Matrix Mechanics & Wave Mechanics

[Varon]

Hi,

When Heisenberg proposed the Matrix Mechanics. It was totally without the concept of waves. It didn't use de Broglie idea of matter waves. In fact, Heisenberg kept fighting about the wave concept. However, Matrix Mechanics is said to be equivalent to the Schroedinger Equation that uses the concept of waves.

How come Matrix Mechanics is successful without waves?

Is the reason the Schrödinger Equation is successful with waves is because the wave concept is only used for Fourier analysis where each component wave stands for the quantum state, and nothing is really waving? Born proposed what is waving is just probability amplitude.

In short. Electron diffraction is said to be proof that matter has wave component. But how come Matrix Mechanics can still work by totally doing away with waves?

Maybe electron diffraction can be explained not by waving but by some method where something is interfering?

 

[eaglelake]

It all depends on what observable you are measuring! If the observable has a discrete eigenvalue spectrum, then the observable is a matrix operator and the experiment is described by a state vector. Spin is such an observable.

But position has a continuous eigenvalue spectrum, as does the momentum for a free particle. The momentum operator is a differential operator and it has a differential eigenvalue equation which yields eigenfunctions. Here we have position state functions and momentum state functions, rather than state vectors.

If you take a course in linear algebra you will see that the linear function spaces of position and momentum are extensions of linear vector spaces. Mathematically, matrix mechanics and wave mechanics are equivalent. Matrix mechanics does not do away with waves.

The deBroglie wavelength is related to momentum and it leads to the concept of a wavefunction. In general, it is not related to spin or to any other observable. Further, as far as we know, there is nothing "waving".

 

[Varon]

It all depends on what observable you are measuring! If the observable has a discrete eigenvalue spectrum, then the observable is a matrix operator and the experiment is described by a state vector. Spin is such an observable.

But position has a continuous eigenvalue spectrum, as does the momentum for a free particle. The momentum operator is a differential operator and it has a differential eigenvalue equation which yields eigenfunctions. Here we have position state functions and momentum state functions, rather than state vectors.

If you take a course in linear algebra you will see that the linear function spaces of position and momentum are extensions of linear vector spaces. Mathematically, matrix mechanics and wave mechanics are equivalent. Matrix mechanics does not do away with waves.

You mean Heisenberg Matrix Mechanics is still used along with Schroedinger nowadays (you mentioned Matrix operator.. is this separate from Heisenberg's formulation or identical)?
I read in wiki "Up until this time, matrices were seldom used by physicists, they were considered to belong to the realm of pure mathematics". Gustav Mie had used them in a paper on electrodynamics in 1912 and Born had used them in his work on the lattices theory of crystals in 1921. While matrices were used in these cases, the algebra of matrices with their multiplication did not enter the picture as they did in the matrix formulation of quantum mechanics."

The deBroglie wavelength is related to momentum and it leads to the concept of a wavefunction. In general, it is not related to spin or to any other observable. Further, as far as we know, there is nothing "waving".

You mean the reason wave is used in QM is because simply of this formula
wavelength = Planck constant/momentum? But this automatically assumes wave as from speed = frequency x wavelength

The derivation being:
(p stands for momentum)

E = m c^2 = (mc) (c) = (p) (c) = (p) (f x wavelength)
Equating E = h f
h f = (p) (f x wavelength)
h/p = wavelength

Here it already assume there is wave. What's strange is that the wave is probability amplitude.. while de Broglie thought it is wave like lightwave... something is not right...

 

[homology]

If you're interested in the history I'd look at primary sources. For example, Schrödinger wrote an article for The Physical Review on http://prola.aps.org/abstract/PR/v28/i6/p1049_1" which gives a lot of interesting insight into his though process. Its quite a good read.

