Teleportation of photons using entanglement
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Sci~Girl - 03 Mar 2005 22:25 GMT On howstuffworks.com... article on photon teleportation... http://travel.howstuffworks.com/teleportation2.htm
The article only goes into a basic description of how entanglement works, and I wasn't able to find a more detailed one anywhere else. I was wondering if anyone knew of any other (accurate) sites that give a detailed description of the method. I'm writing a research paper on it for an independent study project.
The article also mentions teleportation of a laser beam, and information on that would be helpful too.
George Jones - 06 Mar 2005 07:51 GMT > On howstuffworks.com... article on photon teleportation... > http://travel.howstuffworks.com/teleportation2.htm [quoted text clipped - 7 lines] > The article also mentions teleportation of a laser beam, and > information on that would be helpful too. I'll give the theory behind an experiemnt described in the 11 December 1997 issue of Nature. Maybe what I say won't be very helpful. In this issue, Nature has a technical but readable article describing the experiment and a non-technical article describing quantum entanglement.
Consider photons that are either horizontally (h) or vertically (v) linearly polarized, so that state space is 2-dimensional, and suppose that an EPR apparatus creates photons A and B which are in the entangled state
|s> = 2^(-1/2)(|Ah>|Bv>-|Av>|Bh>), and that A is sent to some experimental apparatus, while B propagates freely. Suppose further that an arbitary photon, call it P, with arbitrary state
|P> = a|Ph> + b|Pv>, a^2 + b^2 = 1 arrives (from anywhere) at the experiment apparatus at the same time in and the same place as photon A. The state of the 3-photon system is thus
|u> = |P>|s> = (a/2)|Ph>|Ah>|Bv> - (a/2)|Ph>|Av>|Bh> + (b/2)|Pv|Ah>|Bv> - (b/2)|Pv>|Av>|Bh>.
The experimental apparatus then effectively performs a quantum measurement on the photons P and A in such that they end up in the entangled state
|q> = 2^(-1/2)(|Ph>|Av>-|Pv>|Ah>). But before the measurement, A and B were in an entangled state |s>, so this can't be done without affecting photon B, even though B doesn't interact physically with the apparatus.
The effect of the measurement is represented mathematically by acting on the first two components (P an A) of each term of the initial 3 photon state |u> with the projection operator formed from the final combined (and entangled) state |q> of P and A
Proj = |q><q|
= |Ph>|Av>(<Ph|<Av| - <Pv|<Ah|) - |Pv>|Ah>(<Ph|<Av| - <Pv|<Ah|),
while operating on the third component (the B part) of each term of |u> by the identity operator 1, since no measurement was performed on B.
After some algebra, one arrives at
(Proj x 1)|u> = -(a/2)|Ph>|Av>|Bh> + (a/2)|Pv>|Ah>|Bh> - (b/2)|Ph>|Av>|Bv> + (b/2)|Pv>|Ah>|Bv>,
which factors into
(1/2)(-|Ph>|Av> + |Pv>|Ah>)(a|Bh>+b|Bv>)
Shazzam! Photon B, which, after creation, never came anywhere near the experimental apparatus, now has exactly the state that the arbitrary photon P had before the measurement (the interactions of photons P and A with the apparatus). In effect the identity of photon P has been teleported to photon B! Photon P as single entity has been "destroyed", as it is now completely entangled with A.
Unfortunately, the measurement that created state |s> has 3 other equally probable outcomes, sot this teleportation only happens 25% of the time.
Regards, George
Sci~Girl - 07 Mar 2005 21:49 GMT It helps a little, but I'm really looking for what "entangled" refers to. What does it mean to have photons "entangled?"
Arnold Neumaier - 08 Mar 2005 06:48 GMT > It helps a little, but I'm really looking for what "entangled" refers > to. What does it mean to have photons "entangled?" That the joint wave function of a a pair of photons is not the tensor product of two single-particle wave functions.
Arnold Neumaier
Igor Khavkine - 09 Mar 2005 08:01 GMT >> It helps a little, but I'm really looking for what "entangled" refers >> to. What does it mean to have photons "entangled?" >> > That the joint wave function of a a pair of photons is not the tensor > product of two single-particle wave functions. That is correct and very succinct. But, perhaps a hands on demonstration would be helpful.
A state of a particle is described by a wave function. We can represent it by a function of one variable, say x. If we have two particles, then their state will be described by wave function of two variables, say x and y, one for each particle.
When two particles are independent, their wave function is the product of two individual wave functions:
psi_togehter(x,y) = psi_single1(x) psi_single2(y).
Here's a simple fact: not every possible function of two variables psi_together(x,y) can be written as a product of two individual functions psi_single1(x) and psi_single1(y). A simple example is
psi_together(x,y) = x^2 + y^2.
Real wave functions look very different from the above form, but it is sufficient to demonstrate the point.
We call a state of two particles _entangled_ if the corresponding wave function psi_together(x,y) cannot be written as a product of two single particle wave functions. The discussion above shows that there in fact are entangled states.
If you are at all familiar with linear algebra, you can try an example that is more realistic. A two component (row or column) vector describes the state of a two level quantum system (such as a spin or a photon). If you have two such systems, the joint state will be described by a two by two matrix (four components). Can you write every possible 2x2 matrix as a tensor product of two 2-component tensors (a column vector times a row vector)? If not, then you've just proven that a system of two coupled two level systems exhibits entangled states.
Hope this helps.
Igor
Sci~Girl - 15 Mar 2005 19:21 GMT We call a state of two particles _entangled_ if the corresponding wave function psi_together(x,y) cannot be written as a product of two single
particle wave functions. The discussion above shows that there in fact are entangled states.
