Why matricies?

[gib65]

Why are matricies used in the math behind quantum mechanics unlike the regular numbers and variables that are used in the other sciences? I know this is a loaded question and I'm probably not going to understand the answer, but please try me.

[CPL.Luke]

matrices are used in a large number of sciences, essentially they are used whenever a situation becomes very complex involving a large system of equations.

 

probability theory uses matrices alot, so thats why QM uses them as well.

 

also matrices are used in economics and engineering, GR and by extension of QM, QFT in all of its forms.

[Bignose]

At its most basic, the 'matrix math' and the 'regular math' is exactly the same thing. There is a lot of math that is written that just comes down to notation.

 

y1 = a11*x1 + a12*x2

y2 = a21*x1 + a22*x2

 

is just regular math for

 

y = A*x

 

Well, sure, its easy for 2x2, but writing out 2000x2000? Matrix notation looks better and better.

 

Now, the real power or matrix (and vector/tensor notation) starts to come when you realize the impact of frame-indifference. I don't know GR and QM well, but I can give you an example from fluid mechanics.

 

The fluid does not in any way know if you are trying to describe its motion using Cartesian, cylindrical, spherical or any other coordinate system. But, if it a Newtonian fluid, it has to obey the Navier-Stokes equations http://scienceworld.wolfram.com/physics/Navier-StokesEquations.html, which are really just a description of conservation of momentum. The biggest point is that the Navier-Stokes equations are valid in any coordinate system, and my best guess is that this is exactly the same for QM and SR etc. The photon you are studying should not be aware that you are trying to describe its motion in cylindrical or spherical or any other coordinate system. Nature is indifferent to our description of it.

 

The Navier-Stokes equation has vectors in it (v for the fluid velocity) and tensors (what you call matrices) (T for the stress tensor of the fluid). Eqn (6) on that page looks simple, only 5 terms. But, depending upon the coordinate system, writing out all the terms can take a lot of space, e.g. the equations written out in spherical coordinates takes up almost an entire type-written page, see eqns (23-25) and compare with (6) above.

 

What I am trying to say, is that the tensor/matrix equations typically are the more general equations, with the details left to later. It is often easy to lose understanding if you get caught up with an equation with several terms rather than a few.

 

Finally, at the other extreme, the distinction between functions and matrices become blurred. That is, a vector becomes a finite representation of an infinite dimensional space also known as functions. The clearest example I can think of right now is Fourier expansion where a vector stores the coefficients of the Fourier decomposition. A finite number of sines and cosines of different frequencies summed together can be very close to equal to an arbitrary function, though it would take an infinite number to sum up to be a perfect match, hence the infinite dimensional aspect. You can expand these ideas of function spaces, inner products, self-adjoint operators, etc. But these ideas are pretty far out there and probably not helping answer this question, so I'll stop now.

[ajb]

The basic reason we need to consider matrices is beacuse they can be used to represent linear operators on vector spaces.

 

In quantum mechanics we have the (usually) infinite dimensional space of states, which is technically a Hilbert space. Observables are represented by linear operators on this space, which in turn have the mathematical representation as infinite dimensional matrices.

[Severian]

The real reason is rather simple. Quantum mechanics contains objects which don't commute. By that I mean [math]xy \neq yx[/math].

 

Clearly x and y cannot be ordinary numbers since ordinary numbers do commute. The simplist mathematical objects which don't commute are matrices, so we use them to describe these objects.

 

I should point out that the matrices are just 'representations' of the quantum mechanical 'operators'. They are not really the operators themselves. If we could come up with some other mathematical object which had the same property we could use that too.

[Klaynos]

They're also quite usefull for solving eigenvalue problems...

[5614]

The real reason is rather simple. Quantum mechanics contains objects which don't commute. By that I mean [math']xy \neq yx[/math]

Yep. It was the non-commutive nature that was discovered first, it was only afterwards that scientists realised this meant matrices were required.

[ajb]

Serviant and 5614 make good points. Quantum mechanics stems from a rather simple looking equation

 

[math] [x,p] = x\circ p - p \circ x = i \: \hbar [/math]

 

where {x,p} are the quantum anologue of the position and momentum. Quantum mechanics is basically the study of representations of this equation. It is known as the canonical commutator relations CCR.

 

It was Heisenburg who realised that matrices could be use as a reprsentation of this and Schrodinger who realised that differential operators could be used.

 

von Newmann showed that these two different repesentations are in fact equivalent. (techincal point: this is true provided we only have finite degrees of freedom). I beleive he showed that any representation of the CCR is equivalent to either Heisenburg's matrix representation or Schrodinger's differential operator representation and hence all equivalent.

[5614]

Serviant and 5614 make good points. Quantum mechanics stems from a rather simple looking equation

 

[math] [x' date=p] = x\circ p - p \circ x = i \: \hbar [/math]

 

where {x,p} are the quantum anologue of the position and momentum. Quantum mechanics is basically the study of representations of this equation. It is known as the canonical commutator relations CCR.

 

It was Heisenburg who realised that matrices could be use as a reprsentation of this and Schrodinger who realised that differential operators could be used.

Ah, it was Heisenberg. I remember reading how scientists could not think of anything with non-commutive properties but Heisenberg, who had just come out of university, immediately thought of matrices, having recently studied them. Something like that anyway!

 

ajb: what does the 'o', as in [math]x\circ p - p \circ x[/math], mean?

[ajb]

 

ajb: what does the 'o'' date=' as in [math']x\circ p - p \circ x[/math], mean?

 

 

It is just the product in the algebra. It would be matrix multiplication if we used matrices as a representation.

 

I did hear that Heisnburg did not know that he was using matrices. It was only later that someone pointed that out to him.

[5614]

It is just the product in the algebra. It would be matrix multiplication if we used matrices as a representation

So is that the same as saying a*b if 'a' and 'b' some number? Or a.b if 'a' and 'b' are vectors?

 

By saying [math]x\circ p[/math] you are saying x multiplied by p? And is it the circle, as opposed to a [math]\times[/math] or [math].[/math] because they're matricies?

[Severian]

And is it the circle, as opposed to a [math]\times[/math] or [math].[/math'] because they're matricies?

 

No, that is because ajb is a mathematician.

[ajb]

I wrote down [math]\circ [/math] just to represent some abstract linear, associative binary operation.

 

When we pick a representation for the elements of this algebra we also pick a representation of this operation. So, if we think about matrices, this binary operation is simply matrix multiplication. (And then the multiplication of a matrix with a vector).

 

When we represent this algebra by differential operators the binary operation is composition of the operators, e.g.

 

[math](x \circ p)(f) = x(p(f))[/math]

 

where [math]f[/math] is a function of [math]x[/math] i.e a state vector.