Superposition and Quantum Entanglement

Factual Questions
1

I am reading a popular book on science that is describing superposition and quantum entanglement. What I get out of it is that a particle can be in two states at once and it does not “choose” a state until it is observed, where observe means to interact with some other particle in some manner. Two particles can also be entangled such that the act of observing the state of one will make the other “choose” the opposite state instantaneously.

What I don’t get is how we know that a particle is in superposition? Doesn’t any observation of the particle make it choose a state? It reminds me of Invisible Boy in Mystery men who is invisible as long as no one is looking.

If I have a Queen and a Jack and someone puts them in two different envelopes when I open one of the envelopes I know the value of the other. That doesn’t require the cards to be both a Jack and a Queen at the same time.

2

The simplest explanation is that the math says you’re wrong.

Don’t use logic. Use math. It still won’t make any more sense, but at least you’ll understand why words don’t work. :slight_smile:

3

Wrong about what?

I’ve never seen any decent explanation of QM that says that things at macro level (e.g cards in envelopes) have to obey the same laws as sub-atomic particles.

The problem with most popular science books is that they grotesquely over simplify.

In the case of observer it’s never made clear what, exactly, constitutes an ‘observation’.

The classic, popular, explanation of the Schroedinger’s Cat experiment is complete dingo’s kidneys because it typically talks about the fate of the cat being sealed upon a human opening the box. Yet the temperature of the box would tend to give a good idea of not only if the cat were dead but when it died.

4

Let’s start by talking about one-particle superpositions, and maybe get to entanglement later.

Really, the particle isn’t in two states at once. It’s in one state, but that state is a combination of two other states. The key is what “basis” you use to express the state.

For instance, suppose you have one basis that consists of the states |0> and |1>. (I’m putting the | and > around them because of a notation called bra-ket notation, but that doesn’t matter for this purpose… just think of it as a way of writing a state.)

So a particle might be in the |0> state, or it might be in the |1> state. But it might also be in the |0> + |1> state, or the |0> - |1> state, or other states of the form a|0> + b|1> for some complex numbers a and b.

(Technical note: I’m omitting the normalization factor for simplicity.)

Now let’s say we have a different basis, say consisting of states |x> and |y>, where |x> = |0> + |1>, and |y> = |0> - |1>.

Is |x> a superposition? Not from the perspective of the {|x>, |y>} basis, but it is from the perspective of the {|0>, |1>} basis.

Here’s the thing: a given measurement will distinguish the states in a give basis (not the same basis for every kind of measurement. So if you measure the state with respect to the {|0>,|1>} basis, you’re going to collapse it to either state |0> or state |1>. If the state (pre-measurement) was |0>, the measurement will produce |0>. But if it was |x> = |0> + |1>, then it has a 50% chance of ending up |0>, and a 50% chance of ending up |1>.

Now keep in mind that |x> is a single state just as much as |0> and |1>. In fact, we could have written |0> as a sum of |x> and |y> if we wanted. But nevertheless, |x> is a superposition in the {|0>,|1>} basis, just as |0> is a superposition in the {|x>,|y>} basis.

Now how do we know that a state is a superposition in the {|0>,|1>} basis, if we can only perform measurements in that basis? You point out that we’ll either get |0> or |1>, just as we’d get if it wasn’t a superposition. But prepare a bunch of identical copies of our original state(*), and perform the measurement on each of them, and you’d see that you got |0> part of the time and |1> part of the time, so the state must be a superposition of |0> and |1>.

(*) There’s a little complexity here, since there’s a theorem called the “no cloning” theorem which says you can’t duplicate an arbitrary unknown state, but in this case I’m assuming we know how the state was prepared in the first place.

5

Yes, for many reasons. Schrödinger’s cat is a flawed metaphor at best, and a lousy place to start from when it comes to understanding quantum mechanics.

6

Of course that should be:
where |x> = |0> + |1>, and |y> = |0> - |1>

If an admin wants to change the second + to a - in my original post, I’d appreciate it.

7

Regarding Entanglement (Note – you should read my first post first):

Say we have two particles. There combined state could be something like |0>|0> (meaning the first and second particle are in state |0>, or it could be |0>|1> (meaning the first particle is in state |0> and the second one in state |1>), or it could be |1>|0> or |1>|1>. This defines a basis: {|0>|0>,|0>|1>,|1>|0>,|1>|1>}.

As above, you could also get combinations like: |0>|0> + |0>|1> +|1>|0> + |1>|1>. Note that this combination can be factored out, as (|0> + |1>)(|0> + |1>), or in our other notation |x>|x>. States that can be factored in this way are not entangled.

However, you could also have a state like: |0>|1> + |1>|0>. This state can’t be factored into two one-particle states. So here we say the two particles are entangled.

Now suppose we perform a measurement on this entangled state. Let’s say we measure the left hand particle and get |0>. Then the right hand particle must be in state |1>, because there weren’t any terms where they were both |0>. Likewise, if the left hand particle is |1>, then the right hand particle is |0>. See how that works?

You can also have entangled states like |0>|0> + |1>|1>, where they are either both |0> or both |1>. That’s still just as entangled. The key thing to have the state be entangled is just that it can’t be factored out.

8

You should teach, I’m starting to understand.

9

Using the what tim has described already (i.e. taking two systems each with basis denoted {|1>, |0>}*), any state in the first system can be written as a|1> + b|0> and any state in the second system can be written as c|1> + d|0> where a,b,c,d are complex numbers.

In order to describe the composite system of the two combined systems you use the tensor product ‘⊗’.

The tensor product space contains the individual tensor products of the all the states in each system plus all the linear combinations of these tensor products. Not all states in the tensor product space can be described simply as the tensor product of two states.

The tensor product then of two generic states in the first and second system is (note in bra-ket notation |x> ⊗ |x> = |x>|x>):

a|1> + b|0> ⊗ c|1> + d|0> = ac|1>|1> + ad|1>0> + bc|0>|1> + bd|0>|0>

Any state of the composite system that does not take the form ac|1>|1> + ad|1>0> + bc|0>|1> + bd|0>|0> is said to be entangled as you cannot go back and say the first system is in state a|1> + b|0> and the second system is in state c|1> + d|0> instead you can only consider the state of the composite system.

Hmmm tbh I think Tim did a better job of describing it than me:smack:

*note that |1> and |0> in the first system are totally different vectors from |1> and |0> in the second system, living in different vector spaces