If you are at all familiar with linear algebra, you can try an example that is more realistic. A two component (row or column) vector describes the state of a two level quantum system (such as a spin or a photon). If you have two such systems, the joint state will be described by a two by two matrix (four components). Can you write every possible 2x2 matrix as a tensor product of two 2-component tensors (a column vector times a row vector)? If not, then you've just proven that a system of two coupled two level systems exhibits entangled states.
Hope this helps.
Yes, it does, thanx. All I know in math is basic algebra though, I'm 14 and in 8th grade but in 9th grade math. What's a vector?
Mark Palenik - 16 Mar 2005 09:07 GMT > Yes, it does, thanx. > All I know in math is basic algebra though, I'm 14 and in 8th grade but > in 9th grade math. What's a vector? The most common types of vectors you'll run into are basically arrows. They have direction and length. So, for example, if I were to tell you that something is five miles east, that would be a vector. Similarly, a set of x,y, and z coordinates has direction and magnitude, like [0,0,1] - left zero, forward zero, and up one. This is the definition of a vector that you will learn first in school - that vectors are things with direction and magnitude, and my guess is that most of your teachers won't even know that this isn't actually the *real* definition of a vector.
There is actually a more fundimental definition of a vector space, and in quantum mechanics, the vectors we use actually have neither direction nor magnitude (at least, not in the traditional sense - they can have orthogonality, but there's no need to confuse you with that). For example, the functions Sin(x), Sin(2x), Sin(3x), etc. can be a vectors. You don't really need to worry about how or why, just know that vectors are things that we can do algebra with.
The definitions of entanglement you've been given would probably be difficult for you to understand without an ungodly amount of additional explanation. I'm not even sure if I would have understood them a semester ago (I haven't even done multi-particle QM yet, and just learned the operator/matrix notation about four or five weeks ago - of course, now the 223 students are learning a simplified version of it too, which makes I'd had Thaler when I took that course, but I digress. . .).
At the most simple level (and someone else feel free to jump in if I'm not saying this well), you can think of entangled particles as particles that share some net property, but individually don't have any definite value of that property.
For example, electrons have a property called spin. The spin can be up or down. If you have one electron that is spin up and one that's spin down, your total spin is zero, because up plus down is zero. Understand?
Now, say you have two electrons, and you know that the total spin between the two electrons is zero. But you don't know which one is spin up and which one is spin down. You can actually set up the system so that neither particle is spin up or spin down until it is measured. However, as soon as you measure one, the other one will have the opposite spin. That's an example of entanglement, in my own very non-formal, clumsy words.
I'd like to thanks people for the linear algebra explanation that was posted, because if nothing else, *I* learned something :)
Oz - 16 Mar 2005 17:18 GMT Mark Palenik <markpalenik@wideopenwest.com> writes
>For example, electrons have a property called spin. The spin can be up >or down. If you have one electron that is spin up and one that's spin [quoted text clipped - 5 lines] >up and which one is spin down. You can actually set up the system so >that neither particle is spin up or spin down until it is measured. For a very elementary idea, suitable to carry someone aged 14 onwards without putting too much misinformation into the concept, how about the following:
Take your two particles, one spin up and one spin down.
Individually we have three classical combinations:
up up up down (and down up) down down
Now quantum mechanics (ie the universe) doesn't work quite as simply as this. You can have a pair of particles that is in a very real sense BOTH spin up and spin down. This is NOT the same as having a pair with one spin up and one spin down.
This is a fourth valid description. It only exists in a quantum mechanical world (that as yet you know nothing about).
In many ways the latter behaves as a single particle containing both spin up and spin down (which isn't zero). The devious thing is that if you are very delicate and careful (ie don't look at it too closely) you can separate the two particles, each of which is BOTH spin up and spin down. BUT as soon as you actually measure (ie destroy) one, the other is guaranteed to be the other way up, even if separated by miles.
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a student - 18 Mar 2005 19:41 GMT > At the most simple level (and someone else feel free to jump in if I'm > not saying this well), you can think of entangled particles as particles > that share some net property, but individually don't have any definite > value of that property. I think you have said it very well - 'entanglement' means that the properties of a pair of entangled particles cannot be reduced to properties of the individual particles.
In classical mechanics two non-interacting particles, considered together, are nothing more than a first particle plus a second particle. Each particle has, for example, its own position and velocity, and this is all that is needed to predict any measurement outcome. Moreover, if one has only incomplete knowledge of the positions and velocities, then one can only make statistical predictions (eg, of the average velocity) for the outcome of any particular measurement on the particles, but these predictions are still formulated in terms of the properties of the individual particles. So there is no entanglement.
But quantum particles CAN have entanglement! In particular, if there was NO entanglement, then all properties of a pair of non-interacting particles would be determined completely by the properties of the individual particles. Hence the statistics of any measurements on the pair could be formulated in terms of the properties of the individual particles (eg, their positions and velocities). But, as first pointed out by John Bell, this means that groups of statistical predictions must satisfy certain mathematical conditions (called Bell inequalities). Remarkably, quantum mechanics allows the existence of pairs of non-interacting particles which do NOT satisfy these conditions!! Hence there must be entanglement (if quantum mechanics is correct, and it has never been wrong yet).
It is perhaps worth noting further that, in quantum mechanics, one only ever has incomplete knowledge of the properties of even a SINGLE particle - the mathematics simply doesn't allow, for example, values to exist for both the position and the velocity of a given particle. So, one can only make statistical predictions for the position and velocity (and the statistical errors satisfy the famous Heisenberg inequality).
Hence, it is natural to ask: why doesn't the maths allow complete knowledge of a particle's properties? Is it (i) because while particles actually do have well defined positions and velocities, quantum mechanics is not good enough to describe them (i.e., the theory is incomplete); or (ii) because particles don't have definite positions and velocities (i.e., our classical notion of particles is incomplete).
The generally accepted answer is (ii).
So what rules out choosing option (i) (for most people)? Entanglement! There simply seems little point choosing option (i) - because even if individual particles "really" do have well defined values of position, velocity, etc, these values cannot explain the statistics of PAIRS of non-interacting particles, as noted above.
