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#11
05-04-2007, 01:08 AM
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Re: The Uncertainty Principle
[ QUOTE ]
So particles are really non commuting hermitian operators? PairTheBoard [/ QUOTE ] Not sure if you are being serious but observable quantities are hermitian operators. Particles are describes by normalized wave functions. You can find something like the expectation value of the position by using the postion operator and the wavefunction with its complex conjugate. <W| x | W> is how it would look in the notation most commonly used. 
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#12
05-04-2007, 01:48 AM
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Re: The Uncertainty Principle
[ QUOTE ]
So particles are really non commuting hermitian operators? [/ QUOTE ] As I understand it, the state of a particle (or system) is given by a unit vector in a Hilbert space. The "properties" of the particle (a.k.a. the observables) are modeled by self-adjoint operators from the Hilbert space to itself. If the Hilbert space is finite-dimensional, then a self-adjoint operator can be decomposed as A = \sum L_j P_j, where L_j are the eigenvalues and P_j are the orthogonal projection operators onto the corresponding eigenspaces. This is the spectral decomposition. The translation is that, if you measure observable A and the particle is in state x, then the measurement will result in the outcome L_j with probability ||P_j x||^2. The mean of such a measurement can be calculated and is <Ax,x>, where <.,.> is the inner product in the Hilbert space. The variance can also be calculated and is ||Ax||^2 - |<Ax,x>|^2. The variance is sometimes called the "dispersion" and I have seen the notation D(A|x) used for it. If A and B are two self-adjoint operators, then D(A|x)*D(B|x) >= (1/4)*|<(AB - BA)x,x>|^2. If A and B do not commute, then there are states x for which the right-hand side is strictly positive. In some cases, AB - BA = cI for some constant c. Since x is always a unit vector, the right-hand side then simplifies to (1/4)*|c|^2 for all x. In infinite dimensional Hilbert spaces, the idea is similar but there are serious additional technicalities. In particular, the observables are typically not bounded operators, and their domains are proper subsets of the Hilbert space. Moreover, their spectrum need not be discrete, so their spectral decomposition may involve something called a projection-valued measure.
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#14
05-04-2007, 04:05 AM
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Re: The Uncertainty Principle
[ QUOTE ]
[ QUOTE ] So particles are really non commuting hermitian operators? [/ QUOTE ] As I understand it, the state of a particle (or system) is given by a unit vector in a Hilbert space. The "properties" of the particle (a.k.a. the observables) are modeled by self-adjoint operators from the Hilbert space to itself. If the Hilbert space is finite-dimensional, then a self-adjoint operator can be decomposed as A = \sum L_j P_j, where L_j are the eigenvalues and P_j are the orthogonal projection operators onto the corresponding eigenspaces. This is the spectral decomposition. The translation is that, if you measure observable A and the particle is in state x, then the measurement will result in the outcome L_j with probability ||P_j x||^2. The mean of such a measurement can be calculated and is <Ax,x>, where <.,.> is the inner product in the Hilbert space. The variance can also be calculated and is ||Ax||^2 - |<Ax,x>|^2. The variance is sometimes called the "dispersion" and I have seen the notation D(A|x) used for it. If A and B are two self-adjoint operators, then D(A|x)*D(B|x) >= (1/4)*|<(AB - BA)x,x>|^2. If A and B do not commute, then there are states x for which the right-hand side is strictly positive. In some cases, AB - BA = cI for some constant c. Since x is always a unit vector, the right-hand side then simplifies to (1/4)*|c|^2 for all x. In infinite dimensional Hilbert spaces, the idea is similar but there are serious additional technicalities. In particular, the observables are typically not bounded operators, and their domains are proper subsets of the Hilbert space. Moreover, their spectrum need not be discrete, so their spectral decomposition may involve something called a projection-valued measure. [/ QUOTE ] ok. An observable is modeled by a self adjoint operator on the Hilbert Space. It seems like there would be a lot of self adjoint operators on the Hilbert Space. But there are only a few observables. Are all the self adjoint operators on the Hilbert Space some kind of observables? If not, what do they make of the ones that are not? And how do they decide which ones are and which ones aren't? If all of the self adjoint operators are observables, why are there so few self adjoint operators? PairTheBoard
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#15
05-04-2007, 04:33 AM
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Re: The Uncertainty Principle
[ QUOTE ]
[ QUOTE ] [ QUOTE ] So particles are really non commuting hermitian operators? [/ QUOTE ] As I understand it, the state of a particle (or system) is given by a unit vector in a Hilbert space. The "properties" of the particle (a.k.a. the observables) are modeled by self-adjoint operators from the Hilbert space to itself. If the Hilbert space is finite-dimensional, then a self-adjoint operator can be decomposed as A = \sum L_j P_j, where L_j are the eigenvalues and P_j are the orthogonal projection operators onto the corresponding eigenspaces. This is the spectral decomposition. The translation is that, if you measure observable A and the particle is in state x, then the measurement will result in the outcome L_j with probability ||P_j x||^2. The mean of such a measurement can be calculated and is <Ax,x>, where <.,.> is the inner product in the Hilbert space. The variance can also be calculated and is ||Ax||^2 - |<Ax,x>|^2. The variance is sometimes called the "dispersion" and I have seen the notation D(A|x) used for it. If A and B are two self-adjoint operators, then D(A|x)*D(B|x) >= (1/4)*|<(AB - BA)x,x>|^2. If A and B do not commute, then there are states x for which the right-hand side is strictly positive. In some cases, AB - BA = cI for some constant c. Since x is always a unit vector, the right-hand side then simplifies to (1/4)*|c|^2 for all x. In infinite dimensional Hilbert spaces, the idea is similar but there are serious additional technicalities. In particular, the observables are typically not bounded operators, and their domains are proper subsets of the Hilbert space. Moreover, their spectrum need not be discrete, so their spectral decomposition may involve something called a projection-valued measure. [/ QUOTE ] ok. An observable is modeled by a self adjoint operator on the Hilbert Space. It seems like there would be a lot of self adjoint operators on the Hilbert Space. But there are only a few observables. Are all the self adjoint operators on the Hilbert Space some kind of observables? If not, what do they make of the ones that are not? And how do they decide which ones are and which ones aren't? If all of the self adjoint operators are observables, why are there so few self adjoint operators? PairTheBoard [/ QUOTE ] All self adjoint operators are observable. As to why there are so few of them, my guess would be that many of the ones you could think of would be superpostions of the common ones. For instance the x+y cordinate would be a self adjoint operator so thus an observable but would be pretty useless.
