Struggling with Quantum Mechanics


by Aeropath
Tags: mechanics, quantum, struggling
Aeropath
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#1
Dec9-12, 02:16 PM
P: 5
Hello all. I have recently gotten into Quantum physics, and am having trouble wrapping my head around some of these concepts. I know about the history of it and I have a basic idea of Classical Mechanics, but the main thing I don't really undertand is Quantum Entanglement. Could someone please enlighten me a little bit?

Thanks.
-Aero
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K^2
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#2
Dec9-12, 02:56 PM
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Do you have basic grasp of what superposition is? If so, it's relatively straight forward, so long as you understand that in QM rather than describing states of individual elements, you describe states of the entire system at once. The distinction is academic in classical mechanics, but once you introduce principle of superposition, it becomes important.

So suppose I have two marbles. They can be red or blue. Classically, you can talk about color of each marble separately. In QM, however, there are four distinct states. Red-red, red-blue, blue-red, and blue-blue. System can be in any one of them or any superposition of these. For example, I can have a superposition state of red-blue and blue-red. Naturally, once you look at one of the marbles, you'll discover it to be either red or blue. But what's interesting is that you didn't just learn the state of that marble. You learned the state of the whole system. If you looked at first marble, and it turned out to be red, the the other one has to be blue. And vice versa. This is entanglement.

The reason this is interesting is that the state of each marble is truly undetermined until you decide to check one of them. At that time, the state of the other becomes specific. Furthermore, the change is effectively instant regardless of distance between the two. This, however, does not violate speed of light limit, since no information can be transferred using this.

There are, however, practical applications. Naturally, you need something much smaller than a marble. A very common way to produce entangled particle pairs is electron-positron annihilation, which will most of the time produce a pair of photons with opposite spins. These spins will be entangled as described above. This entangled pair can be used for Quantum Key Distribution or Quantum Teleportation. Both have been performed experimentally.
Aeropath
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#3
Dec9-12, 03:52 PM
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Okay... that makes a little more sense. I guess what I really want to know is how was Einstein's viewpoint proven wrong? Why is it that the state of the "marbles" are not predetermined?

tom.stoer
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#4
Dec10-12, 12:57 AM
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Struggling with Quantum Mechanics


There are some theorems proving that a hige class of theories with "local hidden variables" or "predetermined attributes or states" as one may infer from classical mechanics violate both quantum mechanics (as a theory) and experiment
DiracPool
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#5
Dec10-12, 02:02 AM
P: 492
produce a pair of photons with opposite spins
What does it mean for 2 photons to have opposite spins? I get an electron having 1/2 and -1/2 spin to avoid a Pauli exclusion situation, but photons are bosons and don't have that requirement. Does this mean a photon can have a +1 spin or a -1 spin? And what does that mean physically? How is this opposite photon spin manifested dynamically in their behavior and/or interactions?
tom.stoer
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#6
Dec10-12, 02:17 AM
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You start with a system (e.g. an excited atom or molecule) of total spin S=0. The system may decay into groundstate plus two photons with opposite spin s'=+1 and s'=-1 such that S=S'=0 holds
DiracPool
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#7
Dec10-12, 02:27 AM
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You start with a system (e.g. an excited atom or molecule) of total spin S=0. The system may decay into groundstate plus two photons with opposite spin s'=+1 and s'=-1 such that S=S'=0 holds
OK, I get the idea of the conservation aspect, but other than that, is there any qualitative difference that can characterize or distinguish one from the other? I mean, for the +1/2 spin and -1/2 spin electrons, we can put them through an inhomogeneous magnetic field and get one sailing in one direction and the other sailing on the opposite direction. Is there any homologue for opposite spin photons?
tom.stoer
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#8
Dec10-12, 02:36 AM
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The photons are emitted in opposite directions due to momentum conservation.

As explained you don't need the Pauli principle.

