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Struggling with Quantum Mechanics 
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#1
Dec912, 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 


#2
Dec912, 02:56 PM

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P: 2,470

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. Redred, redblue, bluered, and blueblue. System can be in any one of them or any superposition of these. For example, I can have a superposition state of redblue and bluered. 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 electronpositron 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. 


#3
Dec912, 03:52 PM

P: 5

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?



#4
Dec1012, 12:57 AM

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P: 5,366

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



#5
Dec1012, 02:02 AM

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#6
Dec1012, 02:17 AM

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P: 5,366

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



#7
Dec1012, 02:27 AM

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#8
Dec1012, 02:36 AM

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P: 5,366

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. 


#9
Dec1012, 02:49 AM

P: 534

Ok, thanks



#10
Dec1012, 02:55 AM

P: 79

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. 


#11
Dec1012, 05:10 PM

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#12
Dec1112, 01:42 AM

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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. 


#13
Dec1112, 04:51 PM

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#14
Dec1112, 06:24 PM

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P: 2,470

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. Braket, 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 <ab> is just a sum of corresponding elements of the two matrices. That means <ab> behaves like a dot product of vectors a and b. It also means that <aa> is the square of the norm. Since our state vectors will be normalized, you can assume that <aa> = 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 <rrrr> = 1, but <rrrb> = 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> = xrr> + yrb> + zbr> + wbb>. Requirement that <ψψ> = 1 gives us x² + y² + z² + w² = 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> = <ψrr><rrψ> = <rrψ>² = x² 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 x². Probabilities for rb, br, and bb are y², z², and w² 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_) = <ψMψ> = <rrψ>² + <rbψ>² = x² + y² 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'> = Mψ>/(<ψMψ>)^{1/2}. The division ensures normalization. If applied to our state ψ, we get the following. ψ'> = Mψ>/(<ψMψ>)^{1/2} = (rr><rrψ> + rb><rbψ>)/(<rrψ>² + <rbψ>²)^{1/2} = x/(x²+y²)^{1/2} rr> + y/(x²+y²)^{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 renormalized 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/(z²+w²)^{1/2} br> + w/(z²+w²)^{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 <ψψ> = 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 x²+y² = 0.5. Same for probability of first marble being blue, z²+w² = 0.5. Suppose, first marble was, indeed, found to be red. What is the new state? ψ'> = x/(x²+y²)^{1/2} rr> + y/(x²+y²)^{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'> = <rbrb><rbrb> + <rbbb><bbrb> = 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 2valued quantum measurement, and you'll have it work out exactly the same way. 


#15
Dec1112, 07:10 PM

P: 5

That makes perfect sense! Thanks for the help!



#16
Dec1212, 02:13 AM

P: 534

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 Braket 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. 


#17
Dec1212, 02:48 AM

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P: 2,470

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) = <xψ>. That means you can rewrite the ket state from the function as ψ> = ∫ψ(x)x> dx. Now, say I want to write down <ψψ> given ψ(x). Using the above definition I get. <ψψ> = ∫<yψ^{*}(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 <xz> = δ(xz), we have the following. <ψψ> = ∫∫∫(ψ^{*}(y)<yz><zx> ψ(x))dx dy dz = ∫∫∫(ψ^{*}(y) δ(zy) δ(xz) 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 infinitelydimensional 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. 


#18
Dec1212, 03:00 AM

P: 534




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