Undergrad Entangled system pure state or not?

[bhobba]

So in summary. In an entangled system, prior to tracing out being done and prior to it being in mixed state, the entangled system is in superposition and yet it is never in pure state. Is this correct? So Superposition can occur to both pure state and pure entangled state (before the entangled system is reduced by the density matrix or tracing out (collapse in disguised) done. Correct? (kindly answer by yes or no and not just explanations that don't use yes or no so I won't be confused. Thank you.)

No. There is no prior to being in a mixed state - it is in one. The system is in superposition, but each part acts like its in a mixed state.

Look - it is all explained in the reference I often give:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

See section 1.2.3

Its all there. Stop trying to translate it into English - you will fail.

Now let's get this over with onece and for all - exactly what part of that section is unclear and why? Not in English - but the math.

Thanks
Bill

 

[bhobba]

Oh. How do you define Superposition then?

Its agreed by 100% of physicists - superposition is simply pure states form a vector space.

Thanks
Bill

 

[stevendaryl]

Oh. How do you define Superposition then? (that is agreed by at least 99% of physicists). Because two particle entangled state are non-classical, it's not spin up or spin down entirely but they are somehow merged together (hence I thought is natural to call it superposition)..

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state. It's not |\psi\rangle, and it's not |\phi\rangle, and it's not any superposition of the two. Similarly, the second particle does not have a pure state. But the two-particle system has a pure state. That's what entanglement means: it means that the composite system has a pure state, but the components do not.

If you want to describe the first particle by itself, you would have to use a mixed state. It's in a mixture of |\psi\rangle and |\phi\rangle, but it is not in a superposition.

 

[bluecap]

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state. It's not |\psi\rangle, and it's not |\phi\rangle, and it's not any superposition of the two. Similarly, the second particle does not have a pure state. But the two-particle system has a pure state. That's what entanglement means: it means that the composite system has a pure state, but the components do not.

If you want to describe the first particle by itself, you would have to use a mixed state. It's in a mixture of |\psi\rangle and |\phi\rangle, but it is not in a superposition.

Thanks. You explain much better. Are you a physicist or mathematician?
But isn't the words "Now, supposed that we have two particles" is similar meaning to "If the two-particle system is in state |\Psi\rangle?

 

[stevendaryl]

Thanks. You explain much better. Are you a physicist or mathematician?

I'm nothing. I've studied physics, and I've studied mathematics, but I don't do either for a living.

But isn't the words "Now, supposed that we have two particles" is similar meaning to "If the two-particle system is in state |\Psi\rangle?

Well, one way to have two particles is to have a two-particle system in a pure state. I could also have the two particles in a mixed state, or they could be entangled with yet a third particle (or even more particles). So saying that you have a two-particle system is more specific than just saying that you have two particles.

 

[zonde]

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

There is an error in your reasoning.

A superposition just means a sum of pure states.

Yes of course.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle.

We are not interested if $|a_1\rangle$ and $|b_2\rangle$ are pure states. We need $|a_1,b_2\rangle$ to be pure state in order to talk about superposition of
$$\alpha|a_1,b_2\rangle+\beta|b_1,a_2\rangle$$

So can you say that $|a_1,b_2\rangle$ is a pure state? It does not have fixed relative phase between $|a_1\rangle$ and $|b_2\rangle$. It's sum of their phases that is fixed (or something like that).

 

[bluecap]

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state.

The above statements have logical error. Here's the logic or illogic.
you said |\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle
and you said "If the two-particle system is in state |\Psi\rangle,"
then its like saying "If the two-particle system is in state \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle then the first particle does not have a pure state."

but \alpha |\psi, \phi\rangle is a pure state

But you explained or follow up that "Well, one way to have two particles is to have a two-particle system in a pure state. I could also have the two particles in a mixed state, or they could be entangled with yet a third particle (or even more particles). So saying that you have a two-particle system is more specific than just saying that you have two particles."

Or better yet. Can you give example of a two-particle system in a pure state, a two-particle system in a mixed state (possible)?
Or you seemed to be saying that two-particle system is different to two particles.. please give actual example to avoid confusion. Thanks a lot!