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#16
05-04-2007, 05:15 AM
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Re: The Uncertainty Principle
[ QUOTE ]
[ QUOTE ] [ QUOTE ] [ QUOTE ] So particles are really non commuting hermitian operators? [/ QUOTE ] As I understand it, the state of a particle (or system) is given by a unit vector in a Hilbert space. The "properties" of the particle (a.k.a. the observables) are modeled by self-adjoint operators from the Hilbert space to itself. If the Hilbert space is finite-dimensional, then a self-adjoint operator can be decomposed as A = \sum L_j P_j, where L_j are the eigenvalues and P_j are the orthogonal projection operators onto the corresponding eigenspaces. This is the spectral decomposition. The translation is that, if you measure observable A and the particle is in state x, then the measurement will result in the outcome L_j with probability ||P_j x||^2. The mean of such a measurement can be calculated and is <Ax,x>, where <.,.> is the inner product in the Hilbert space. The variance can also be calculated and is ||Ax||^2 - |<Ax,x>|^2. The variance is sometimes called the "dispersion" and I have seen the notation D(A|x) used for it. If A and B are two self-adjoint operators, then D(A|x)*D(B|x) >= (1/4)*|<(AB - BA)x,x>|^2. If A and B do not commute, then there are states x for which the right-hand side is strictly positive. In some cases, AB - BA = cI for some constant c. Since x is always a unit vector, the right-hand side then simplifies to (1/4)*|c|^2 for all x. In infinite dimensional Hilbert spaces, the idea is similar but there are serious additional technicalities. In particular, the observables are typically not bounded operators, and their domains are proper subsets of the Hilbert space. Moreover, their spectrum need not be discrete, so their spectral decomposition may involve something called a projection-valued measure. [/ QUOTE ] ok. An observable is modeled by a self adjoint operator on the Hilbert Space. It seems like there would be a lot of self adjoint operators on the Hilbert Space. But there are only a few observables. Are all the self adjoint operators on the Hilbert Space some kind of observables? If not, what do they make of the ones that are not? And how do they decide which ones are and which ones aren't? If all of the self adjoint operators are observables, why are there so few self adjoint operators? PairTheBoard [/ QUOTE ] All self adjoint operators are observable. As to why there are so few of them, my guess would be that many of the ones you could think of would be superpostions of the common ones. For instance the x+y cordinate would be a self adjoint operator so thus an observable but would be pretty useless. [/ QUOTE ] Do the standard observables with meaningful physical interpretation form some kind of Basis for the space of all self adjoint operators on the hilbert space? PairTheBoard
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#17
05-04-2007, 09:50 AM
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Re: The Uncertainty Principle
op,
suppose u got a little photon bouncing around in a box and u want to find out it's V(t) and r(t). now conventionally measuremeants in systems of our scale we can measure by taking a ruler and reading off the position of something with our eyes. But how do our eyes work? they recieve information from photons that have bounced off the system in question. Obviously, if we are dealing with a 1 photon in a box system, by the time the light has hit your eyes, the process of measuremeant (i.e. photon interactoin) has already changed the state of the system. That is where the uncertainty comes from in this example.