As for matrices in QM, they're certainly not extinct. In a very loose sense every operator in quantum is a matrix in the sense that if \mid \lambda\rangle are a basis and if P is an operator then \langle\lambda'\mid P \mid\lambda\rangle is an element of the operator which, even if the basis is uncountably infinite, reminds us of the ordinary matrices. Of course, as said by eaglelake, if the basis is countably infinite then you can write down part of the operator in matrix form (component by component).

 

[dextercioby]

Hi,

When Heisenberg proposed the Matrix Mechanics. It was totally without the concept of waves.

However, it still had E=h\nu in it which had been proposed for light waves.

It didn't use de Broglie idea of matter waves. In fact, Heisenberg kept fighting about the wave concept. However, Matrix Mechanics is said to be equivalent to the Schroedinger Equation that uses the concept of waves.

True.

How come Matrix Mechanics is successful without waves?

Because it makes the same predictions as Schroedinger equation.

Is the reason the Schrödinger Equation is successful with waves is because the wave concept is only used for Fourier analysis where each component wave stands for the quantum state, and nothing is really waving? Born proposed what is waving is just probability amplitude.

Yes.

In short. Electron diffraction is said to be proof that matter has wave component.

I'd say that <matter has wavelike behavior> using an analogy with the classical theory of electromagnetic or mechanical waves.

But how come Matrix Mechanics can still work by totally doing away with waves?

Because describing physics by infinite-dimensional matrices, or by 'wavefunctions' is really the same thing.

 

[unusualname]

Both Heisenberg and Schrödinger discovered the same thing, the existence of a complex probability amplitude, admittedly in Heisenberg's formulation it was not so obvious as it was expressed as Fourier coefficients for experimentally observed intensities in huge matrix form.

Fourier representations of smooth functions is not really more deep than the fact that e^x=1+x+x^2/2! + ...

So since the complex probabilities are represented as e^i.theta you can get an infinite Fourier series representation of wave functions and a nice mathematical Hilbert Space for talking about QM (which may just be convenient for us)

 

[qsa]

Hi,

When Heisenberg proposed the Matrix Mechanics. It was totally without the concept of waves. It didn't use de Broglie idea of matter waves. In fact, Heisenberg kept fighting about the wave concept. However, Matrix Mechanics is said to be equivalent to the Schroedinger Equation that uses the concept of waves.

How come Matrix Mechanics is successful without waves?

Is the reason the Schrödinger Equation is successful with waves is because the wave concept is only used for Fourier analysis where each component wave stands for the quantum state, and nothing is really waving? Born proposed what is waving is just probability amplitude.

In short. Electron diffraction is said to be proof that matter has wave component. But how come Matrix Mechanics can still work by totally doing away with waves?

Maybe electron diffraction can be explained not by waving but by some method where something is interfering?

you can get the history of all of physics up to 1930 from E._T._Whittaker in an easy to read book

http://www.google.com.kw/#hl=en&sou...0l1l1l0l0l0l0l206l206l2-1&fp=a9e9f9109d96774b


http://en.wikipedia.org/wiki/E._T._Whittaker

 

[A. Neumaier]

It all depends on what observable you are measuring! If the observable has a discrete eigenvalue spectrum, then the observable is a matrix operator and the experiment is described by a state vector. Spin is such an observable.

But position has a continuous eigenvalue spectrum, as does the momentum for a free particle. The momentum operator is a differential operator and it has a differential eigenvalue equation which yields eigenfunctions. Here we have position state functions and momentum state functions, rather than state vectors.

Heisenberg essentially worked in a representation where a harmonic oscillator Hamiltonian such as H=(p^2+q^2)/2 is diagonal. Since such a Hamiltonian has a discrete spectrum, he got discrete matrices as linear operators. Schroedinger essentially worked in a representation where free motion Hamiltonian such as H=p^2/2 is diagonal. Since such a Hamiltonian has a continuous spectrum, he got differential and integral operators as linear operators.

Both representations describe the same Hilbert space, whence it is a matter of taste and convenience which one is chosen.