It's not |\psi\rangle, and it's not |\phi\rangle, and it's not any superposition of the two. Similarly, the second particle does not have a pure state. But the two-particle system has a pure state. That's what entanglement means: it means that the composite system has a pure state, but the components do not.

If you want to describe the first particle by itself, you would have to use a mixed state. It's in a mixture of |\psi\rangle and |\phi\rangle, but it is not in a superposition.

 

[bluecap]

No. There is no prior to being in a mixed state - it is in one. The system is in superposition, but each part acts like its in a mixed state.

Look - it is all explained in the reference I often give:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

See section 1.2.3

Its all there. Stop trying to translate it into English - you will fail.

Now let's get this over with onece and for all - exactly what part of that section is unclear and why? Not in English - but the math.

Thanks
Bill

Yes. My favorite is section 1.2.3. but I don't know how to type the symbols and I can't take scan the page so I'll first learn how to type the symbols here. Wait half a day. I have some critical questions about it.

 

[bhobba]

So can you say that $|a_1,b_2\rangle$ is a pure state? It does not have fixed relative phase between $|a_1\rangle$ and $|b_2\rangle$. It's sum of their phases that is fixed (or something like that).

Its a pure state - it simply means system 1 is in pure state |a> and system 2 is in pure state |b>. The vector space is the joining of the vector spaces of the system. Phases have nothing to do with it. In fact for pure states phases are meaningless because states really are operators and the operator of a pure state is |a><a| which is easily seen to be the same regardless of phase. The real writing of |a>|b> is |a>|b><b|<a| where again phase means nothing.

Thanks
Bill

 

[Simon Phoenix]

The above statements have logical error. Here's the logic or illogic

There's no error that I can see.

Let's keep things very simple and think of qubits - which are just 2 level systems. We're going to imagine that Clive prepares 2 qubits in the pure state that is described (up to a normalization constant) as |01> + |10>.

Clive puts one particle in a box and gives it to Alice and puts the other particle in a box and gives it to Bob.

So in Alice's lab there is this box with one 2 level system in it, and in Bob's lab there is a box with one 2 level system in it. Now Alice and Bob are set to work and asked to devise experiments to figure out the properties of their particles - let's assume they've been locked in their labs and can't talk to one another or otherwise communicate.

The question is can we describe the properties of the particle in Alice's lab using the mathematical object we've called a pure state? More operationally we can ask whether experiments conducted by Alice on her particle, independently of anything done by Bob, are best described mathematically by a pure state or by something else?

It's probably easier to talk about ensembles of identically prepared systems - so we imagine that Clive has furnished Alice and Bob N boxes each - all derived from N copies of the 2 qubits prepared in the pure state |01> + |10>.

So Alice has N boxes to work with. She can perform any measurement, or set of measurements, she likes on her N boxes, but she has no access whatsoever to Bob's boxes, or any information about those boxes, or what Bob is doing to his boxes.

Alice's measurement results (whatever she does) are not consistent with assuming that her qubits are in pure states. They are consistent with assuming they are in a statistical mixture of states.

Another way of seeing what's going on is to imagine that Clive fools Alice and only tells her that her particles have come from an original pure state |01> + |10>. What he's actually done is to choose, at random for each box, whether to give her a |0> qubit or a |1> qubit. The question is whether there is any set of experiments Alice can do on the qubits she has been given that will tell her whether Clive is trying to fool her, or whether he really has given her qubits that come from entangled pairs?

There are no such experiments that Alice can perform. In other words - when we look at the properties of just one qubit from an entangled pair, it's exactly the same as looking at a single qubit prepared in a statistical mixture - which is not a pure state.

So for this entangled 2 qubit system we have
total system : pure state
either qubit looked at on their own : mixed state

I'm not sure whether that helps or just confuses things further - apologies if it's the latter.

 

[bluecap]

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state. It's not |\psi\rangle, and it's not |\phi\rangle, and it's not any superposition of the two. Similarly, the second particle does not have a pure state. But the two-particle system has a pure state. That's what entanglement means: it means that the composite system has a pure state, but the components do not.

If you want to describe the first particle by itself, you would have to use a mixed state. It's in a mixture of |\psi\rangle and |\phi\rangle, but it is not in a superposition.

