Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

On studying quantum states of the hydrogen atom, I came with a question: what is the precise definition of a quantum state? Particularly, when solving the Schroedinger eq for the H using separation of variables, the quantum numbers and wavefunctions for the states are clear. However, mathematically speaking, any combination (linear) of wavefuntions is still a solution for the H atom; is this solution a state? Moreover, what are the quantum numbeers of this solution? It blows my mind, since this solution has 6 quantum numbers, instead of 3.

More generally, my question is: is there a precise definition of state of system?

share|improve this question

2 Answers 2

A state in quantum mechanics is any vector in the Hilbert space describing the system. Sometimes one requires that a state should be normalized, that is, of unit norm. Since vectors can be added, the sum of two states is another state (or can be normalized to be, if normalization is taken as part of the definition.)

The states you find when you find the energies of the hydrogen atom are a basis for the Hilbert space. They are a particularly convenient basis because they are eigenvectors of three operators: the Hamiltonian ($= $ the energy) $H$, the magnitude of angular momentum $L^2$ and the $z$-component of angular momentum $L_z$. The eigenvalues are labeled by the quantum numbers $n$, $l$ and $m$. When you see "quantum number" in a text, you should think "eigenvalue of some observable".

A linear combination of states, such as $$|\psi\rangle = 2^{-1/2} ( |n=1,l=0,m=0\rangle + |n = 2, l = 1, m=0\rangle )$$ is still a state, but it is not an eigenstate of the operators $H$ or $L^2$ anymore. In this state, the energy and magnitude of angular momentum do not have definite values. The possibility of such states arises because in quantum mechanics states are vectors and can be added together.

share|improve this answer
    
Note that if you multiply the vector by a phase factor it is still the same state. –  jjcale May 4 at 16:57

I'm afraid states are not vectors, but the so-called 'state representatives' are. States are typically rays in a complex separable Hilbert space. Thus the sum of 2 states doesn't yield a state, the space of all states is no longer a vector space. Among the infinity of vectors in a ray, one may choose a generic one which is then called 'state representative' and can be added with other vectors (state representatives) as normal vectors do. The Schrödinger equation describes the evolution of state representatives.

There's of course the von Neumann's description: states are trace-class linear operators onto some complex separable Hilbert space which are called density operators.

While H-atom's bound state representatives form a countable basis in $L^2(R^3)$, they are not complete, because the scattering states representatives have been left out. When you add the latter, you will then have a real basis, albeit in a rigged Hilbert space.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.