In fact, the only way to 'save' option (i) is to claim that either (a) 'non-interacting particles' are, in fact, interacting ! (it turns out that these interactions must be instantaneous and hence travel faster than the speed of light - there is a theory along these lines called Bohmian mechanics after David Bohm); or (b) there is a conspiracy of nature which denies experimenter's the choice of which property to measure, implying that the measurement statistics are not really random in the sense that is required for the Bell inequalities (I don't think anyone takes this seriously); or (c) quantum mechanics and its prediction of entanglement is wrong (but all experiments to date suggest the opposite!).
Arnold Neumaier - 16 Mar 2005 17:18 GMT > All I know in math is basic algebra though, I'm 14 and in 8th grade but > in 9th grade math. What's a vector? A list of numbers written below each other. For example the x,y, and z coordinate of a point in a 3-dimensional coordinate system. Physicists write the three coordinates as x_1, x_2, x_3 and combine it to a vector simply called x. - - | x_1 | x = | x_2 | (The parentheses look a bit awkward in ascii.) | x_3 | - - The same for a list of n numbers. This gives a vector x with n coordinates x_1,...x_n, and is thought of as a point in a space with n dimensions.
Two vectors are added or subtracted or multiplied by a number just by adding, subtracting or multiplying their entries. Then there is the inner product of two vectors x dot y = sum x_i*y_i which is a number and not a vector.
Once you master vectors you need to understand matrices. These are rectangular arrays of numbers.
If you want to learn more about quantum physics and really understand you need to learn first how to do calculations with vectors and matrices. This is essential for quantum information theory (and entanglement is relevant mainly there). Look in your local library for math books, about 'linear algebra' or 'analytic geometry'. You may have to try several before you find one suitable at your level.
Maybe at first it is better to get math schoolbooks from your older peers. Good school books are written in a way that they can be used for self study. If you are motivated it can be very exciting!
Arnold Neumaier
Igor Khavkine - 18 Mar 2005 19:42 GMT > All I know in math is basic algebra though, I'm 14 and in 8th grade but in > 9th grade math. What's a vector? In addition to good explanations by Arnold Neumaier and Mark Palenik, here's a slightly more high brow one.
In modern mathematics vectors are an abstract but simple concept. There are many objects that can rightfully be called vectors. The only requirement is that they satisfy certain axioms, that is certain operations are possible with them.
Here's a boost of confidence for you. You say you know basic algebra. Do you know how to add? Do you know how to multiply? Then you can understand what vectors are. Suppose U and V are vectors and a and b are numbers. Then you can do the following:
1) Add: U + V is also a vector 2) Multiply: a U (should be read "a times U") is also a vector 3) Combine the two operations: a U + b V is also a vector (such an expression is called a linear combination of U and V) a (U + V) = a U + b V is also a vector (a + b) U = a U + b U is also a vector (both of these are called distributivity)
The examples that you've seen so far are:
1) Geometric vectors These are arrows of fixed length and direction. You've probably already learned how to add them, just stick the second arrow at the tip of the first one and draw an arrow from the base of the first arrow to the tip of the second one. Multiplying them by numbers is also easy, just extend or shrink the arrow along its direction.
2) Column vectors These are columns of numbers usually written with either square [ ] or round ( ) brackets around them. To add two column vectors with the same number of entries, just add them one by one. To multiply by a number, just multiply each entry in the column individually.
3) Functions If I have two functions f(x) and f(y), then their sum is also a function. If I multiply f(x) by a number b, then b f(x) (b times f(x)) is still a function.
It turns out that if you only care about the abstract properties of vectors that I described above, then many different kinds of vectors can be treated as identical. For example, working with column vectors are very convenient, so often people translate their vector problems purely into the language of column vectors. In fact, they are so convenient to work with, that some people forget that there are other kinds of vectors. Which is unfortunate, because there are some things that are not particularly convenient to do in the language of column vectors.
In quantum mechanics, almost every classical state is promoted to a vector. If X denotes the state of a particle being at point x and Y denotes the state of a particle being at point y, then in quantum mechanics any state of the form a X + b Y is also allowed, even though X and Y are exclusive classically. If you've ever heard of Schroedinger's cat, it is an example of so called superposition of states. If D represents a state of the cat being dead and A represents the state of the cat being alive, then in quantum mechanics a state of the form a D + b A is as natural as a vector a N + b E that points somewhere in between North and East. Don't pay attention to all the brouhaha surrounding Schroedinger's cat, remember that it's just an illustration.
Linear algebra can be quite fun to study, especially since it's so useful. I will second other suggestions in this thread that you go to the library and pick up a book on elementary linear algebra. There you will learn about things such as vector spaces, linear equations, functions between vector spaces (commonly called matrices), linear forms, bilinear forms, etc.
Finally, I'll expand on my explanation of entanglement. First a note about tensor products. It is sometimes convenient (or even necessary) to introduce a product between vectors themselves (note that this operation was not defined for abstract vectors I described above). However, usually if you take the tensor product (or just product) of two vectors, you don't get the same kind of vector back, but a different kind. The example of multiplication of functions from my previous post is a good one. A function f(x) is one kind of vector, a function g(y) is another kind of vector, while their product h(x,y) = f(x) g(y) is yet a third kind of vector. The first two were functions of one variable (but different variables), while the third one is a function of both variables at the same time.
Back to quantum mechanics. Consider a light bulb, if classically it only has two states N = on and F = off, then the quantum analog will have many possible states, each of the form a N + b F where a and b are some numbers. If we have two light bulbs let N1 and F1 describe the states of the first one and N2 and F2 describe the sates of the second one. Then to consider possible states of both bulbs together we must have the states N1 N2, N1 F2, F1 N2, and F1 F2. You can look at the joint states as products of individual states. For example, N1 N2 can be read as "N1 times N2" or "N1 tensor N2". Now, the quantum version of these light bulbs will in general be a linear combination of these four sates
a N1 N2 + b N1 F2 + c F1 N2 + d F1 F2.