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#18
05-04-2007, 10:41 AM
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Re: The Uncertainty Principle
[ QUOTE ]
ok. An observable is modeled by a self adjoint operator on the Hilbert Space. It seems like there would be a lot of self adjoint operators on the Hilbert Space. But there are only a few observables. Are all the self adjoint operators on the Hilbert Space some kind of observables? If not, what do they make of the ones that are not? And how do they decide which ones are and which ones aren't? If all of the self adjoint operators are observables, why are there so few self adjoint operators? [/ QUOTE ] I think this discussion could generate some semantic problems. The word "observable" means some kind of observable property of the particle or system. But it also means a self-adjoint operator on the underlying Hilbert space. (This fact alone demonstrates that they are generally believed to be in one-to-one correspondence.) To try to be clear, I will call the physical properties "real observables" and the operators either "observables" or just "self-adjoint operators." A self-adjoint operator determines a mapping that takes a state x to a probability distribution on the real line. So your question amounts to this: given such a mapping, can we cook up an experiment whose outcome has the right distribution for each state x? I think it is generally believed that this is true. So yes, every self-adjoint operator corresponds to a real observable. In practice, maybe one can cook up examples of self-adjoint operators whose corresponding real observables are presently unknown. But maybe not. For instance, suppose one is not concerned with position, momentum, etc., but simply wants to measure spin. The appropriate Hilbert space for this is C^2. (C is the complex numbers.) The self-adjoint operators are simply 2x2 matrices A with complex entries such that A is equal to its own conjugate transpose. There are three "primitive" real observables: the spin in the x, y, and z directions. Their matrices are given by S_x = [0 1; 1 0], S_y = [0 -i; i 0], and S_z = [1 0; 0 -1]. (There is actually a constant of h/4pi in front of each of these, where h is Planck's constant.) We can generate other (real) observables by taking real linear combinations of these operators. For example, S_x + S_y would correspond to spin the direction of the vector (1,1,0). (Or, perhaps more accurately, this measures sqrt{2} times the spin in that direction.) In this way, we can generate a set of observables which is a 3-dimensional vector space over the reals. Think, however, about how to construct an arbitrary self-adjoint operator on C^2. To build such a matrix, one selects any two real numbers for the main diagonal, then any complex number for the upper right entry. The lower left entry is then determined. So the entire space of observables is a 4-dimensional vector space over the reals. It follows that not all observables can be realized as real linear combinations of the spin matrices. The problem is resolved by noting that we are not restricted to taking linear combinations. We may also take functions of these matrices. For example, we could measure 2^{x-spin}. This is a real observable. Its corresponding self-adjoint operator is 2^[0 1; 1 0] = [5/4 3/4; 3/4 5/4]. Notice that this operator is not a real linear combination of the spin matrices. In general, every self-adjoint operator on C^2 can be written as a real-valued function of a linear combination of the spin matrices. Which function and which linear combination to use are not hard to compute. In this way, every self-adjoint operator can be associated to a real observable. I do not know if something similar can be done for more complicated systems. Edit: By "more complicated systems," I mean more complicated Hilbert spaces. For example, once you want to measure position and momentum, you must work in an infinite dimensional Hilbert space. The nature of self-adjoint operators in infinite dimensions is very different from finite dimensions.
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#19
05-04-2007, 12:05 PM
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Re: The Uncertainty Principle
jason1990's answer is pretty much the technical reason for the uncertainty principle. If people are happy thinking about wavefunctions, then the special case of the position-momentum uncertainty principle (and yes, ChrisV, there are others, and I think the only criterion is that they be noncommuting, as I think jason mentioned) can be viewed as just a result about Fourier transforms, which is probably more familiar to most people. The analogy in frequency-time land: if you want a pure tone, it has to last forever, or a short pulse in time has huge frequency bandwidth. Position-momentum uncertainty is really just the same thing.
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#20
05-04-2007, 01:12 PM
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Re: The Uncertainty Principle
Usually, I am only interested in the mathematical aspects of quantum mechanics. I leave the physical interpretations to the physicists. But PairTheBoard's question has me thinking a little more physically, at least for the moment.
What I am wondering is this. If we have two self-adjoint operators A and B, then A + B is self-adjoint. (This is actually not always true. It is always true in finite dimensions, but not always true in infinite dimensions. A sufficient condition for this to be true in infinite dimensions is that the domain of A + B is dense and at least one of A or B is a bounded operator. But let us ignore this for now and assume that A + B is self-adjoint.) Suppose we have physical interpretations for A and B. Does that give us a physical interpretation for A + B? Here is an example. Again, consider spin and let us look at the observables S_x = [0 1; 1 0], S_z = [1 0; 0 -1], A = 0.6 S_x + 0.8 S_z, and B = 2^S_x. Each of these has a physical interpretation. The first two are spin in the x and z directions. The third is spin in the direction of v = (0.6, 0, 0.8). The fourth represents the result of measuring spin in the x-direction and then computing 2^{x-spin}. Since x-spin is either 1 or -1, this means we simply measure it and then map it to 2 or 1/2, respectively. Now, what about C = S_z + B? What is the physical interpretation of this? One can verify that C = r^A - (1/r)*I, where r = (5 + sqrt{41})/4 and I is the 2x2 identity matrix. In other words, the physical interpretation of C is that it is the result of measuring spin in the v-direction and then computing f(v-spin), where f(x) = r^x - 1/r. Since v-spin is either 1 or -1, this means we simply measure it and then map it to 5/2 or 0 respectively. Could we have obtained the physical interpretation of C directly from the interpretations of S_z and B?
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Linear Mode
