The Uncertainty Principle

[deleteduser]

Ive been trying to understand what exactly it is, and from my perspective it is a principle that explains the more accurately you try to measure a particles position the less accurately you will be able to measure its speed.

Also that: '
the uncertainty of a particles position X the uncertainty of a particles speed X the uncertainty of a particles mass = a number always greater than planks constant.


So does this mean we cannot measure a particles position and velocity at one given time? But we know it will be greater than planks constant?


Also what is "a position of a particles" what references are used in determining the position?

[AdamL]

On the last question - the position of a particle is the location upon measurement. It's actualization.

[Silent A]

the uncertainty principle is:

uncertainty in momentum * uncertainty in position >= plank's constant > 0

It means you cannot know a particles exact position and exact momentum. The more you know about one, the less you must know about the other.

And strictly speaking, this isn't just an issue about measurement. At the fundamental level, the uncertainty princple says that the particle in question doesn't have an exact position or momentum.

[ChrisV]

I think there are pairs of properties other than position/momentum to which the uncertainty principle also applies. As originally stated though it is as given above.

[ QUOTE ]
And strictly speaking, this isn't just an issue about measurement. At the fundamental level, the uncertainty princple says that the particle in question doesn't have an exact position or momentum.

[/ QUOTE ]

This is important. A lot of explanations imply that it is the act of measurement (for instance, bouncing a photon off a particle) which causes the uncertainty. But in fact the uncertainty is intrinsic to the particle itself; it really does not have a defined momentum or position. Another way to say this is the familiar statement that particles have wavelike characteristics.

[Silent A]

[ QUOTE ]
I think there are pairs of properties other than position/momentum to which the uncertainty principle also applies. As originally stated though it is as given above.

[/ QUOTE ]

Another description is:

Uncertainty of energy * uncertainty of time >= plank's constant > 0

this actually implies that conservation of energy can be violated for small time periods. Or, at least, that's how it was taught to me in high school.

[Max Raker]

[ QUOTE ]
I think there are pairs of properties other than position/momentum to which the uncertainty principle also applies.

[/ QUOTE ]

Yes, there are also energy time and one for different components of angular momentum.

[Neuge]

Nit time: It isn't exactly ">= Planck's constant" either. For all pairs of non-commutative operators, the product of the uncertainty in the measurement of their corresponding observables is >= a number proportional to Planck's constant. For the position/momentum and energy/time complementary pairs, the product of their uncertainty is actually >= h/4*pi.

[Neuge]

[ QUOTE ]
[ QUOTE ]
I think there are pairs of properties other than position/momentum to which the uncertainty principle also applies.

[/ QUOTE ]

Yes, there are also energy time and one for different components of angular momentum.

[/ QUOTE ]
These are the most common, but the strict definition of complementary observables is "observables of two non-commutative quantum mechanical operators." I think there are others, but I can't remember any off the top of my head.

[Max Raker]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I think there are pairs of properties other than position/momentum to which the uncertainty principle also applies.

[/ QUOTE ]

Yes, there are also energy time and one for different components of angular momentum.

[/ QUOTE ]
These are the most common, but the strict definition of complementary observables is "observables of two non-commutative quantum mechanical operators." I think there are others, but I can't remember any off the top of my head.

[/ QUOTE ]

The other common one is angular position and angular momentum in the same component. But you are right, any non commuting hermitian operators will have an uncertainity realation.

[PairTheBoard]

So particles are really non commuting hermitian operators?

PairTheBoard

[Max Raker]

[ 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.

[jason1990]

[ 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.

[daryn]

isn't it not just h but h/4pi or h-bar/2? or are you guys just simplifying it?

[PairTheBoard]

[ 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

[Max Raker]

[ 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.

[PairTheBoard]

[ 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

[UptownExpress]

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.

[jason1990]

[ 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.

[gumpzilla]

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.

[jason1990]

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?

[thylacine]

[ QUOTE ]
[ 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.

[/ QUOTE ]

I don't follow. Why not just add I=[1,0;0,1] to your list?

[jason1990]

[ QUOTE ]
I don't follow. Why not just add I=[1,0;0,1] to your list?

[/ QUOTE ]
Aha... Why didn't I think of that? So, for example,

2^S_x = (3/4)S_x + (5/4)I.

Both observables amount to measuring spin in the x-direction, which is 1 or -1, and mapping it to 2 or 1/2 respectively. This actually looks like an interesting way to derive a lot of matrix identities.

Does this answer my other question:

[ QUOTE ]
Could we have obtained the physical interpretation of C directly from the interpretations of S_z and B? [B = 2^S_x]

[/ QUOTE ]
Given the above, we can write

C = S_z + 2^S_x
= S_z + (3/4)S_x + (5/4)I
= (5/4)(0.6 S_x + 0.8 S_z) + (5/4)I.

From here it is clear that C corresponds to measuring v-spin, which is 1 or -1, and mapping it to 5/2 or 0 respectively.

[jason1990]

[ QUOTE ]
This actually looks like an interesting way to derive a lot of matrix identities.

[/ QUOTE ]
So, in general,

f(S_x) = [(f(1) - f(-1))/2]*S_x + [(f(1) + f(-1))/2]*I.

I thought this was interesting, but maybe it is trivial. I have to think on it some more.

[jason1990]

[ QUOTE ]
I don't follow. Why not just add I=[1,0;0,1] to your list?

[/ QUOTE ]
Okay, here is one more comment on this. PairTheBoard wants a physical interpretation. Given an operator A, he wants to know the real observable quantity being measured. Without I on the list, the linear combinations look like

aS_x + bS_y + cS_z.

In this case, the quantity being measured is spin in a certain direction. With I on the list, one has for example

aS_x + bI.

The experiment you would perform that corresponds to this operator is one in which you first measure the x-spin, and then you apply the function ax + b to your measurement. So if you add I to the list and want to talk about physical interpretations, you must still introduce the idea of a function of a measured quantity, and not a measured quantity itself. Maybe there is no real distinction between the two, but my non-physicist's intuition makes me feel like there is.

Granted, even aS_x is a*(x-spin). But we can change units so that this is x-spin itself. We cannot do any unit changing trick to make a*(x-spin) + b anything other than a function of x-spin.

[PairTheBoard]

[ QUOTE ]
[ 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.

[/ QUOTE ]

wow. That is great stuff. I always wondered what they were talking about with those "Operators". So it looks like they construct the Hilbert Space according to the Primitive Physical Observables they know about. I was thinking they had some general apriori Hilbert Space and then discovered the Primitive Physical Observables they could apply to it. But of course. How would they know what that Hilbert Space was to begin with.

PairTheBoard

[m_the0ry]

Even though the uncertainty principle is most often applied when comparing position and speed, it applies to any of the conjugate pair variables.

It also says frequency and time measurements have a minimum uncertainty, mass and velocity, etc. etc...

[Metric]

[ QUOTE ]
isn't it not just h but h/4pi or h-bar/2? or are you guys just simplifying it?

[/ QUOTE ]
Yeah, people always simplify this kind of stuff. The factor of 2 isn't the important thing (I often don't remember factors of 2 myself) -- the important thing is the concept that "simultaneously measuring two non-commuting observables" is exactly analogous mathematically to "simultaneously diagonalizing non-commuting matrices." This was one of the most profound realizations in 20th century physics.