One way to produce joint quantum states is two take two individual light bulb states and multiply them like we did for the classical states (just following standard rules of multiplication):
(e N1 + f F1) (g N2 + h F2) = eg N1 N2 + eh N1 F2 + fg F1 N2 + fh F1 F2.
Now, here's a puzzle for you. Can you find a joint quantum state described by numbers a, b, c, d that *cannot* be written as a product of two individual states, say (e N1 + f F1) and (g N2 + h F2)? If yes, then you've found an entangled state of the two light bulb system.
Hope this helps.
Igor
whopkins@csd.uwm.edu - 23 Mar 2005 01:38 GMT >From Sci~Girl <palmtree...@juno.com> >All I know in math is basic algebra though, I'm 14 and in 8th grade but >in 9th grade math. I describe the situation in Bell's theorem in clear terms. news://sci.physics 2005, January 5 Alain Aspect Non-Locality/Entanglement experiment http://groups-beta.google.com/groups?q=whopkins+entanglement+DIFFERENT+questions
Oz - 23 Mar 2005 21:52 GMT whopkins@csd.uwm.edu writes
>I describe the situation in Bell's theorem in clear terms. >news://sci.physics >2005, January 5 >Alain Aspect Non-Locality/Entanglement experiment >http://groups-beta.google.com/groups?q=whopkins+entanglement+DIFFERENT+questions Gosh, so you do.
I don't know how I missed this at the time because its very simple and clear.
===========And Hartley said: Independent particles are like independent coin flips. Entangled particles are like coin flips that are not independent.
We use the word "independent" in both quantum theory and probability, but when it is absent we use two different words "entangled" and "correlated". I'm not sure why (there is *some* logic to it). ============
So it would seem to me that my visualisation of an entangled particle as a *single* particle is not unreasonable. Its just a very extended particle but I have no problem with that since I see extended particles all over the place and particularly photons with 2000km wavelengths. An entangled pair is thus just a particularly interesting wavefunction with TWO localised maxima of increasing separation. And why not?
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Ralph Hartley - 24 Mar 2005 08:10 GMT > So it would seem to me that my visualisation of an entangled particle as > a *single* particle is not unreasonable. Its just a very extended > particle but I have no problem with that since I see extended particles > all over the place and particularly photons with 2000km wavelengths. An > entangled pair is thus just a particularly interesting wavefunction with > TWO localised maxima of increasing separation. And why not? No! Two particles, one distribution (wavefunction).
If I tossed two coins at the same time, and it was guaranteed that they either would land both heads, or both tails (e.g. because I was cheating), would you say there was only one coin?
Of course not. You would say that the coin flips were not independent.
Two photons are two photons, not one.
I can describe the probabilities of a coin flip with a function of one (boolean) variable P(X).
I can use a linear table:
Table (1) X | P(X) ------------- Heads | 0.5 Tails | 0.5
Suppose I toss a second coin and call the result Y. I can describe that the same way:
Table (2) Y | P(Y) ------------- Heads | 0.5 Tails | 0.5
But does that tell you everything you might want to know? No it doesn't. You might want to know about combinations of the two flips. For that I would need a function of two variables P(X,Y).
I would need a square table:
Table (3) Y P(X,Y)| Heads | Tails ---------------------- X Heads | 0.25 | 0.25 Tails | 0.25 | 0.25
The table I just gave is a special case where P(X,Y) = P(X)P(Y). When that is true we call X and Y independent. Instead of table (3) I could have given you tables (1) and (2) and told you that X and Y are independent.
But suppose instead that the probabilities were given by
Table (4) Y P(X,Y)| Heads | Tails ---------------------- X Heads | 0.5 | 0.0 Tails | 0.0 | 0.5
If you look at X and Y separately, you will see that Tables (1) and (2) are still correct, but you would be unwise to bet me that the two coins would be different.
Two photons are just the same. They *always* have *two* positions, spins etc. Sometimes they may be independent. Then you can get away with describing them separately. You still can use the full joint distribution, but you don't *have* to.
If they are entangled, You *do* have to use the full joint distribution, because describing the particles separately doesn't tell you everything you want to know.
Ralph Hartley
Oz - 25 Mar 2005 12:08 GMT Ralph Hartley <hartley@aic.nrl.navy.mil> writes
>Oz wrote: >> So it would seem to me that my visualisation of an entangled particle as [quoted text clipped - 9 lines] >either would land both heads, or both tails (e.g. because I was >cheating), would you say there was only one coin? Of course.
>Of course not. You would say that the coin flips were not independent. No, I would claim that that was indistinguishable from a body rigidly holding the pennies both facing the same way.
>Two photons are two photons, not one. Not if they are not independent.
>I can describe the probabilities of a coin flip with a function of one >(boolean) variable P(X).
>Table (1) > X | P(X) >------------- >Heads | 0.5 >Tails | 0.5 snip
>Table (3) > Y > P(X,Y)| Heads | Tails > ---------------------- >X Heads | 0.25 | 0.25 > Tails | 0.25 | 0.25 snip
>But suppose instead that the probabilities were given by > [quoted text clipped - 8 lines] >are still correct, but you would be unwise to bet me that the two coins >would be different. Quite. Note though the structural similarity to table (1) Its IDENTICAL except I redefine head = (head+head) and tail = (tail+tail)
>Table (1) > X | P(X) >------------- >Heads | 0.5 >Tails | 0.5 These are thus (in probability terms) identical and so the particle number (ie divisibility) is the same.
What happens after is not important. By chopping the pennies in table one into two identical halves AFTER the throw, I can have two pairs of identical particles and I could in fact do many such choppings up (for a circularly symmetrical penny).
