Why we need completeness of eigenfuntions for QM to be internally consistent?

[Happiness]

For quantum mechanics to be internally consistent, the following must be true:

The eigenfunctions of an observable operator are complete: Any function (in Hilbert space) can be expressed as a linear combination of them.

In other words, it is not alright for the solution space (to the Schrödinger equation) to span only a subspace of the Hilbert space. It has to span the entire Hilbert space. But why? What's inconsistent about it?

I find this quite unbelievable, because it's saying any function in Hilbert space is a solution to Schrondinger equation. But when we solve an equation, we don't expect everything to turn out to be a solution.

Reference: Introduction to Quantum Mechanics, 2nd ed., David J. Griffiths, p 102.

 

[A. Neumaier]

Why? Because it is a provable theorem, valid in each Hilbert space. It is called the spectral theorem.

Note that linear combinations of eigenstates are generally not eigenstates again.

 

[PeroK]

For quantum mechanics to be internally consistent, the following must be true:In other words, it is not alright for the solution space (to the Schrödinger equation) to span only a subspace of the Hilbert space. It has to span the entire Hilbert space. But why? What's inconsistent about it?

I find this quite unbelievable, because it's saying any function in Hilbert space is a solution to Schrondinger equation. But when we solve an equation, we don't expect everything to turn out to be a solution.

Reference: Introduction to Quantum Mechanics, 2nd ed., David J. Griffiths, p 102.

What is important about the Schrödinger equation is the time evolution of states. In principle , any wave function could be a starting point, but how it evolves is really the solution.

This is not that different from the 1D wave equation, where the wave can, in theory, have any shape.

 

[vanhees71]

Well, there are some exceptions from the arbitrariness of initial states, the socalled superselection rules, which are closely related to symmetries. E.g., there should be no superpositions of states with half-integer and integer spin in order to have a proper (ray) representation of the rotations as symmetry group of both Galilei and Minkowski spacetime. In non-relatistic QT, there should also be no superpositions of states with different mass.

 

[dextercioby]

The time-independent Schroedinger equation is a spectral equation whose solutions, if the self-adjoint Hamiltonian has a pure-point spectrum, span the whole Hilbert space. This is plain, not simple mathematics. If the Hamiltonian is not self-adjoint, there is no guarantee that the whole spectrum has no complex numerical value. We cannot measure 5+4i J, right?

 

[martinbn]

If the eigenfunctions of some observable operator don't span the whole space, and you have a system in a state given by a wave function in the complement of the span, and you make a measurement of that observable, then what result can you get!?

 

[Happiness]

If the eigenfunctions of some observable operator don't span the whole space, and you have a system in a state given by a wave function in the complement of the span, and you make a measurement of that observable, then what result can you get!?

You get the eigenvalue of the eigenfunction closest to the wave function. Would this make QM consistent?

 

[A. Neumaier]

You get the eigenvalue of the eigenfunction closest to the wave function. Would this make QM consistent?

No, it's wrong. You get any eigenvalue, with a probability decreasing with that distance.

 

[Happiness]

No, it's wrong. You get any eigenvalue, with a probability decreasing with that distance.

Ok, that answers @martinbn’s question.

 

[martinbn]

Ok, that answers @martinbn’s question.

It wasn't a question, it was an answer. It seemed than the given answers didn't satisfy you, so I tried to give you a more intuitive answer. I'll try again. When you measure an observable the possible results are eigenvalues of the corresponding operator with probabilities given by the Born rule. If the eigenvectors of the operator do not span the whole space, and your system is in a state that is in the orthogonal compliment of that span, when you perform a measurement the probability to obtain any of the eigenvalues will be zero. And after the measurement the state of the system will be reduced to the zero vector. That is absurd, hence the span of the eigenvectors should be the whole space.

 

[vanhees71]

Indeed, it violates the basics of probability theory: If you have a probability distribution for a well-defined random experiment (and according to quantum mechanics all physics experiments are in a well specified sense random experiments) the probabilities must add up to 1, i.e., when measuring an observable (with an ideal measurement device) you always get a well-defined result, though the measured observable may not have had a determined value before the measurement. That's implied by any probabilistic description of random experiments. If this is not the case it's ill-defined, i.e., you have to find a better description.

That's, by the way, the formal reason, why observables (usually) are described by self-adjoint operators (and Hermitean operators is too weak an assumption). The important point is not so much that the possible measurement outcomes, i.e., the eigenvalues of the operator, are real but that self-adjoint operators have a complete spectral decomposition, i.e., there exists a complete set of orthonormal generalized eigenvectors (including the case of continuous eigenvalues, where you have to generalize the concept of eigenvectors in the sense of the theory of generalized functions, the most elegant formulation being the "rigged Hilbert space").

 

[Happiness]

In principle , any wave function could be a starting point, but how it evolves is really the solution.

Is there an existence (and uniqueness) theorem for a bivariate second-order linear partial differential equation?

In other words, does the equation $$a(x,t)\frac{\partial^2}{\partial t^2}y(x,t)+b(x,t)\frac{\partial}{\partial t}y(x,t)+c(x,t)y(x,t)+d(x,t)\frac{\partial^2}{\partial x^2}y(x,t)+e(x,t)\frac{\partial}{\partial x}y(x,t)+f(x,t)\frac{\partial^2}{\partial x\partial t}y(x,t)=g(x,t)$$ always have a (unique) solution $y(x,t)$ for any given $y(x,0)$, $\frac{\partial y}{\partial t}(x,0)$, $\frac{\partial y}{\partial x}(x,0)$, $a(x,t)$, $b(x,t)$, $c(x,t)$, $d(x,t)$, $e(x,t)$, $f(x,t)$ and $g(x,t)$?

($y(x, t)$ is in Hilbert space, just like wave functions are.)

 

[martinbn]

Is there an existence (and uniqueness) theorem for a bivariate second-order linear partial differential equation?

In other words, does the equation $$a(x, t)\frac{\partial^2}{\partial t^2}y(x, t)+b(x, t)\frac{\partial}{\partial t}y(x, t)+c(x, t)y(x, t)+d(x, t)\frac{\partial^2}{\partial x^2}y(x, t)+e(x, t)\frac{\partial}{\partial x}y(x, t)+f(x, t)\frac{\partial^2}{\partial x\partial t}y(x, t)=g(x, t)$$ always have a (unique) solution $y(x, t)$ for any given $y(x, 0), a(x, t), b(x, t), c(x, t), d(x, t), e(x, t), f(x, t)$ and $g(x, t)$?

The question is too vague for an answer. It is second order in $t$ and includes hyperbolic equations, say the wave equation, so you need the initial value for the derivative too. It also includes elliptic equations, say Poison equation, for which the Cauchy problem is ill posed. Also you don't specify what space the given functions belong to. It is like asking does the equations $x^3-2=0$ have a unique solution. In rationals, no, in reals it has on, in complex numbers three.

 

[Happiness]

The question is too vague for an answer. It is second order in $t$ and includes hyperbolic equations, say the wave equation, so you need the initial value for the derivative too. It also includes elliptic equations, say Poison equation, for which the Cauchy problem is ill posed. Also you don't specify what space the given functions belong to. It is like asking does the equations $x^3-2=0$ have a unique solution. In rationals, no, in reals it has on, in complex numbers three.

$y(x, t)$ is in Hilbert space, just like wave functions are.