If you manage to prepare a system X* that decays into X plus two electrons you may use the same reasoning with fermions.
DiracPool
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#9
Dec10-12, 02:49 AM
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Ok, thanks
nanosiborg
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#10
Dec10-12, 02:55 AM
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Quote Quote by Aeropath View Post
Hello all. I have recently gotten into Quantum physics, and am having trouble wrapping my head around some of these concepts. I know about the history of it and I have a basic idea of Classical Mechanics, but the main thing I don't really undertand is Quantum Entanglement. Could someone please enlighten me a little bit?

Thanks.
-Aero
Quantum entanglement refers to the mathematics of QM. There's no 'concept' or 'understanding' of 'quantum entanglement' other than that as far as I'm aware. You might be tempted to think that it's based on common origins or interactions, but the mainstream view is that that's not the basis of quantum entanglement. So, if the mainstream view is correct, then quantum entanglement remains a mystery without explanation or understanding.

Nobody at PF or anywhere else can tell you what quantum entanglement is, in a physical sense. But they can help you with the math if you have any questions about that.
Aeropath
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#11
Dec10-12, 05:10 PM
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Quote Quote by nanosiborg View Post
Quantum entanglement refers to the mathematics of QM. There's no 'concept' or 'understanding' of 'quantum entanglement' other than that as far as I'm aware. You might be tempted to think that it's based on common origins or interactions, but the mainstream view is that that's not the basis of quantum entanglement. So, if the mainstream view is correct, then quantum entanglement remains a mystery without explanation or understanding.

Nobody at PF or anywhere else can tell you what quantum entanglement is, in a physical sense. But they can help you with the math if you have any questions about that.
Actually, I would like to know more about the math.
K^2
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#12
Dec11-12, 01:42 AM
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Quote Quote by Aeropath View Post
I guess what I really want to know is how was Einstein's viewpoint proven wrong? Why is it that the state of the "marbles" are not predetermined?
Bell's Theorem. There is a class of conditions known as Bell Inequalities, and some of them have been confirmed experimentally. To be satisfied, hidden parameter interpretation must be false. So we basically know that the states are not predetermined.

Quote Quote by DiracPool
What does it mean for 2 photons to have opposite spins? I get an electron having 1/2 and -1/2 spin to avoid a Pauli exclusion situation, but photons are bosons and don't have that requirement. Does this mean a photon can have a +1 spin or a -1 spin? And what does that mean physically? How is this opposite photon spin manifested dynamically in their behavior and/or interactions?
Yes, photons can only have spin +1 or -1. That corresponds to angular momentum carried, being [itex]\pm \hbar[/itex], as well as polarization of the photon. Left or right circularly polarized photon carries angular momentum. Linear or elliptical polarization is due to superposition, of course.

A lot of quantum mechanics measurements experiments can be carried out having in your possession nothing more than a light source, some polarizers, and a light intensity meter.

Quote Quote by Aeropath
Actually, I would like to know more about the math.
There is nothing particularly tricky about that math. How is your linear algebra? Do you understand how a projection matrix works?
Aeropath
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#13
Dec11-12, 04:51 PM
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Quote Quote by K^2 View Post
Bell's Theorem. There is a class of conditions known as Bell Inequalities, and some of them have been confirmed experimentally. To be satisfied, hidden parameter interpretation must be false. So we basically know that the states are not predetermined.


Yes, photons can only have spin +1 or -1. That corresponds to angular momentum carried, being [itex]\pm \hbar[/itex], as well as polarization of the photon. Left or right circularly polarized photon carries angular momentum. Linear or elliptical polarization is due to superposition, of course.

A lot of quantum mechanics measurements experiments can be carried out having in your possession nothing more than a light source, some polarizers, and a light intensity meter.


There is nothing particularly tricky about that math. How is your linear algebra? Do you understand how a projection matrix works?
My calculus is good.. Algebra's a little rusty. I have only worked with regular matrices before.
K^2
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#14
Dec11-12, 06:24 PM
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If you understand matrix products, I can work with that.

First of all, it's much easier to work with QM like that using Dirac notation.