>Two photons are just the same. They *always* have *two* positions, spins >etc. Sometimes they may be independent. Then you can get away with [quoted text clipped - 4 lines] >because describing the particles separately doesn't tell you everything >you want to know. Yes, but the result is a single particle (OK its a TAD more) and behaves like one. A 'double photon'.
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whopkins@csd.uwm.edu - 29 Mar 2005 01:49 GMT > All I know in math is basic algebra though, I'm 14 and in 8th grade but > in 9th grade math. What's a vector? "Mark's Elements" Me news://sci.math.research I. June 22, 1996 II. June 24, 1996 III. June 25, 1996 http://groups-beta.google.com/groups?q=%22Mark%27s+Elements%22
(Part III is an axiomatization of Minkowski space, the earlier parts deal with the formal definition of a vector space)
An updated (and corrected and formatted, the definition of Hilbert space needs to be redone) copy of this will appear in the archive I mentioned previously ... along with more accessible writeups which show how vector algebra and vector calculus works on the context of formulating and solving geometric or trigonometric problems.
(For those already familiar with Calculus: the development of Complete Metric spaces in Mark's Elements uses NO EPSILON-DELTAS. Instead, an alternative concept -- the "Zeno sequence" is employed, which is more transparent and even constructive.)
For those interested in gauge theory, there will also be material which provides a substantial reworking of the theory if principal fibre bundles, using a new concept (the quotient operation) to significantly simplify the subject and render it transparent.
Also of interest: "The Absolute: Intro to Relativity" Me June 13 - August 24, 1995 news://sci.physics http://groups-beta.google.com/groups?q=%22The+Absolute%22+%22Intro+To+Relativity%22
"TIME IS SPACE -- Part 2/5: The Hudgin Axioms" Me June 13, 1995 news://sci.physics http://groups-beta.google.com/groups?q=%22The+Hudgin+Axioms%22
Early precursor to the material in Part III of Mark's Elements.
bjflanagan - 18 Mar 2005 19:29 GMT You might have a look at 'Mathematics of Classical & Quantum Physics,' by Byron & Fuller. Meanwhile, let me try to provide a bit of motivation by pointing out that, in quantum theory, every physical thing can be described, in principle, by a wave function that is (more or less) mathematically equivalent to a vector in what is called Hilbert space, after the profoundly influential David Hilbert, who, along with Felix Klein, assembled one of the most dazzling array of minds ever known, in Gottingen in the late 19th and early 20th centuries. (See the masterful 'Hilbert' by Constance Reid.)
This is how Byron & Fuller present the standard wisdom: "Any physical system is completely described by a normalized vector (the state vector or wave function) in Hilbert space. All possible information about the system can be derived from this state vector by rules..."
Here are some more references:
http://www.geometry.net/scientists_bk/hilbert_david.html
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Klein.html)
___________________________
Here's an excerpt from a fairly recent article by Asher Peres (who is one of the elder statesmen of physics today), which ties this issue into "EPR" and the famous Einstein-Bohr debates. Einstein said that the entire future history of physics would revolve around the resolution of this issue. In this connection it is quite exciting to note that three of our most prominent contemporary theorists -- Hartle, Smolin and 't Hooft -- have recently revived the "hidden variables" debate.
"That inequality was originally derived for solving a completely different problem: is quantum theory compatible with an underlying pseudo-classical "subquantum" back-ground (deterministic or possibly stochastic)? Such a post-quantum theory would pre-sumably involve additional "hidden" variables, and the statistical predictions of ordinary quantum theory would be reproduced by performing suitable averages over these hidden variables. Bell [4] was the first to show that if the constraint of locality is imposed on the hidden variables (namely, if the hidden variables of two distant quantum systems are themselves separable into two distinct subsets), then there is an upper bound to the correlations of results of measurements that can be performed on the two distant systems."
arXiv:quant-ph/9707026 v1
(JS Bell was a theorist at CERN who also believed that there "must be hidden variables" (private communication)).
_____________________
EPR:
http://plato.stanford.edu/entries/qt-measurement/
Sci~Girl - 18 Mar 2005 19:39 GMT > "At the most simple level (and someone else feel free to jump in if I'm > not saying this well), you can think of entangled particles as particles > that share some net property, but individually don't have any definite > value of that property."
> The definitions of entanglement you've been given would probably be > difficult for you to understand without an ungodly amount of additional > explanation. I was able to pull out that there's some similarity between the particles, but that was about it. That helps a lot, although of course I have questions as always. Unfortunately I don't think I'd be able to understand the answers.
Vincent N. Virgilio - 22 Mar 2005 09:21 GMT > I was able to pull out that there's some similarity between the > particles, but that was about it. That helps a lot, although of course > I have questions as always. Unfortunately I don't think I'd be able to > understand the answers. Beware that I am an amateur in this forum . . .
Entanglement reminds me of correlated random variables. In very plain words: if you learn something about one variable (photon), you learn at least a little about the other. I think this is not so with uncorrelated variables (or dis-entangled photons); whatever you know about one of the variables (photons) tells you nothing about the other. At least in this context. (Side note: Dependence and independence is a very important idea in math and science; even in the linear algebra others have referred to in this thread. Example, vectors can be dependent or independent with respect to each other.)
It is as you say, "Similarity between the particles"; and a measureable one at that.
Corrections are welcome.
Vince Virgilio
Ralph Hartley - 23 Mar 2005 11:33 GMT >>I was able to pull out that there's some similarity between the >>particles, but that was about it. That helps a lot, although of course >>I have questions as always. Unfortunately I don't think I'd be able to >>understand the answers.
> Entanglement reminds me of correlated random variables. In very plain > words: if you learn something about one variable (photon), you learn at > least a little about the other. I think this is not so with > uncorrelated variables (or dis-entangled photons); whatever you know > about one of the variables (photons) tells you nothing about the other. This is exactly correct!