Simon Phoenix.. Thanks a lot for your description. It is very clear and it described the latter of Stevendaryl statement "where If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state".

But in the sentences before it. Stevendaryl mentioned :
"Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle"

Here the first particle is in the pure state |\phi\rangle. This is not obviously the copy of Alice or Bob in your example. So what example of Alice and Bob can you give where the first particle (Alice copy) is in pure state?? Thanks so much for the help guys!

 

[Simon Phoenix]

This is not obviously the copy of Alice or Bob in your example

Yes, it is. Let's rewrite what Stevendaryl correctly wrote, but in terms of qubits in the hope that this lends some clarity for you :

"Let |0,1> be the state in which the first particle is in the pure state |0> and the second particle is in the pure state |1>. Similarly we let |1,0> be the state in which the first particle is in the pure state |1> and the second particle is in the pure state |0>

Now we have two distinct pure states |0,1> and |1,0>

We can form a superposition of these two different pure states as a|0,1> + b|1,0>"

For this superposition, is the first particle in the pure state |0> or is it in the pure state |1>?

It is in neither of these pure states - but a statistical mixture of these states.

If the combined state of both particles is a pure state AND particle 1 is in a pure state then the second particle must also be in a pure state. Furthermore the total state of both particles can be described by a product state |particle1>|particle2>

If two particles are entangled - so of the form a|0,1> + b|1,0> then there is NO possible way to decompose this state into a product of 2 individual pure states like this.

The statement combined state = pure, particle 1 = pure, automatically implies we can't have the entangled (superposition) state above.

 

[vanhees71]

Simon Phoenix.. Thanks a lot for your description. It is very clear and it described the latter of Stevendaryl statement "where If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state".

But in the sentences before it. Stevendaryl mentioned :
"Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle"

Here the first particle is in the pure state |\phi\rangle. This is not obviously the copy of Alice or Bob in your example. So what example of Alice and Bob can you give where the first particle (Alice copy) is in pure state?? Thanks so much for the help guys!

Stevendaryl is correct. The state of particle 1 is defined by tracing out particle 2 in the statistical operator of the two-particle system, which in this case is a projection operator, because the two-particle state is in a pure state,
$$\hat{\rho}_{12}=|\Psi \rangle \langle \Psi |=|\alpha|^2 |\psi \phi \rangle \langle \psi \phi|+\alpha^* \beta |\psi \phi \rangle \langle \phi \psi| +\alpha \beta^* |\phi \psi \rangle \langle \psi \phi| + |\beta|^2 |\phi \psi \rangle \langle \phi \psi |.$$
Now for simplicity we assume that $|\phi \rangle$ and $|\psi \rangle$ are orthonormal vectors. Then we can take them to be part of a complete orthonormal set (Schmidt decomposition theorem) and take the partial trace over this set to find
$$\hat{\rho}_{1}=\mathrm{Tr}_2 \hat{\rho}_{12} = |\alpha|^2 |\psi \rangle \langle \psi|+|\beta|^2 |\phi \rangle \langle \phi|,$$
which is not a projection operator and thus represents a mixed state.

 

[bluecap]

Yes. I understood now. Thanks to all the folks for all valuable help. Now let me apply this to atoms as this is my main motivation. In the atoms. Electrons are entangled to the nucleus. Now is the nucleus alone or electrons alone in pure states? Or in statistical mixture? But then electrons are quantum objects, nucleus are quantum objects.. by quantum objects I mean describable as ray of trace 1 or pure state. Therefore if the nucleus is in pure state, and electrons are pure state, then can we say the electrons are entangled to the nucleus or the electrons and nucleus are (or atom) in pure state? (Let's assume the atom (nucleus and electrons) are isolated and not in interaction with its environment (no photons or thermal sources to perturb the atom). Thank you.

 

[vanhees71]

Take as the most simple example the hydrogen atom first. For a thorough discussion about the state of the proton and the electron in this case (in the realm of non-relativistic QM) see

https://arxiv.org/abs/quant-ph/9709052

 

[stevendaryl]

We are not interested if $|a_1\rangle$ and $|b_2\rangle$ are pure states. We need $|a_1,b_2\rangle$ to be pure state in order to talk about superposition of
$$\alpha|a_1,b_2\rangle+\beta|b_1,a_2\rangle$$

I'm not sure I understand what you're saying. |a_1,b_2\rangle is a pure state of the composite system. |b_1,a_2\rangle is a pure state of the composite system. The sum \alpha |a_1,b_2\rangle + \beta |b_1, a_2\rangle is a pure state of the composite system.