<a| - This is a row matrix 1xN, also known as a bra.
|a> - This is a column matrix Nx1, also known as a ket.

Bra-ket, get it? Anyways. The dimension N corresponds to total number of states you are working with. For the most basic pair of entangled particles, you need at least 4 states, so we'll be dealing with 1xN, Nx1, and NxN matrices.

Naturally <a|b> is just a sum of corresponding elements of the two matrices. That means <a|b> behaves like a dot product of vectors a and b. It also means that <a|a> is the square of the norm. Since our state vectors will be normalized, you can assume that <a|a> = 1, unless otherwise specified.

Finally, |a><b| is the outer product. The result is an NxN matrix. So if you have vectors <a| = (1, 0, 0, 0), <b| = (0, 1, 0, 0), <c| = (0, 0, 1, 0), and <d| = (0, 0, 0, 1), then |a><a| + |b><b| + |c><c| + |d><d| = I, the identity matrix. That tells you that vectors a, b, c, and d form a complete orthonormal basis.


Now, I'm going to keep using marble analogy for the rest of this, but all the math is exactly the same with whatever four states you end up having. The four states of the system in the conventional basis are going to be |rr>, |rb>, |br>, and |bb>. The labels on each one denote the colors of the first and second marbles in that order. The states themselves are orthonormal, so that <rr|rr> = 1, but <rr|rb> = 0. In other words, dot product with itself is 1, dot product with any of the other three is zero. Then the general state can be written as some |ψ> = x|rr> + y|rb> + z|br> + w|bb>. Requirement that <ψ|ψ> = 1 gives us x + y + z + w = 1. Otherwise, these can be any number. (In general, a complex number, but we don't need that for this example.)

To find probability that both marbles are red, we must project the state onto the basis. This is trivial here, because the basis corresponds to our state, but hopefully it's still clear.

P(rr) = <ψ| (|rr><rr|) |ψ> = <ψ|rr><rr|ψ> = <rr|ψ> = x

The |rr><rr| is the NxN projection matrix. It is placed between row and column versions of our state vector, and gives us the probability of finding system in that state, which is x. Probabilities for rb, br, and bb are y, z, and w respectively. Since all these add up to 1, total probability is conserved, which is good.

Now, suppose I want to know what the probability is of finding just the first marble being red? The corresponding projection matrix is M = |rr><rr| + |rb><rb|. Probability is computed as before.

P(r_) = <ψ|M|ψ> = <rr|ψ> + <rb|ψ> = x + y

Again, this makes perfect sense. But the more interesting question is, what is the state of the system after the measurement if we indeed found the first marble to be red? The system must collapse to a state that corresponds to the result. In other words, the state of the system after measurement |ψ'> = M|ψ>/(<ψ|M|ψ>)1/2. The division ensures normalization. If applied to our state ψ, we get the following.

|ψ'> = M|ψ>/(<ψ|M|ψ>)1/2 = (|rr><rr|ψ> + |rb><rb|ψ>)/(<rr|ψ> + <rb|ψ>)1/2 = x/(x+y)1/2 |rr> + y/(x+y)1/2 |rb>

In other words, we get a superposition of the two states corresponding to the measurement result weighted in same proportion as originally, but re-normalized so that total probability is one. Naturally, if measurement of the first marble came up blue, instead, the resulting state vector would change accordingly.

|ψ'> = z/(z+w)1/2 |br> + w/(z+w)1/2 |bb>


So that's just measurement in QM. Lets look at entanglement. Specifically, suppose that the system was prepared in the following state.

|ψ> = [itex]\frac{1}{\sqrt{2}}[/itex](|rb> + |br>)

Note that <ψ|ψ> = 1 as required. Furthermore, probabilities of state rb and br are 0.5 each, and probabilities of rr and bb are 0.0 each. For simplicity, you can also just look up all of the previous results with x=w=0 and y=z=[itex]\small 1/\sqrt{2}[/itex]

What is probability of measuring the first marble to be red? As before, it works out to be x+y = 0.5. Same for probability of first marble being blue, z+w = 0.5.

