## Wednesday, 20 April 2016

### Fun measurements for pairs of qubits

This is part of a project to get people involved with quantum error correction. See here for more info.

If you have one qubit, you can do fun stuff. With two, you can do all that stuff twice. But you can also do more fun stuff, that's only possible with two. This can create entanglement: a special kind of correlation only possible for quantum objects. In this post we'll look at two qubit measurements that can create this valuable resource. Then in this post, we use them for teleportation.

You should probably familiarize yourself with one qubit before we move on to two. You can do that by checking out our post on the maths of qubits. Or you can also look at the summary at the beginning of the post with the teleportation.

Two qubits are better than one

Now we are up to speed with the maths of one qubit, lets move on to two.

Each qubit has two possible states: up and down. So two qubits have four possible states: both up, both down, the first one up and the second one down and vice-versa. Let's call these uu, dd, ud and du.

The state of a qubit can be any superposition of these four states. So it has what we'll call an uuness, a ddness, an udness and a duness. Some state T for two qubits can then be written

T = uuness x uu + udness x ud + duness x du + ddness x dd .

We can go through all the stuff about overlaps from the last post and find that, again

uunees2 + udness2 + duness2 + ddness2 = 1

And, again, these numbers turn out to be the probabilities that we get these results if we measure whether the two qubits are up or down.

We can also think of our two qubits as having the four states ll, lr, rl and rr, where l is left and r is right. Our state T can also be written using these.

T = llness x ll + lrness x lr + rlness x rl + rrness x rr

Describing two qubits as two qubits

Suppose we have a qubit, that we call qubit A. It is in a state we'll call SA. We also have qubit B in state SB.

SA = (upness of SA) x up + (downness of SA) x down
SB = (upness of SB) x up + (downness of SB) x down

What is the combined state of the two qubits? How do we combine the upness and downness of the two to get the uuness, etc?

Two options seem sensible. One is to add them.

uuness = upness of SA + upness of SB,   etc.

The other is to multiply them.

uuness = upness of SA x upness of SB,   etc.

We need only look at a one simple example to see which makes more sense. Suppose both qubits are in state up. So (upness of SA) = (upness of SA) = 1, and (downness of SA) = (downness of SA) = 0. It's clear then that the state of the two should be uu, because both are up. This means that the uuness should be 1, and the udness, duness and ddness should all be zero.

If we go by the adding method, we get

uuness = upness of SA + upness of SB = 2
udness = upness of SA + downness of SB = 1
duness = downness of SA + upness of SB = 1
ddness = downness of SA + downness of SB = 0

That's almost completely wrong. The uuness is too high, and the udness and duness are not zero when they should be. How about the multplying method?

uuness = upness of SA x upness of SB = 1
udness = upness of SA x downness of SB = 0
duness = downness of SA x upness of SB = 0
ddness = downness of SA x downness of SB = 0

This is completely right. So multiplying is definitely the way to do things.

If we want to measure our two qubits, we could ask them both if they are up or down. Or we could also then if they are left or right. Or we could ask one about up and down and the other about left or right. They are all valid options, but they are also pretty boring. Can we do something more interesting with two qubits than just a couple of single qubit measurements?

Measuring both the qubits tells us about both the qubits, obviously. But it also tells us
everything about them. Is there a way to measure the pair that only gives us a bit of information, and allows it to retain some air of mystery?

One possibility is to ask whether the state of the two qubits is the same, or different. But that's not yet a well defined question. Qubits can be thought of as being in the states up or down, or thought of as being in the states left and right. Which ones are we talking about?

First, let's use up and down. What we want is a measurement that tells us 'same' if it sees uu, dd or any superposition of them. And also, it should not give is any more information than that. For the state ud, du or any superposition thereof, we should similary get the result 'different'.

It's important to think about how a measurement like this might actually be possible. One way to do it is to get another qubit that is simply in the state up. Qubits like this are just to help us do other stuff are called ancillas. To do the measurement we get the first qubit to interact with the ancilla. This interaction should do nothing if the first qubit is up, but make the ancilla flip between the up and down states if the first qubit is down. Then we interact the second qubit with the ancilla in the same way.

If the pair of qubits is in the state uu, the interation won't have made the ancilla flip at all. If the state is dd, the ancilla will have flipped twice: from up to down, and then down to up again. Either way, the ancilla ends up in the state down. The same is true for a superposition of uu and dd.

