# Operator Equation

1. Mar 9, 2009

### chrisphd

In physics at university, I was told that quantum mechanics was based on a number of postulates, two of which were:
1. For every particle moving in a conservative field of force there exists an associated wave function, psi, which determines every property which can be physically measured. In general this is a complex number and represents the state of the system.

2. To every physical observable O, there exists an associated operator O such that Opsi = o*psi, and this equation is solvable for o and a corresponding value of psi.

Can anyone please tell me how the second postulate was "thunk up" as it appears to be completely random. I understand that the postulates are justified as they explain physical observations, but how would one even come to develop such a bizaar postulate as postulate 2.

2. Mar 9, 2009

### alxm

Historically it wasn't just 'thunk up'. First the wave-like properties of matter were discovered empirically, then the question arose whether an equation similar to the wave equation could describe it, which lead to the Schrödinger equation, which was later formalized into more basic postulates.

That said, it wasn't entirely bizarre and random. This operator formalism and so on goes back to Lagrangian and Hamiltonian mechanics.

3. Mar 9, 2009

### Peeter

I've just finished the first half of Bohm's book Quantum Theory. He has a very nice motivation of the QM postulates in there (chapters 3-9). The first two chapters also provide historical context that may be of interest, but IMO are a bit more difficult than the straight QM content that follows.

4. Apr 4, 2010

### mpkannan

2. To every physical observable O, there exists an associated operator O such that Opsi = o*psi, and this equation is solvable for o and a corresponding value of psi.

Is it right to state that "such that Opsi = o*psi, and this equation is solvable for o and a corresponding value of psi"?. I think it is not correct, since the wave function psi need not be an eigen function of the operator of every observable.

5. Apr 4, 2010

### chrisphd

If the wavefunction isn't an eigenfunction of the operator O, then Opsi will not equal o*psi. That is, Opsi will not be simply a number multiplied by psi. Therefore only eigenfunctions can be solutions to the equation Opsi = o*psi. In fact, this forms the definition of the eigenfunctions of operator O.

6. Apr 5, 2010

### mpkannan

Re: Operator Equation
2. To every physical observable O, there exists an associated operator O such that Opsi = o*psi, and this equation is solvable for o and a corresponding value of psi.

The above postulate (such that Opsi = o*psi,) gives an impression that the wave function of a system is always an eigenfunction of the operator of every observable of the system. But this is untrue.

7. Apr 5, 2010

### chrisphd

Maybe it gives that impression however that isn't what I intended. I state that the equation Opsi = o*psi can be solved for o and psi. Meaning that the possibilities of o and the possibilities of psi are both dependent on O.

That is, I am not saying that for any given psi, that the equation is true, but rather you can solve the equation for certain forms of psi, corresponding to certain values of o for a given operator O. In other words, eigen-functions psi and eigenvalues o are both deduced from the operator O.

8. Apr 5, 2010

### Fredrik

Staff Emeritus
I think it started with the idea that it may not be possible to measure position and momentum at the same time. If we assume that observables are represented mathematically by bounded self-adjoint operators, we can prove the uncertainty relation as a mathematical theorem.

9. Apr 5, 2010

### SpectraCat

Actually, I am pretty sure it was Heisenberg who first came up with the insight of only considering physically observable quantities while developing his matrix formulation of quantum mechanics. This is also the context in which he discovered that position and momentum could not be precisely-defined quantities within his formulation (which made him think it must be wrong/crazy at the time). Thus I agree that it seems likely that the uncertainty principle is related to this.

The wiki page on "Matrix mechanics" give some pretty cool historical context to think about.

10. Apr 5, 2010

### dx

I don't know the exact history, but here's a plausible line of thought: de Broglie discovered that to a particle with energy E and momentum p, there corresponds a wave of the form

χ = cos(2π(ft - σx + φ)). (E = hf, p = hσ)

To get equations of motion for χ in the Hamiltonian form, the function χ(x) on space is not enough, and its time derivative (∂χ/∂t)(x) is also needed. These two functions were combined into a complex function called the quantum state ψ(x)

ψ(x) = e2πi(ft - σx + φ)

In this representation, clearly

(ih/2π)(∂/∂x)ψ(x) = pψ(x)

From this, we can generalize away from the position representation by replacing ψ(x) with the abstract quantum state |ψ> and replacing (-ih/2π)(∂/∂x) by the abstract operator P, and we get

P|ψ> = p|ψ>

Generalizing this equations for obervables other than momentum, we get the general eigenvalue equation H|h> = h|h>. Of course the real history is no doubt far more complex and subtle.

Last edited: Apr 5, 2010
11. Apr 7, 2010

### mpkannan

In physics at university, I was told that quantum mechanics was based on a number of postulates, two of which were:
1. For every particle moving in a conservative field of force there exists an associated wave function, psi, which determines every property which can be physically measured. In general this is a complex number and represents the state of the system.

2. To every physical observable O, there exists an associated operator O such that Opsi = o*psi, and this equation is solvable for o and a corresponding value of psi.
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Here, I find the following problem:
The state of a system is described by a state function, also called wave function, usually represented by the symbol 'psi'. This state function/wave function 'psi', is the energy eigen function, i.e., the eigen function of the Hamiltonian operator (H^).

Though it is possible to discover some eigen function (say 'phi') for any operator (say O^) and write the eigen value equation

O^ phi = a constant. phi (1)

the wave function, 'psi' need not be an eigen function of the operator.
In fact, an eigen value equation of the operator, as given in your statement, is usually not included in the operator postulate.

Look at a commonly seen statement of the operator postulate:

"Every observable A is associated with a linear Hermitian operator A^, which can be constructed using the following rules.
(i) The operator form of the position coordinate x is x itself, i.e. x^ = x (similarly for y, z)
(ii) The operator form of linear momentum component px is [h/2(pi)i] d/dx, i.e.,
px^ = [h/2(pi)i] d/dx (similarly for py, pz)
(iii) The operator form of any observable A is obtained by first writing its classical expression in terms of the basic properties x, y, z and px, py, px and then replacing them by their respective operators given above."

12. Apr 7, 2010

### Fredrik

Staff Emeritus
The wavefunction is usually not an eigenvector of the Hamiltonian.

I'm not sure what you're trying to say with this post, but don't forget that immediately after a measurement of an observable represented by a self-adjoint operator A, the system is in an eigenstate of A.

This is not an axiom of QM. It's a recipe for how to construct quantum theories from classical theories. Note that this "quantization" procedure is ambiguous, because p and x don't commute. Your book should include a discussion of that fact.

13. Apr 7, 2010

### enotstrebor

What if : for every physical observable O, there does NOT exists an associated operator O such that Opsi = o*psi, and this equation is solvable for o and a corresponding value of psi

or

what if: for every physical observable O, there exists an associated operator O such that Opsi = o*psi, and this equation is NOT solvable for o and a corresponding value of psi.

or what if any other NOT combination is true, what use would the theory be?

14. Apr 8, 2010

### mpkannan

15. Apr 8, 2010

### Fredrik

Staff Emeritus
Any function f of the form f(x,t)=exp(-iHt)u(x), where u is a twice differentiable function, solves the Schrödinger equation. You need to perform an energy measurement to change u into a function that satisfies Hu=Eu for some u.