Suppose, first marble was, indeed, found to be red. What is the new state?

|ψ'> = x/(x+y)1/2 |rr> + y/(x+y)1/2 |rb> = |rb>

This is where things get interesting. What is now the probability of finding second marble to be blue?

<ψ'|(|rb><rb| + |bb><bb|)|ψ'> = <rb|rb><rb|rb> + <rb|bb><bb|rb> = 1 + 0 = 1.

So P(_b) is now 1, and P(_r) is now 0. What if the first marble was measured to be blue? Then the new state |ψ'> = |br> and P(_b) = 0 and P(_r) = 1.

You can go through this and check what happens if you made measurements on the second marble first. You'll find exactly the same thing. It will have probability or red or blue coming up at 1/2 and 1/2, but following that, the other marble will have 100% of being the opposite.

This is the mathematics of entanglement. You can replace labels of red and blue marble with "spin up" and "spin down", or any other 2-valued quantum measurement, and you'll have it work out exactly the same way.
Aeropath
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#15
Dec11-12, 07:10 PM
P: 5
That makes perfect sense! Thanks for the help!
DiracPool
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#16
Dec12-12, 02:13 AM
P: 492
First of all, it's much easier to work with QM like that using Dirac notation.
Wow, nice summary K^2. That's what they call goin the extra mile. I like it. Anyway, just a quick question. I see two different approaches to QM in the literature and on the web. One is a Shrodinger based approach and the other is the Dirac notation approach with the Bras and Kets. I'm not sure if one is as good as the other, one is superior than the other, or if you need to know both, etc. My thinking is that, historically the Dirac notation was an offshot of the Pauli and Heisenberg matrix algebra formulation of QM that was developed before Shrodinger's wave equation-calculus formulation. Am I correct on this?

I've read that the formulations are identicle but just approach QM from different mathematical points of view. I don't know. I could be wrong in this analysis. I guess to summarize, what do you get by going at these problems with the Bra-ket notation that you couldn't get without going that direction. Because if it doesn't provide an advantage other than some ostensible sense of notational convenience, then I'm not interested, because I HATE linear algebra.
K^2
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#17
Dec12-12, 02:48 AM
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Functional notation is a special case, where the Hilbert space has infinite dimension.

Consider ψ(x) that is a solution to Schrodinger Eqn. It is actually a representation of the ket vector |ψ> in the basis of |x>. Specifically, ψ(x) = <x|ψ>. That means you can re-write the ket state from the function as |ψ> = ∫ψ(x)|x> dx.

Now, say I want to write down <ψ|ψ> given ψ(x). Using the above definition I get.

<ψ|ψ> = ∫<y|ψ*(y) dy ∫ψ(x)|x> dx

That doesn't look immediately helpful until you realize that you can insert a unit matrix in there. Specifically, I want ∫ |z><z| dz in there by analogy with the finite case. Keeping in mind that <x|z> = δ(x-z), we have the following.

<ψ|ψ> = ∫∫∫(ψ*(y)<y|z><z|x> ψ(x))dx dy dz = ∫∫∫(ψ*(y) δ(z-y) δ(x-z) ψ(x))dx dy dz = ∫ ψ*(z) ψ(z) dz

Which should be a familiar result.

So ultimately, it's all the same stuff. You are doing linear algebra either way, whether you are using operators in Dirac notation or linear functionals in functional form. Derivatives and integrals can be written in matrix form, so long as you don't mind infinitely-dimensional matrices. And you really need to be comfortable working with it in either form and going between one and another. If linear algebra gives you trouble - practice. Learn a bit more theory behind it. Quantum Mechanics is built completely upon linear algebra. So are many branches of applied and computational mathematics. Pretty much any serious science you end up doing, you'll have to deal with linear algebra.
DiracPool
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#18
Dec12-12, 03:00 AM
P: 492
Pretty much any serious science you end up doing, you'll have to deal with linear algebra.
Awe, Fiddlesticks! Alright, I'll get right on it. Thanks.


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