So can you say that $|a_1,b_2\rangle$ is a pure state?

Yes. The superposition of pure states produces a pure state.

 

[bluecap]

No. There is no prior to being in a mixed state - it is in one. The system is in superposition, but each part acts like its in a mixed state.

Look - it is all explained in the reference I often give:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

See section 1.2.3

Its all there. Stop trying to translate it into English - you will fail.

Now let's get this over with onece and for all - exactly what part of that section is unclear and why? Not in English - but the math.

Thanks
Bill

Ok let me address section 1.2.3. I don't know how to type the symbols. But how is equation 1.21 derived or

p1=|psi1> < psi1| = 1/2 (|arrow upz>< arrowupz| + |arrowupupz><arrowdownz|+|arrow down z> <arrowupz|+|arrowdownz><arrowdownz|)??

and prior to that is written "We prepare the superposition |psi1> = 1/sqrt (2) (|arrowupz> + |arrowdown z>)

Why is there a square root of 2 and why are there many ><... what does the symbol >< mean and all the arrows?

Hope you can just address the above and not let me read a whole book about the math. And what kind of math are they called?

 

[bhobba]

Ok let me address section 1.2.3. I don't know how to type the symbols. But how is equation 1.21 derived or

p1=|psi1> < psi1| = 1/2 (|arrow upz>< arrowupz| + |arrowupupz><arrowdownz|+|arrow down z> <arrowupz|+|arrowdownz><arrowdownz|)??

and prior to that is written "We prepare the superposition |psi1> = 1/sqrt (2) (|arrowupz> + |arrowdown z>)

Why is there a square root of 2 and why are there many ><... what does the symbol >< mean and all the arrows?

Hope you can just address the above and not let me read a whole book about the math. And what kind of math are they called?

I can't give you a course in the math of QM here. See Susskind:
https://www.amazon.com/dp/0465062903/?tag=pfamazon01-20

But a few answers to the many you rase.

1. 1/Root2 is for normalisation.
2. The symbol |a><b| is an operator invented by Dirac. If you apply it to the bra |u> you get the bra |a><b|u> (in fact it's a projection operator.)

As a beginner you asked a question where even the terms you use (entanglement and pure state) are at least intermediate concepts, probably even advanced.

No wonder you are finding it hard.

You want more advanced answers, then you must learn the more advanced concepts.

Thanks
Bill

 

[stevendaryl]

Ok let me address section 1.2.3. I don't know how to type the symbols. But how is equation 1.21 derived or

p1=|psi1> < psi1| = 1/2 (|arrow upz>< arrowupz| + |arrowupupz><arrowdownz|+|arrow down z> <arrowupz|+|arrowdownz><arrowdownz|)??

and prior to that is written "We prepare the superposition |psi1> = 1/sqrt (2) (|arrowupz> + |arrowdown z>)

Why is there a square root of 2 and why are there many ><... what does the symbol >< mean and all the arrows?

Hope you can just address the above and not let me read a whole book about the math. And what kind of math are they called?

I think that what you're really asking for is a course in multiparticle quantum mechanics. "Why is there a square-root of 2?" is best answered as part of a course. The short answer is that states are usually normalized, so that the sum of the squares of their components (in an orthonormal basis) add up to 1. (\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 = 1/2 + 1/2 = 1

 

[zonde]

why are there many ><... what does the symbol >< mean

This might help: http://www.mathpages.com/home/kmath638/kmath638.htm

 

[bluecap]

Yes, it is. Let's rewrite what Stevendaryl correctly wrote, but in terms of qubits in the hope that this lends some clarity for you :

"Let |0,1> be the state in which the first particle is in the pure state |0> and the second particle is in the pure state |1>. Similarly we let |1,0> be the state in which the first particle is in the pure state |1> and the second particle is in the pure state |0>

Now we have two distinct pure states |0,1> and |1,0>

We can form a superposition of these two different pure states as a|0,1> + b|1,0>"

What is this term a|0,1> exactly called?
Is this a|0,1> the infamous branch or Worlds (of Many Worlds) that we Laymen are mostly familiar about. If it is not. then what is the mathematical expression of the branches or Worlds?