Quantum theory is just the square root of probability theory.
Most people have some idea what a probability is. A quantum "amplitude" is very similar, but it can be negative (footnote 1). You square an amplitude to get a probability, but always do that as the *last* step in a calculation. All the other rules for calculating probabilities are the same! (footnote 2)
Independent particles are like independent coin flips. Entangled particles are like coin flips that are not independent.
We use the word "independent" in both quantum theory and probability, but when it is absent we use two different words "entangled" and "correlated". I'm not sure why (there is *some* logic to it).
Just to make things trickier, the term "linearly independent" uses the word "independent" in *another* sense. Sometimes in the same sentence.
Entangled systems can do things that don't make sense in terms of correlated probabilities, because positive and negative amplitudes can cancel out when they are added. That can't happen with probabilities because they are all positive.
1) In physics quantum amplitudes are usually complex numbers. That makes the equations simpler, because every number has a square root etc., but makes no difference at this level of detail.
2) The basic rules are:
Divide all the ways things can happen into independent and/or mutually exclusive cases (there may be more than one way to do that, but it won't affect the answer).
For independent events A and B, to get the probability (amplitude) of A *and* B happening, *multiply* the probabilities (amplitudes).
For mutually exclusive events C and D, to get the probability (amplitude) of C *or* D happening, *add* the probabilities (amplitudes).
Ralph Hartley
Sci~Girl - 18 Mar 2005 19:39 GMT >In many ways the latter behaves as a single particle containing both >spin up and spin down (which isn't zero).
>You can have a pair of particles that is in a very real sense BOTH >spin up and spin down. This is NOT the same as having a pair with >one spin up and one spin down.
That's... hard to comprehend... both up and down but not zero? If it wasn't zero, either the up or the down would have to be dominant and then it would be called just up or just down... Well, I know that's not right, but I have no other way to think until I know more.
In order to truly understand quantum mechanics I'd need to understand calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS. It would be okay to just borrow a textbook if it was, say, next year's geometry I was trying to get to, but it's much farther away. I'd have to read the geometry book, then the algebra 2 book, then the advanced math and trigonometry book, before I even arrived at differential calculus. And then there's integral... and I don't even know what that means. I'm sure it would be exciting and I've thought about doing it before, but it just doesn't seem like something within my capacity to do.
Arnold Neumaier - 19 Mar 2005 09:00 GMT (was Teleportation of photons using entanglement)
> In order to truly understand quantum mechanics I'd need to understand > calculus. You can understand some part of quantum mechanics already without calculus, namely everything involving entanglement, quantum cryptography, and the like. This only needs linear algebra, which may be easier. (On the other hand, calculus is not really difficult either, once one gets used to it.) Linear algebra (i.e., vectors and matrices) is more fundamental to quantum mechanics than calculus.
> I HAVE THREE YEARS TO GO BEFORE CALCULUS. It would be okay to > just borrow a textbook if it was, say, next year's geometry I was [quoted text clipped - 4 lines] > sure it would be exciting and I've thought about doing it before, but > it just doesn't seem like something within my capacity to do. If you like math it is much less work than you might think, and it is fun! Just start with next years textbook and read it in your spare time! I started reading math beyond my age when I was 12, and didn't regret it.
With the right motivation, you can learn 10 times as fast as when you just wait till the subject comes up in school!
You don't need to do all the exercises but just enough that you think you know how it works. Go back to practicing more if you need it. This speeds up things a lot. And you don't need to read everything in the order it is in the book - just go where your curiosity leads you, and if you encounter things you don't know yet, go back to where it was introduced. In this way you get the idea of what is happening long before you understand it thoroughly, and it will be a motivation to learn the missing things.
Learning math and physics is a life-long challenge (so much intersting stuff accumulated over the centuries...), and you can't start early enough.
And at any time in life there will be parts you understand well, parts you understand partly or superficially only, and parts where you know little more than a few buzz words. So you need not aim at understand everything fully on first acquaintance, but learn whatever you can in whatever order you pick it up. The stuff to be practiced and learnt well is only the part that comes up over and over again, and when you notice that you know what to learn, and you quickly see how to do it!
Arnold Neumaier
John Forkosh - 20 Mar 2005 08:40 GMT : (was Teleportation of photons using entanglement) : [quoted text clipped - 17 lines] : > sure it would be exciting and I've thought about doing it before, but : > it just doesn't seem like something within my capacity to do. You might try taking a look at, Quantum Mechanics in Simple Matrix Form, Thomas E. Jordan John Wiley & Sons, 1986, ISBN 0-471-81751-1 In its preface the book says "it assumes only basic algebra", which is true, although it assumes a facility with basic algebra that may require a bit of effort in 8th grade. Matrices (and vectors) are explained from scratch, so you needn't worry about that. The book was reviewed on page 1154 of the American Journal of Physics, Volume 54, Number 12, December 1986.
: If you like math it is much less work than you might think, : and it is fun! Just start with next years textbook and read it [quoted text clipped - 28 lines] : : Arnold Neumaier  Signature John Forkosh ( mailto: j@f.com where j=john and f=forkosh )
Ralph Hartley - 22 Mar 2005 20:44 GMT >>In order to truly understand quantum mechanics I'd need to understand >>calculus. [quoted text clipped - 6 lines] > Linear algebra (i.e., vectors and matrices) is more fundamental > to quantum mechanics than calculus. This is one of my biggest problems with how mathematics is taught (at least in the US, is it the same elsewhere?).
Calculus (even multi variable) is taught first, often in high school nowadays, and *then* linear algebra. This results in many people learning calculus, but no linear algebra.
I think this is backwards, because linear algebra is both easier and more useful than calculus.
Quantum mechanics is not the only field in which calculus is used but linear algebra is fundamental.