If the pair of qubits is the state ud, du or some superposition, the interactions will have caused the ancilla to flip exactly once. This means it'll end up in the state down rather than up. So by measuring the ancilla and seeing if it is up or down, we learn about whether our pair of qubits had the same or different up/down states.

What if we have the more general state

T = (uuness x uu + ddness x dd) + (udness x ud + duness x du) ?

This is a superposition of a part where everything is the same, and a part where everything is different. When we make the measurement, it must decide which to be.

Suppose the qubits decide to be the same. The state after measurement would then just be the bits that are consistent with that1
uuness x uu + ddness x dd

and for different, the state would become

udness x ud + duness x du

We'll be using this measurement a lot in this post, so we'd probably be better off giving it a name. Let's call it a ZZ measurement. Why? Well, we won't get into that. But its a nice name, so let's give it a go.

Another interesting question we could ask of two qubits is to look at whether they are left or right, and tell us if they are the same or different. That would then just be the same as the above, but with left and right. We'll call this an XX measurement.

Complementary and commuting measurements

If we have a single qubit, the measurement of whether it is up or down, and the measurement of whether it is left or right, do not play well with each other.

The only way to have a state that we are sure is going to have the result up for an up/down measurement is one with upness2 = 1. The ony way to always get right for a right/left measurement is to have a state with a rightness2 = 1. There is no superposition we can write down for which both is true, so no state can be certain to be both up and right.

Also, suppose we have a qubit in state up. If we do an up/down measurement, we will always get the result up. If we then do a right/left measurement, we will randomly get the result right or left.

But suppose we did the right/left measurement first. We would again get a random result, and the state afterwards would be either left or right. Since both are a superposition of up and down, the following up/down measurement would also give a random result. So the results we get depend on the order in which we do the measurements.

Measurements like this, which block each others certainty and mess each others results up, are called complementary. It seems like an odd name, because they seem more likely to insult each other than complement. But that's nevertheless what they are called.

Some types of measurement do play together nicely. For example, with two qubits we could do an up/down measurement for both or we could do an ZZ measurement. Since both are based on whether qubit states are up or down, by preparing the states uu, ud, du or dd we can have definite outcomes for both measurements. Also, no matter what state we use, it won't matter which order we do them in.

Measurements that work like this are called commuting. This has nothing to do with how they travel to work. It's just a mathsy way of saying that the order doesn't matter.

ZZ and XX commute!

The ZZ measurement is built on a foundation of the up and down, and XX is built on a foundation of right and left. Since these are complementary, you'd probably expect ZZ and XX to be complementary too. But they aren't! They commute, and that commutation allows us to do some quantum magic.

If we want a state that is certain to come out with the answer 'same' when we make a ZZ measurement, its state will be

uuness x uu + ddness x dd

for some choice of the uuness and ddness.

For a state that is certain of getting the result 'same' for an XX measuremnt, the state will be

llness x ll + rrness x rr .

If I'm right that these measurements aren't complementary, there should exist a state that is certain of both. So there should be some choice of uuness, ddness, llness and rrness such that

uuness x uu + ddness x dd = llness x ll + rrness x rr .

Let's not do all the maths needed to solve this. Instead I'll just tell you the answer and we'll show that it's right. The state we need is the one with

uuness = ddness = llness = rrness = √½ .

Which means

½ x uu + √½ x dd = √½ x ll + √½ x rr = √½ x (uu + dd) = √½ x (ll + rr).

In this, the whole multplying by √½ thing turns out to be just something we need to do to make the probabiities work. The important thing is that

uu + dd = ll + rr .

To see if this is really true, let's rewrite the left hand side of the equation using the up and down states.

The state rr is the state of two qubits that are both in the right state. We need to write this as a superposition of ups and downs. The first step towards this is to remember how right itself can be written with up and down

right = √½ x up + √½ x down .

And we can also use what we found out earlier, about how uuness and so on can be calculated

uuness = upness of SA x upness of SB,   etc

With this we find that

uuness = √½ x √½ = ½,

and the same for udness, duness and ddness. So

rr = ½ x uu + ½ x ud + ½ x du + ½ x dd = ½ x (uu + ud + du + dd) .

Doing the same trick with ll means that we have to contend with a couple of minus signs, but in the end we get

ll = ½ x uu + (-½) x ud + (-½) x du + ½ x dd = ½ x (uu - ud - du + dd) .