For this superposition, is the first particle in the pure state |0> or is it in the pure state |1>?

It is in neither of these pure states - but a statistical mixture of these states.

If the combined state of both particles is a pure state AND particle 1 is in a pure state then the second particle must also be in a pure state. Furthermore the total state of both particles can be described by a product state |particle1>|particle2>

If two particles are entangled - so of the form a|0,1> + b|1,0> then there is NO possible way to decompose this state into a product of 2 individual pure states like this.

The statement combined state = pure, particle 1 = pure, automatically implies we can't have the entangled (superposition) state above.

Your message finally makes things very clear. Thanks a lot.

 

[zonde]

I'm not sure I understand what you're saying. |a_1,b_2\rangle is a pure state of the composite system. |b_1,a_2\rangle is a pure state of the composite system. The sum \alpha |a_1,b_2\rangle + \beta |b_1, a_2\rangle is a pure state of the composite system.

Let's say that relative phase between |a_1\rangle and |b_2\rangle is not fixed. As I understand relative phase between separate states becomes phase of combined state: |a_1\rangle \otimes e^{i\theta}|b_2\rangle = e^{i\theta}|a_1,b_2\rangle
then if we say that $\theta$ is not fixed it is something like pure but incoherent state. But I'm not sure if it's meaningful. Say is it possible to have ensemble that is (classical) combination of $|a_1\rangle$ and $i|a_1\rangle$? Are they the same ray mathematically, or different rays?

But certainly relative phase matters when we talk about state:
\frac{1}{\sqrt{2}}(|a_1,b_2\rangle+e^{i\theta}|b_1,a_2\rangle)
because it changes predicted result. So I don't see how phase can be considered unphysical as bhobba stated.

And you don't have to answer. I'm just explaining my confusion.

 

[vanhees71]

I don't understand where your problem is. For sure if the phase is not determined, then you don't have a superposition, but you have a mixed state. For your example it's something like the following. First we define the pure states,
$$|\psi(\theta) \rangle=\frac{1}{\sqrt{2}} (|a_1,b_2 \rangle + \exp(\mathrm{i} \theta) |b_1,a_2 \rangle).$$
Then with a probability distribution $\Theta(\theta)$ for the phase you have
$$\hat{\rho}=\int_0^{2 \pi} \mathrm{d} \theta \Theta(\theta) |\psi(\theta) \rangle \langle \psi(\theta)|.$$
If your phase is totally indetermined, you have $\Theta(\theta)=1/(2 \pi)=\text{const}$.

 

[Simon Phoenix]

What is this term a|0,1> exactly called?

OK, forgive me if I'm wrong, but it looks to me like you're trying to run before you can walk. You're interested in all the funky stuff like entanglement (and quite rightly so, because it is fascinating) but it would seem that you don't have the basic framework in place on which to hang these funky ideas.

In classical physics we describe things in terms of concepts like position, momentum, field strength, etc and each of these ideas has some intuitive appeal. If an apple falls on my head I get an immediately intuitive feel for where it is and what its momentum is

So the notion of 'state' - the state of an object - is to some extent an intuitive thing classically. We write down a list of properties and call that a state because it contains all the things we need to describe the object. Then the laws of physics tell us how that state will change. So if we apply a force to something we'll change its state of motion (i.e. its state) and that change and its subsequent motion can be figured out by solving those laws of motion.

In quantum mechanics we kind of lose this immediately intuitive feel for what is being described. It's not that we can't build up an intuition about what happens or how things evolve, but developing that intuition requires a bit of re-wiring of our mental pictures. A large part of this failure of our 'classical' intuition is because in QM we have to describe the 'state' of an object in a rather abstract way - as a vector in a complex Hilbert space (there's a bit more to it than that but that's a good enough place to start).