In an ideal world, one would learn both, but seriously, what percentage of college graduates *ever* use calculus again? Many of those, I suspect, would have a use for linear algebra, if they had ever learned any. How many of those would *actually* apply math to a problem that needs it? I don't know. You can lead a horse to water ...
I should confess that I have spoken about this to a woman who succeeded in learning calculus, but failed to learn linear algebra, so your mileage may vary.
If you are going to read math "outside" class (we know you are actually inside class when you do it, no one is fooled, we remember), read linear algebra.
Easy, useful, your competitors aren't learning it.
Ralph Hartley
Gerard Westendorp - 23 Mar 2005 01:38 GMT You might want to check out this site:
http://www.phys.uu.nl/~thooft/theorist.html
It is 't Hooft's attempt to help people like you.
Gerasd
Sci~Girl - 25 Mar 2005 23:50 GMT If you like math it is much less work than you might think, and it is fun! Just start with next years textbook and read it in your spare time! I started reading math beyond my age when I was 12, and didn't regret it.
With the right motivation, you can learn 10 times as fast as when you just wait till the subject comes up in school!
Yep that's how I learned physics, and how I'm learning German. Maybe I'll try. I read the rest of this year's textbook, so I know linear algebra... I think.
Thomas Palm - 23 Mar 2005 11:34 GMT "Sci~Girl" <palmtree117@juno.com> wrote in news:1111019108.240443.118670 @o13g2000cwo.googlegroups.com:
> In order to truly understand quantum mechanics I'd need to understand > calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS. It would be okay to [quoted text clipped - 5 lines] > sure it would be exciting and I've thought about doing it before, but > it just doesn't seem like something within my capacity to do. You could try to read Feynmann's "QED The strange theory of light and matter". It was written based on a series of lectures he held trying to explain quantum mechanics to an audience of laymen and contains essentially no math and yet outlines quantum mechanics at a rather deep level. The problem with it is that the level is so deep that it is hard to connect it to the elementary stuff you encounter elsewhere. But it's a very short book so give it a try.
Mike Stay - 23 Mar 2005 11:35 GMT > >In many ways the latter behaves as a single particle containing both > >spin up and spin down (which isn't zero). [quoted text clipped - 7 lines] > then it would be called just up or just down... Well, I know that's not > right, but I have no other way to think until I know more. That's because in "Hilbert space", "spin up" and "spin down" are at right angles to each other! When you measure a quantum system, all the possible answers are at right angles to each other. Here's a physical example:
You can measure the polarization of photons with polarized sunglasses. Light that reflects off the surface of a pool ("glare"), for instance, is polarized parallel to the surface. Polarized sunglasses are aligned such that only light perpendicular to the surface passes through (well, assuming you're upright when you wear them ;) so you can see down into the pool.
The two results, parallel and perpendicular, are at right angles. These form what's called a "basis". By taking different combinations of parallel and perpendicular (the "basis vectors"), you can get diagonally polarized light, and even left- and right-handed polarizations. The set of all the different possible polarizations is called the "Hilbert space".
When you have 45-degree polarized light, a photon will be absorbed half of the time and transmitted half of the time. To figure out what the probability is for other angles, you'll need to do some trig. The light itself lays on the hypotenuse of a triangle. The hypotenuse is one unit long; the other two sides are on the basis vectors. You just square the length of the sides to figure out what the probability is that it will be transmitted or absorbed.
For example, if you have light at 36.8 degrees from parallel (a 3-4-5 triangle), then the probability that it gets absorbed is (4/5)^2 = 0.64 and the probability that it gets transmitted is (3/5)^2 = 0.36 .
Spin and polarization are two names for the exact same thing. "Spin up" and "spin down" are at right angles, just like parallel and perpendicular. To measure the polarization (or "spin") of an electron, you use a magnet instead of a polarized filter. The terms "up" and "down" come from the model of an electron as a spinning ball, which turns out to have some serious problems. You could just as well talk about the electron being polarized parallel or perpendicular to the magnetic field.
To give an example of entanglement, consider a "polarizing beam splitter" (PBS): this is a mirror that's reflective to parallel light, but transparent to perpendicular light.
Now, if we send a diagonally polarized photon into the PBS, then there's a chance it will go through the PBS and a chance it'll reflect off of it. If you detect it behind the PBS, then you also know that it's perpendicularly polarized without even measuring it. The photon's position and polarization are entangled.
Now comes the really wierd part. Consider a pair of photons. There are *four* different basis vectors for this pair of photons:
para*para para*perp perp*para perp*perp
You can combine any of these in the same way that you combined parallel and perpendicular to get diagonal. For instance, we can take
para*para + para*perp
in equal amounts, and say that the first photon is polarized parallel, while the second one is 45-degree diagonal. These two photons are "separable", because we can describe them independently, i.e. we could write them as
para*(para+perp)
Two 45-degree photons use all of the basis vectors:
(para+perp)*(para+perp) = para*para + para*perp + perp*para + perp*perp
But we can also consider a pair of photons whose polarizations are entagled:
para*perp + perp*para
There's no way to write this as (a)*(b), so these are entangled. You might get either polarization when you measure one of them, but whatever you got, you know what the other one will be.
There's a special crystal, called a "down-conversion" crystal that takes an ultraviolet photon and emits two green photons that have polarizations related like this. -- Mike Stay
Sci~Girl - 25 Mar 2005 23:37 GMT [ Quotation and references modified by moderator. Please quote and attribute properly in the future. -ik ]
> That's because in "Hilbert space", "spin up" and "spin down" are at > right angles to each other! When you measure a quantum system, all > the possible answers are at right angles to each other. Here's a > physical example: Holy cow. Ok now it makes a little more sense. To go back to the entanglement though - so how did they teleport a whole laser beam? Entangling all the photons?