So now lets add these up to see what rr + ll is

rr + ll = ½ x (uu + ud + du + dd + uu - ud - du + dd)
= ½ x ( 2 x uu + 2 x dd)
= uu + dd

See, its true! So the state

½ x uu + √½ x dd

Is exactly the same as the state

½ x ll + √½ x rr

They are just two different ways of writing the same thing. If we have this state S, and do an ZZ measurement, it'll look at the ups and downs and see they are the same. If we do an XX measurement, it'll also look at the lefts and rights and see they are the same. This state is sure of the answers to both. It's a special state which deserves a special name. We'll call it Φ, because Greek letters are cool.

It has three friends

Φ' = √½ x uu + (-√½) x dd = √½ x lr + √½ x rl
Ψ = √½ x ud + √½ x du√½ x ll + (-√½) x rr
Ψ' = √½ x ud + (-√½) x du√½ x lr + (-√½) x rl .

Φ' will always answer 'same' for a ZZ measurement but 'different' for XX, and so on. Both Ψ and Ψ' will answer 'different' for ZZ, but Ψ answers 'same' for XX and Ψ' answers 'different'.

All these states have correlations between the two qubits. With Φ, for example, if you look at either qubit on its own, they don't know whether to be up or down. But they do know that whenever one is up, the other is up too. And when one is down, so is the other. The left and right states are also correlated at the same time. This is a kind of correlation that only quantum systems can do. It's entanglement!

Footnotes
1. The udness and duness should also be divided by the square root of the probability of 'different', so that their squares add up to 1 again. But who can be bothered with that?

## Tuesday, 19 April 2016

### Entanglement with the simplest maths possible

This is part of a project to get people involved with quantum error correction. See here for more info.

You have probably heard that quantum mechanics is weird. You may also know that it has some strange correlations, known as entanglement, which are a big contribution to that weirdness.
In this post we will try to show that it's not all that weird, and that you can understand it without complicated maths. We'll also show how you can use entanglement to do something that no boring normal correlations can do: Teleportation!

This is a sequel to my earlier, and surprisingly successful, post on the maths of qubits (also known as 'Quantum mechanics with the simplest maths possible'). It also provides some of the maths behind my recent posts on quantum error correction. Nevertheless, I'll try to keep it pretty self-contained.

The story so far...

We are dealing with the simplest kind of quantum system. Some call them qubits. Some call them spin-1/2s. It doesn't matter what names we use though, so we'll just choose the ones that make our life easy.

A qubit can be in one of two states: up or down. You can ask it whether it's in state up or down: we call this a measurement. When it is up, it is not down. When it is down, it is not up. If it were neither, it would not be a qubit. So far, so normal.

But a qubit is quantum. It can be undecided about whether to be up or down. This is described by two numbers, which we call the upness and downness. When it is definitely up, its upness is 1 and its downness is 0. When its down, its upness is 0 and its downness is 1. When its undecided, both are somewhere in between. These are called quantum superpositions.

If we measure whether it is up or down, it will be forced to choose. The probability of choosing up is the upness squared (i.e. upness × upness = upness2). The probability of down is the downness squared. Since it must be up or down, adding up the probabilites for the two must give us 1. So

upness2 + downness2 = 1 .

In some sense, these probabilities are the true measures of how biased the qubit is towards up and down. So the actual upness and downness can be a bit crazy as long as they still make sense as probabilities when squared. The consequence of this is that the upness and downness can actually be negative numbers if they want, because negative numbers still square to positive numbers.

For example, imagine a qubit with an upness of 1, and one with an upness of -1. What's the difference? Well, the one with an upness of 1 is just the state up. For the other, if we ask whether it's up or down it'll answer 'up' with probability (-1)= 1, which makes it a certainty. And if it is certain that its own state is up, with no regard to the strange minus sign, who are we to argue?

To do maths we need some way of writing the superpositions down. If we have some state that we want to call S, we write it down like this.

S = (upness of S) x up + (downness of S) x down

Here we have written quantum states in an equation as if they were numbers, which might be a bit confusing. We have to remember what is a number and what isn't. So, from now on, all quantum states will be underlined.

S = (upness of S) x up + (downness of S) x down

That's better! This includes all the relevant information, and looks nice and mathsy.

This equation has a x and a + in there, which look like multiplication and addition. However, unlike normal multiplication, the one here doesn't multiply numbers together. Instead it multiplies a number with a quantum state, whatever that means. Nevertheless, when we do algebra type stuff with them, they follow the same rules as multiplication and addition. So let's just pretend they make sense.