You'll see on here that there is still lots of heated debate about what this QM state actually means. Is it describing some objective thing or is it a mathematical device that just allows us to predict the right things? Right now we can't really say one way or another - neither of these ways of viewing the quantum state is wholly satisfactory to everybody.

So the first thing you need to do, if you want to understand more advanced things like entanglement, is to get a good feel for the basic mathematical machinery of QM. You can go quite a long way with just a few ideas from complex numbers and linear algebra and it doesn't need to involve really difficult maths.

When a state in QM is written as |a> for example, this is just notation for one of these vectors - it's a very nice notation invented by Dirac. I find it amazingly useful. But essentially when you see it think 'vector' to begin with. So like any vector it 'lives' in a vector space and it has certain properties. Adding any 2 vectors gives us another vector - or put it another way we can 'decompose' or 'expand' any given vector into a sum of other vectors. This decomposition or expansion is not unique and for a given vector there will generally be more than one way to expand it (and often there are an infinite number of different ways).

So if we walk from A to B we can label our journey as AB, but suppose we walk from A to B via C (all in straight lines), then we can see that AB is 'made up' of the journey AC followed by CB. This is nothing more than the principle of superposition so that the vector AB is equal to the sum of the vectors AC and CB. On another day we might decide to go via D so that our journey is AD plus DB. In all these cases we go from A to B - but this single journey can be 'expanded' in lots of different ways.

In principle this is also what we have in QM. When we write a state as |s> = |r> + |t> then it means just the same thing. In fact we could go daft and use the Dirac notation to describe our walking journey. So the vector AB we could write as |s>, the vector |r> would be AC and the vector |t> would be CB - and then we have just used a Dirac notation applied to real vectors to describe AB = AC + CB.

Honestly, apart from some more technical details, this is really all that superposition in QM is. Because these quantum states are vectors - they have all the usual vector properties. So mathematically at least, if you know linear algebra, you'll understand much more about the basic maths behind some things in QM.

Where things get more tricky is in the meaning. It's easy to visualize the journey through real space described by AB - you could even do it, that is walk from A to B in a straight line. The vectors in QM don't live in this real space and so it's a bit more difficult to describe them in such simple terms. They are, nevertheless, simply vectors, albeit in a complex Hilbert space.

When we write |0,1> in the above discussion, for example, this is just another vector, but it's a single vector describing 2 things - in this case 2 qubits. Just like 'normal' vectors in 'real' 3D space we can form superpositions so that |0,1> + |1,0> is another perfectly good vector.

I may have pitched all that way too low - and I apologize if I have been inadvertently patronizing. If you need a bit more technical flesh on the bones then Susskind's stuff (either his lectures on youtube, or his book 'the theoretical minimum') goes into more detail and are a truly excellent place to begin.

 

[zonde]

I don't understand where your problem is.

Say is it possible to have ensemble that is (classical) combination of $|a_1\rangle$ and $i|a_1\rangle$? Are they the same ray mathematically, or different rays?

Say is it meaningful to write something like:
$$\hat{\rho}=\int_0^{2 \pi} \mathrm{d} \theta \Theta(\theta) e^{i\theta}|a_1 \rangle e^{-i\theta}\langle a_1|$$
Result seems the same as for pure state $|a_1 \rangle \langle a_1|$ as the phase cancels out. But there would't be any interference if we would perform double slit experiment with such a state (assuming it's meaningful).

 

[vanhees71]

As is very easy to see, what you get then is the pure state
$$\hat{\rho}=|a_1 \rangle \langle a_1|$$
since
$$\int_0^{2 \pi} \Theta(\theta)=1,$$.
and the $\exp$ factors cancel. I think, you have to learn the formalism (linear algebra) first, before you go further to quantum theory.

 

[stevendaryl]

Say is it meaningful to write something like:
$$\hat{\rho}=\int_0^{2 \pi} \mathrm{d} \theta \Theta(\theta) e^{i\theta}|a_1 \rangle e^{-i\theta}\langle a_1|$$
Result seems the same as for pure state $|a_1 \rangle \langle a_1|$ as the phase cancels out. But there would't be any interference if we would perform double slit experiment with such a state (assuming it's meaningful).

Overall phases are not important, but relative phases (between different elements of a superposition) are important.

