

I am quite skeptical…

## Let’s start with the real thing. The quantum language of arrows.

The quantum language is formulated in term of arrows! More precisely, the state of a quantum system is represented by an arrow. The mathematical term for these arrows is “unitary vectors of a Hilbert space (usually of finite dimension)”.

You theorists are such Barbarians!

Well yeah… I mean… Whatever… Each of these words is important, but basically this “barbarian” expression means “arrow of length 1”. So in more precise terms, a quantum state is (represented by) an arrow of length 1. This length restriction is here because the state is related with probabilities – as we will see in a moment – and probabilities add up to 1.

To simplify I will only talk about two-level systems. These are the simplest quantum systems you can find. They correspond to systems that have at most two possible outcomes when measured. For example for the spin of an electron, when we measure it, we can only measure that the spin is up or down, nothing more. We call such systems “qubits” (short for quantum bits).

Instead of using “dead” or “alive”, “up” or “down”, I will use 0 and 1 by analogy to the values that can take a bit in computer science. To specify that I am talking about arrows I will write a state as follows: |name of the state>. For example $|0\rangle$ is a state, $|1\rangle$ is a state.

In this picture we see the state $|0\rangle$ and $|1\rangle$ and another state in between called V. Because the set of all states is the set of arrows of length 1, to every point of the blue circle (of radius 1) corresponds the ending of an arrow, and therefore each of these points corresponds to an arrow. The states $|0\rangle$ and $|1\rangle$ are orthogonal, i.e. they form a right angle. Having a right angle like this means, in the language of quantum mechanics, to be “incompatible”, while being in between two “incompatible” arrows is said to be in superposition of those arrows. To understand why we need to understand how quantum measurements work.

We first need to choose a measurement, that’s to say we need to choose what we want to measure. For example, if you wanted to “measure” a person, you need to decide whether you want to measure his/her height or his/her weight. In quantum mechanics the possible measurements are determined by the pairs of orthogonal vectors, in other words to choose a measurement we need to choose two arrows that form a right angle. We then say that we measure in the basis formed by these vectors. For example one measurement is characterized by the arrows $|0\rangle$ and $|1\rangle$ , so let’s describe in more details this particular measurement.

Let say that we want to measure the arrow $|V\rangle$ that is between $|0\rangle$ and $|1\rangle$ (see image above). What we do is that we look at the projection of $|V\rangle$ on the two arrow that represent the measurement…

Whaaaattt?

Again, the projection is like a shadow, but let me draw it:

You see here we draw a green line parallel to the arrow $|0\rangle$ from the tip of the arrow $|V\rangle$, and a red line parallel to the arrow $|1\rangle$ from the tip of the arrow $|V\rangle$. This allows us to get two new but shorter arrows, the green and the red. We say that the green arrow is the projection of the arrow $|V\rangle$ on the arrow $|1\rangle$ , and the red arrow is the projection of $|V\rangle$ on $|0\rangle$ . It looks a bit like for the 3D object from the last section, where the arrow $|V\rangle$ plays the role of the 3D object, the dashed lines are the rays of light, the projection are the shadows, and the basis of the measurement (here $|0\rangle$ and $|1\rangle$ ) are the walls.

One of the major characteristics of quantum mechanics is that the outcome of a measurement is random. What happens during the measurement is that the arrow $|V\rangle$ will randomly become either the arrow $|0\rangle$ or the arrow $|1\rangle$ with probability given by the length of the projections (in fact the square of their length): The length of the red arrow determines the probability that the arrow $|V\rangle$ becomes the arrow $|0\rangle$ after the measurement, while the length of the green arrow determines the probability that the arrow $|V\rangle$ becomes the arrow $|1\rangle$ after the measurement. When the arrow $|V\rangle$ becomes $|0\rangle$ we get outcome 0, and when it becomes $|1\rangle$ we get 1. You can also see the big difference with the shapes of the previous section, here after the measurement the state is not the same as before. When you get outcome 0, the state after measurement is the arrow $|0\rangle$. On the contrary if you project light on the weird 3D shape above, it does not suddenly become a circle or rectangle.

Hmm, that’s a lot to process… And why is it random?

It is true that this is a lot, but try to play with different arrows, and think about the analogy of the shadows. After a while you will get a good grasp on how to get the probability of getting outcome 0 or 1 out of a measurement. For the randomness the language of arrows does not tell us why, or how the projection process works.

Hmm, ok.
And what happens if I measure the arrow $|0\rangle$ in the basis formed by the arrows $|0\rangle$ and $|1\rangle$?

Imagine you rotate the arrow $|V\rangle$ so that it becomes really close to the arrow $|0\rangle$. You see that the closer to the arrow $|0\rangle$ the bigger the red arrow becomes. And when the arrow $|V\rangle$ touches the arrow $|0\rangle$, then the red arrow becomes the arrow $|0\rangle$ itself, and has therefore length 1. On the other the closer the arrow $|V\rangle$ is to $|0\rangle$ the smaller the green arrow is. When it touches the arrow $|0\rangle$ the green arrow has length 0. This means that if you measure the arrow $|0\rangle$ in the basis formed by the arrows $|0\rangle$ and $|1\rangle$ , then with probability 1 the arrow $|0\rangle$ will stay $|0\rangle$ and you’ll get outcome 0, and you will never get outcome 1.

That is why the arrows $|0\rangle$ and $|1\rangle$ are “incompatible” or “mutually exclusive”, if you try to measure $|0\rangle$ (in the $|0\rangle$$|1\rangle$ basis) you will always get 0 and never 1, and on the contrary if you try to measure $|1\rangle$ you will always get 1 and never 0. For any other arrow in the middle, like $|V\rangle$ for example, you can get 0 or 1 with non-zero probability! That’s somehow why we say it is in superposition, it is not completely “incompatible” with either of the two basis arrows, it contains some properties of $|0\rangle$ and some of $|1\rangle$ , it has “both shadows”.

So remember that, while you probably cannot be both dead and alive at the same time, an arrow might be able to be somewhere in between!

Jérémy Ribeiro is a theorist at QuTech. He specialises in quantum cryptography: the use of quantum principles to design secure protocols for communication and other cryptographic tasks. In his free time (and sometimes during work hours) he enjoys practicing his Yo-yo skills and talking about open source software and undecidable problems.

## 3 thoughts on “Dead or Alive: Can you be both?”

1. shihan. says:

great piece. thanks.

2. wambui julien wanjiru says:

I don’t think that makes sense either…..it’s impossible to be dead and alive at the same time….

3. Entangled Guy says:

Great post! Such a nice analogy…