Two of our favourite states are called left and right. The state called right is a superposition of up and down that has the same upness as downness.

right = √½ × up + √½ × down,

Left is the same, but it has a negative downness.

left = √½ x up + (-√½) x down

It turns out that these are as different from each other as up and down. This is because of the minus sign, which does interesting things with the maths when part of a superposition as we found out last time. And because these states are completely different, we are also allowed to measure whether our qubit is left or right.

What if we have a qubit that is in the up state and we measure whether it is left and right? Well, just as left is a superposition of up and down, up can be thought of as a superposition of left and right.

up = √½ × left + √½ × right

So if we do the left/right measurement, the qubit will have to decide one way or the other.

It's not just the up state can be thought of as a superposition of left and right. Anything can

S = (leftness of S× left + (rightness of S) × right
= (upness of S) x up + (downness of S) x down

Entanglement

In the original draft of this post, there was a bunch of stuff about how to describe pairs of qubits,  and some of the details about some of the more interesting measurements for them work. But that made everything far too long. All that stuff was moved to its own post: Fun measurements for pairs of qubits. So if you want the full experience, check that out before you continue.

I'll assume you're skipping it, so let's take a few paragraphs for a very brief overview of the world of two qubits. For this we need to know what the possible states of two qubits are. Each qubit has two possible states: up and down. So two qubits have four possible states: both up, both down, the first one up and the second one down and vice-versa. Let's call these uu, dd, ud and du. The kind of superpositions that you can get for two qubits then look like this

uuness x uu + udness x ud + duness x du + ddness x dd .

If we simply have something like uu, the two qubits are in their own state and don't really care what the other is doing. This is also true for the state

½ uu + ½ ud + ½ du + ½ dd

Despite having lots of fancy superpositions going on, it turns out that this is just another way of writing that both qubits are in the state left. Again, they are both doing their own thing, without any relation to the other. There are lots of states like this. They are called separable states.

Separable states are not so intesting. More interesting would be something like

√½ x uu + √½ x dd

Here the qubits are in a superposition of being up and down, but they will always be the same. So if one gets measured and decides to be up, the other will somehow instantly know about it. Then if it gets measured, it can give the same result. States that are correlated like this are more interesting than separable states. They are called entangled. In the rest of this post we'll look at how the kind of measurements that can create entanglement, and see how it can be used to teleport.

This will by no means tell you everything you need to know about entanglement. As a quantum information theorist, I find it best to try and understand entanglement by looking at what it can do. Teleportation is part of that, but we'll also look at other things like non-locality in the future.

Bell Measurements

What kind of measurements can we do for two qubits? Well we could just measure both to see if they are up or down. But there's another possibility, that's a bit less disruptive. This looks at the up and down states, but only tells you whether these states for the two qubits are the same or different. So if they are in the state uu or dd, it gives you the result same. Crucially, it would do this also for any superposition of uu and dd, without messing up the superposition. Similarly, ud, du or any superposition thereof would give the result 'different'.

This measurement is called ZZ. It has a friend, XX, which does the same thing but with the states left and right. These measurements are also ones that play nicely together. Suppose you have a state for which you know what the outcome of the ZZ measurement would be, but instead measure XX. This might force a decision out of the qubits and change the state. But whatever changes happen, the outcome for the ZZ measurement will stay the same. This sounds quite normal, given that this is exactly how measurements work in our world of big things. But in quantum mechanics, measurements working well together like this is a rarity.

After doing a ZZ measurement, we know exactly how the up/down states of our two qubits are correlated. After doing an XX we know the same for the left/right states. Because they play nicely together, after doing both of these measurements we know about both types of correlation at the same time. This gives us a state of two qubits that are correlated state in an interesting way. The combination of an ZZ and XX measurement must therefore be quite an interesting measurement. For that reason, it gets its own name: a Bell measurement.

Suppose the results to both XX and ZZ came out 'same'. Because they play nicely together, we know that they'd definitely give the same results if we did the same measurements again. There is only one state that give these definite answers. It's one with a lot of quantum entanglement, and is called Φ
Φ = √½ x uu + √½ x dd = √½ x ll + √½ x rr

For the other three possible combinations of answers we would get three similar states, called Φ', Ψ or Ψ'.