Let's look at a specific case: the spin state \frac{1}{\sqrt{2}} (|U\rangle - e^{i \theta} |D\rangle). This corresponds to the density matrix: \rho(\theta) = \frac{1}{2} |U\rangle \langle U| + \frac{1}{2} e^{i \theta} |D\rangle \langle U| + \frac{1}{2} e^{-i \theta} |U\rangle \langle D| + \frac{1}{2} |D\rangle \langle D|. Now, if \theta is unknown, then we average that density matrix over all possible values of \theta:

\rho = \frac{1}{2\pi} \int d\theta \rho(\theta) = \frac{1}{2} |U\rangle \langle U| + \frac{1}{2} |D\rangle \langle D|

 

[bhobba]

OK, forgive me if I'm wrong, but it looks to me like you're trying to run before you can walk. You're interested in all the funky stuff like entanglement (and quite rightly so, because it is fascinating) but it would seem that you don't have the basic framework in place on which to hang these funky ideas.

Exactly.

You need to know a LOT more about QM before venturing into deep waters like this.

Thanks
Bill

 

[zonde]

I cleared my confusion. And I would say that notation is confusing.
Say we write a vector
$|\psi\rangle=|a_1,b_2\rangle$ but actually we mean a ray (all the vectors that are the same modulo phase factor)
and then we write
$|\phi\rangle=\frac{1}{\sqrt2}(|a_1,b_2\rangle+e^{i\theta}|b_1,a_2\rangle)$ and here $|a_1,b_2\rangle$ actually is a vector with particular phase as otherwise we couldn't speak about certain relative phase. But combined vector again is meant as a ray.

So in one case $|a_1,b_2\rangle$ means a ray (even so we write a vector) but in other case it is a vector.

 

[stevendaryl]

I cleared my confusion. And I would say that notation is confusing.
Say we write a vector
$|\psi\rangle=|a_1,b_2\rangle$ but actually we mean a ray (all the vectors that are the same modulo phase factor)
and then we write
$|\phi\rangle=\frac{1}{\sqrt2}(|a_1,b_2\rangle+e^{i\theta}|b_1,a_2\rangle)$ and here $|a_1,b_2\rangle$ actually is a vector with particular phase as otherwise we couldn't speak about certain relative phase. But combined vector again is meant as a ray.

So in one case $|a_1,b_2\rangle$ means a ray (even so we write a vector) but in other case it is a vector.

Right, it is a little confusing. In doing QM calculations, you start with a specific representation of basis states, and you fix the phases. It's pretty arbitrary how you assign the relative phases of the different basis states, but you need to pick some convention. Then in terms of the basis states, you can describe a pure state of a system as a ray. The basis states themselves are not rays, they are vectors. It doesn't make any sense to make a superposition of rays.

The main point of phases is to take into account interference. To illustrate, let's consider a two-state system with basis states |U\rangle and |D\rangle. Fix a time period T, and let:

\psi_{UU} = \langle U| e^{-iHT} |U\rangle
\psi_{UD} = \langle U| e^{-iHT} |D\rangle
\psi_{DU} = \langle D| e^{-iHT} |U\rangle
\psi_{DD} = \langle U| e^{-iHT} |U\rangle

\psi_{ij} is the probability amplitude that a system initially in state j will be found in state i a time T later. You square the amplitude to get the probability. But let's suppose we started instead in a state |\chi\rangle = \alpha |U\rangle + \beta |D\rangle. Then what's the probability amplitude that it will be found in state |D\rangle a time T later? Well, it's \alpha \psi_{DU} + \beta \psi_{DD}. You square that to get the probability:

P = |\alpha|^2 |\psi_{DU}|^2 + |\beta|^2 |\psi_{DD}|^2 + 2 Re(\alpha^* \psi_{DU}^* \beta \psi_{DD})

The term 2 Re(\alpha^* \psi_{DU}^* \beta \psi_{DD}) is an interference term, and it is an observable effect of relative phases. Multiplying the state \alpha \psi_{DU} + \beta \psi_{DD} by an overall phase will make no change to this term. But changing \alpha to \alpha e^{i \theta_1} and changing \beta to \beta e^{i \theta_2}, where \theta_1 \neq \theta_2, will change the interference term.