# Vector representation of a Quantum State

1. Feb 15, 2015

### Goodver

In Griffith books of introduction to QM, it say that Quantum State, mathematically represented as a vector.

My problem is with understanding what are the components of such a vector. Do I understand it correctly, that, say, in case of a particle in a box, Quantum State, as a vector is a superposition of all singular possible in particle in a box wave functions which constitute a components of a quantum state vector?

If so, then, if Quantum State is purely a wave function corresponding to second energy level say, then the Quantum State vector will be

{0, sqrt(2/a), 0, 0, ... 0}

or vector component can be either only either 1 or 0? Or value of a component is a Fourier coefficient?

* "a" is a width of a potential well

Thank you.

2. Feb 15, 2015

### Staff: Mentor

I answered a similar question in another thread that I will repeat here.

I will tell you what the more advanced books say - which is correct - others may not be quite right. Its from Ballentine - QM - A Modern development:
https://www.amazon.com/Quantum-Mechanics-Modern-Development-Edition/dp/9814578584

I suggest if you are interested in such things get a hold of the book and read the first three chapters.

Axiom 1
Associated with each measurement we can find a Hermitian operator O, called the observations observable, such that the possible outcomes of the observation are its eigenvalues.

Axiiom 2 - called the Born Rule
Associated with any system is a positive operator of unit trace, P, called the state of the system, such that expected value of of the outcomes of the observation is Trace (PO).

Note - the state of a system is not an element of a Hilbert space - it is an operator. How they come into it I will explain.

A state is called pure if its of the form |u><u|. A state is called mixed if its the convex sum of pure states ∑ pi |ui><ui|. It can be shown all states are either pure or mixed. Its the pure states that can be mapped to the Hilbert space - the u in |u><u| can be mapped to normalised vectors but there is an ambiguity in doing so because |cu><cu| = |u><u| if c is simply a phase factor. It can be extended further for all vectors in the Hilbert space by |u/||u||><u/||u||| - hence the states become rays in the Hilbert space - but the length is on no consequence. However always bear in mind the state is really an operator - this is simply a mapping.

Like any vector the basis you choose to write the vector in is entirely arbitrary - it purely depends on utility - and in QM that often depends on the observable you are interested in.

Given an Hermitian operator O by the spectral theorem it can be expanded in terms of its eigenvalues and eigenvectors O = ∑yi |yi><yi|. Since the |yi> form an orthonormal basis any vector |v> can be expanded in terms of that basis |v> = ∑vi |yi>. The vi is called the representation of the vector v in terms of the observable O. Without proving it, it turns out, from the Born rule, given a representation of an observable O, vi, then if you observe it with the observation associated with O the probability of outcome i is |vi|^2.

Now in the particle in the box you use Schroedingers equation which is written in terms of the wave function which is the expansion of the state in terms of the position eigenvalues. You get the eigenvalues and eigenvectors in terms of the energy operator. The only allowable states are superpositions of those eigenvectors. Let |u> = ∑ wi |ei>, where i labels the possible energies ie the eigenvalues of the energy operator, be on such supposition. The 'weight' wi in that superposition gives via |wi|^2 the probability of the system being in energy i if you were to observe it.

Thanks
Bill

Last edited: Feb 15, 2015
3. Feb 15, 2015

### Goodver

Thanks bhobba,

However, your answer is a little bit too broad for me. Back to the problem of a particle in a box. What is a basis of a quantum state vector in a particle in a box system? All singular allowed wave functions, without superposition of them?

4. Feb 15, 2015

### jus01

In short, yes. The components of the quantum state vector represent the weight of eigenvector's (in this case allowed energy wavefunctions) of the operator (observable, the Hamiltonian in this case). The quantum state then is the superposition of those eigenvector wavefunctions with the weights given in the state vector, normalized of course. The basis is all eigenvectors of the operator.

5. Feb 15, 2015

### Staff: Mentor

The solutions of the Schroedinger equation is a basis of the allowable states.

Thanks
Bill

6. Feb 16, 2015

### DrDu

Hm, I wouldn't say axioms, here, as this implies mathematical rigor. E.g. you most of the physically relevant operators like momentum and position don't have eigenvalues.
Furthermore, also the mixed states generate elements of a hilbert space e.g. via famous GNS construction.