Bell measurements are pretty useful. Suppose we take some boring state like uu, which is not entangled. If we then we do a Bell measurement, they will be forced to be one of these four entangled states. They are boring no longer.

Bell measurements are named after J. S. Bell, who was the first to really show that entanglement could do things beyond the abilities of normal correations. The four states Φ, Φ', Ψ or Ψ' are call the Bell states, or Bell pairs.

Quantum Teleportation

Now we have Bell measurements we can do fun things. One of those fun things is teleportation!

Here's the situation. There are these two guys: Alice and Bob. They met ages ago, at a conference or something. They did a Bell measurement on a couple of qubits and got the state Φ. Then they went home to their own labs, each taking one of the qubits of the Bell pair.

Much later, Alice gets a spare qubit. She doesn't know what the state is. Maybe its even entangled to some other stuff. She doesn't know. She doesn't care. She doesn't want it.

But Bob does. He really wants it. I don't know why. Alice could just send it to Bob, but sending it so it doesn't get messed up by noise is really expensive. He doesn't want it that much! Is there a cheaper way from Alice to send it to Bob?

Perhaps Alice could measure it, work out what it is and just tell Bob. Unfortunately, it cannot be so simple. What measurement would she do? If she did an up/down measurement, it would have to choose to be up or down. If it was actually left or right, we wouldn't know. If it was entangled to something else, that entanglement would be destroyed. So she won't have given Bob what he wanted. She would have messed it up. But Alice knows her quantum mechanics, so she doesn't do this.

What could she do instead? Maybe they could use that Bell pair they've been sitting on.

Before we look at how to do this, lets just recap a little. Alice and Bob have three qubits between them, one is a some state S and the other two are a Bell pair. Alice has two (the spare one and half of a Bell pair) and Bob has one (half of a Bell pair).
Alice wonders what would happen if she forced her to qubits to become a Bell pair, by doing a Bell measurement. Then there would still have a Bell pair, but now Alice would have it all. So what would Bob have? Alice thinks that, to make everything nice and symmetric, he must end up with the state of the spare qubit.

Bob is skeptical. He thinks that the Bell measurement might end up extracting some information about the spare qubit. As we saw before, that would be bad and mess it up.

Alice can easily counter this. Her half of the Bell pair is randomly either up or down, and no-one knows which. This randomness will make the outcome of a ZZ measurement random too, whatever the state of the spare qubit is. The same is true for an XX measurement. In neither case do we find out anything about the state of the spare qubit.

Bob is still not convinced. He thinks that it might cause the state of the spare qubit to just be destroyed. This perhaps seems reasonable. His half of the Bell pair has never even been near the spare qubit, so how could it magically acquire the spare qubit's state? And information gets lost all the time in our everday human world. Why wouldn't qubits suffer the same.

Alice nevertheless sticks to her guns, and demands they do the maths to see who is right. For that we'll to work out how to describe three qubits. But that's not much different to two qubits, which isn't too much different to one qubit, so we'll dive straight in.

For the unknown state S of Alice's spare qubit, we'll need to write it in the most general way possible.

S = (upness of S) x up + (downness of S) x down

The Bell pair is in the state
Φ = √½ x uu + √½ x dd

Now lets write the state of all three together, and do so as some superpoistion of the three qubit states uuu, uud, udu, etc. For this we'll think of the first qubit as Alice's spare one, the next one as Alice's half of the Bell pair and the last as Bob's half of the Bell pair. So the state uud would mean that both of Alice's are up, and Bob's qubit is down.

To write down the correct three qubit state, we'll need to calculate the uuuness, the uudness and all the rest. We can get the uuunes by multiplying the upness of the spare qubit and the uuness of Φ

uuuness = (upness of S) x √½ ,

And so on for all the others. Because

uddness = (upness of S) x √½ ,
duuness = (downness of S) x √½ ,
dddness = (downness of S) x √½ ,

All the rest are zero because they would need the two qubits of the Bell pair to be different, and Φ does not allow that.
All in all, we get the three qubit state

(upness of S) x √½ x uuu + (upness of S) x √½ x udd
+ (downness of S) x √½ x duu + (downness of S) x √½ x ddd .

If Alice makes a ZZ measurement on the first two qubits, there's some parts of this state that will give the result 'same' (namely uuu and ddd), and some parts that'll give 'different' (udd and duu). This means the qubits have to decide one way or the other when she makes the measurement. Let's suppose they decide to be the same. The state after measurement will then be

(upness of S) x uuu + (downness of S) x ddd .