Mike Stay - 29 Mar 2005 09:21 GMT > > That's because in "Hilbert space", "spin up" and "spin down" are at > > right angles to each other! When you measure a quantum system, all [quoted text clipped - 4 lines] > entanglement though - so how did they teleport a whole laser beam? > Entangling all the photons? Yep. They shone one UV laser through the down-conversion crystal I mentioned, and got two entangled lasers coming out.
-- Mike
Oz - 19 Mar 2005 09:06 GMT Sci~Girl <palmtree117@juno.com> writes
>>In many ways the latter behaves as a single particle containing both >>spin up and spin down (which isn't zero). [quoted text clipped - 6 lines] >wasn't zero, either the up or the down would have to be dominant and >then it would be called just up or just down... Quite. Quantum mechanics is a wonderful thing. Its the sort of thing that keeps physics really interesting, even to people who have studied it for a lifetime.
There is no particular reason why the universe should behave intuitively, after all we see only a tiny subset of the universe. This can be roughly described as that of billions of particles going very slowly. One should not be surprised that a few particles going very fast behaves rather differently.
Lucky you, you will be taught by people who have been taught by a generation of teachers very familiar with it.
>Well, I know that's not >right, but I have no other way to think until I know more. In many ways its best to accept these (and many other magical) things just DO happen that way, and let your head get round it in its own time. The results of a well chosen selection of experiments helps here. Do not be surprised if some seem contradictory, quantum mechanics can handle them. Just be thankful a couple of generations of physicists racked their brains trying to make sense of it for you.
>In order to truly understand quantum mechanics I'd need to understand >calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS. Sure. You will find it fun.
>It would be okay to >just borrow a textbook if it was, say, next year's geometry I was >trying to get to, but it's much farther away. I'd have to read the >geometry book, then the algebra 2 book, then the advanced math and >trigonometry book, before I even arrived at differential calculus. And >then there's integral... and I don't even know what that means. Yup. Loads of really scary stuff designed to put the fear of god into the uninitiated. Not so hard when you get there though. The really good thing is that for an ignorant antique like myself it seems to never end. I can't understand most of what gets posted here, but it sure is fascinating. You, at least, have a lifetime to get stuck into these new mathematical tools. In ten years (with some work) most of the esoteric stuff you read here will seem quite ordinary, accurate and precise. I'm jealous.
>I'm >sure it would be exciting and I've thought about doing it before, but >it just doesn't seem like something within my capacity to do. At 14? Well you could. Unfortunately you could really easily confuse yourself and make something easy into something difficult. It would take quite a bit of your spare time up. It might be better to spend some of that spare time 'socialising' or playing football.
 Signature Oz This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions]. BTOPENWORLD address has ceased. DEMON address has ceased.
Sci~Girl - 26 Mar 2005 09:42 GMT At 14? Well you could. Unfortunately you could really easily confuse yourself and make something easy into something difficult. It would take quite a bit of your spare time up. It might be better to spend some of that spare time 'socialising' or playing football.
Make something easy into something difficult? Ugh, I've done that so many times it drives me crazy. I already have independent study projects to do in a couple classes so I don't know if I will have the time.
Sci~Girl - 26 Mar 2005 09:43 GMT >Yep that's how I learned physics, and how I'm learning German. Maybe >I'll try. I read the rest of this year's textbook, so I know linear algebra... I think.
Oh wait, no, not linear algebra, just plain linear functions from basic algebra (had to correct myself) No vectors yet...
whopkins@csd.uwm.edu - 29 Mar 2005 09:22 GMT > In order to truly understand quantum mechanics I'd need to understand > calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS. "The Untold Story of Calculus" Me June 17, 2002 news://sci.math http://groups-beta.google.com/groups?q=%22The+Untold+Story+Of+Calculus%22
"A Trig Primer" Me January 14, 2002 news://sci.math http://groups-beta.google.com/groups?q=%22A+Trig+Primer%22+whopkins
"A Complete Introduction to Calculus" Me October 27, 1995 news://sci.math http://groups-beta.google.com/groups?q=%22Complete+Introduction+To+Calculus%22
.. or, if you're willing to wait, I'm in the process of establishing an extensive archive which will have better typeset versions of all these, as well as what will probably end up being upwards of 1000 papers, monographs and expositories from me, in mathematics and physics and other areas. Stay tuned for future developments...
whopkins@csd.uwm.edu - 29 Mar 2005 09:25 GMT > All I know in math is basic algebra though, I'm 14 and in 8th grade but > in 9th grade math. What's a vector? "Mark's Elements" Me news://sci.math.research I. June 22, 1996 II. June 24, 1996 III. June 25, 1996 http://groups-beta.google.com/groups?q=%22Mark%27s+Elements%22
(Part III is an axiomatization of Minkowski space, the earlier parts deal with the formal definition of a vector space)
An updated (and corrected and formatted, the definition of Hilbert space needs to be redone) copy of this will appear in the archive I mentioned previously ... along with more accessible writeups which show how vector algebra and vector calculus works on the context of formulating and solving geometric or trigonometric problems.
(For those already familiar with Calculus: the development of Complete Metric spaces in Mark's Elements uses NO EPSILON-DELTAS. Instead, an alternative concept -- the "Zeno sequence" is employed, which is more transparent and even constructive.)
For those interested in gauge theory, there will also be material which provides a substantial reworking of the theory if principal fibre bundles, using a new concept (the quotient operation) to significantly simplify the subject and render it transparent.
Also of interest: "The Absolute: Intro to Relativity" Me June 13 - August 24, 1995 news://sci.physics http://groups-beta.google.com/groups?q=%22The+Absolute%22+%22Intro+To+Relativity%22
"TIME IS SPACE -- Part 2/5: The Hudgin Axioms" Me June 13, 1995 news://sci.physics http://groups-beta.google.com/groups?q=%22The+Hudgin+Axioms%22
Early precursor to the material in Part III of Mark's Elements.
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