Here we've also removed a the √½ for each state. This is just something we need to do after measurements, to make sure the numbers that should add up to 1, still do.

The state we get here is quite interesting. We now have all three qubits entangled together, and the information about the state of the spare qubit is now spread over all three. It's as if all three qubits are now working together to be the spare qubit. If you wanted to measure whether it was up or down, you could do it by measuring any of the three.

Now we are part way there. Bob has a share in the spare qubit state, and could even do an up/down measurement if he wanted. But he might want to do other, more interesting things. For that he needs complete control of the spare qubit state, so Alice needs to somehow give up her part.

By doing an XX measurement she will not get any information about the spare qubit state, as we discussed earlier. She will also force her two qubits to be a Bell pair. This is because she will know definitely the outcomes of both and XX and ZZ measurement, and only Bell pairs can do that. So she'll have no information about the spare qubit, either in her measurement results or her qubits. So Bob must have it all.

Let's make sure by doing the maths. For that we need to know a little more about the state Φ'. Unlike Φ, which gives the result 'same' for both an ZZ and XX measurement, Φ' will always say 'same' for ZZ and 'different' for XX. In the mathsy way, it looks like this

Φ' = √½ x uu + (-√½) x dd = √½ x lr + √½ x rl

Using this, and the mathsy form of Φ from before, we can see that uu and dd can be though of as a superpositon of Φ and Φ'

uu = √½ x Φ + √½ x Φ'
dd = √½ x Φ - √½ x Φ'

So the state that we got after the ZZ measurement can be rewritten

(upness of S) x uuu + (downness of S) x ddd
= (upness of S) x √½ x (Φ u + Φ' u)
+ (downness of S) x √½ x (Φ d - Φ' d)

Here Φ u describes the state where Alice's two qubits are in state Φ and Bob's one is in state u, etc.

When Alice makes the XX measurement of the her qubits, she gets the result 'same' or 'different'. The same result is consistent only with Φ, and the different result is consistent only with Φ'. So if she gets 'same', the state afterwards is

(upness of S) x Φ u + (downness of S) x Φ d

Again we've removed each √½, because that's what you do after measurements. In this state we find that Alice's qubits always have the state Φ. So all the superposition stuff is actually just because Bob's state is

(upness of S) x u + (downness of S) x d

This is exactly the state of the spare qubit, just as he wanted.
It worked! Alice was right! If she makes the ZZ and XX measurements, and gets the result 'same' for both, Bob's qubit magically acquires the state that the spare qubit used to have. Even though his qubit never went anywhere near the spare one!

But what if Alice gets different results? For the other three options for her Bell test, Bob will get the one of the following states

(upness of S) x up + ( - downness of S) x down
(downness of S) x up + (upnness of S) x down
(downness of S) x up + (-upness of S) x down

For the first one the downness has got a minus sign in front of it. But that's okay, Bob just needs to do something called a phase flip to get his qubit into the state it should be. For the second we have the upness and downness the wrong way around. But Bob can use a bit flip to correct that. For the last, Bob just needs to do both
But it is important that Bob knows exactly which flip to do. So he needs Alice to tell him the result of her measurement. She just gets on the phone, tells him and then he does it. Then he always has the state of the spare qubit, just as he wanted.

Without this information, all he knows is that he randomly has one of the four possibilities. That doesn't do him any good. In fact, there's no way to tell the difference between that and his half of the Bell pair, so he can't even tell whether or not Alice has done the measurement yet. That's good, because this would allow Alice and Bob to communicate instantly, which the theory of relativity does not allow. So without knowing anything about the speed of light, the maths we've used to describe quantum states have nevertheless made sure that we can't communicate faster than it.

When Bob does get Alice's result (which is just some random outcome that gives no information about the spare qubit state) and combines it with his qubit (which never met the spare qubit) somehow that combine to form the spare qubit state. That's that magic of entanglement!

I think that one interesting thing here is that quantum mechanics had a choice. It could either allow information to get deleted from the universe, or allow states of qubits to magically be transported far away Star Trek style. It chose Star Trek. Quantum mechanics is weird, but it's good weird.

Footnotes
1. This is the same as asking whether there are an even or odd number of downs, as we do in this post on quantum error correction. If there are even, it means uu or dd, so they are the same. Odd means ud or du, so they are different.