# Time As a Basis Vector

1. Jan 13, 2010

### sol66

I was explaining basis vectors to my brother, I said that in quantum mechanics that when you have a number of dimensions, each dimension being an eigenket in vector space, that every dimension is independent of all the other basis vectors. It is however interesting to think that if this is the case, then time would not qualify as an eigenbasis but as rather transform or observable on your eigenkets. Is time suppose to be a unitary operator or something? This would mean time cannot be the 4th dimension. I suppose this would effect relative quantum mechanics given that all states of a particle must be in the same time state making time negligible as an eigenbasis. I suppose my question is, what is time considered to be(a dimension or unitary operator) and how is it treated in quantum relativity ( a course I have yet to take ).

2. Jan 13, 2010

### Fredrik

Staff Emeritus
Neither time nor space has anything to do with the basis vectors of the Hilbert space. Space and time are "dimensions" in the classical description of spacetime, but not in the mathematical representation of quantum mechanical "states". The symmetries of that classical theory do however correspond to symmetries of the quantum theory, in a way that ensures the existence of certain operators in the mathematical representation of "observables". The one that corresponds to translations in time is the Hamiltonian (energy) and the ones that correspond to translations in space are the momentum components. (The time translation operator is unitary and can be written as exp(-iHt) where H is the Hamiltonian).

The existence of a position operator isn't guaranteed by this, but one can usually be constructed. It depends on the details of the specific quantum theory of matter that we're considering. The theory of photons is especially problematic. There's no position operator for photons (or other massless particles).

3. Jan 13, 2010

### sol66

Oh, ok .. I suppose that makes a bit of sense, its just that in my classes it seemed that position was described as a state vector |x> and so I thought it was indeed part of a basis vectors of Hilbert Space. I guess what |x> really is is a projection onto a basis of vectors in Hilbert Space. And as you mentioned the Hamiltonian being the time evolution, though unitary not effecting probabilities of a particular state, seems to effect properties of that state while preserving probabilities; in the case that I am referring to the components of |x> eigenkets are changed.

Hmmm, I'll be honest I've never taken a class in this stuff and everything I've done up to now is self taught, I'm taking my first quantum class this upcoming semester. I'm still dumb in this stuff but I feel like I'm getting somewhat of a grasp on it. If hopefully my interpretation of the material is now correct. Thanks.

4. Jan 13, 2010

### Fredrik

Staff Emeritus
If we ignore some technical difficulties, it is essentially correct to say that the position operator has eigenvectors that we can write as |x>, and that the set of such position eigenvectors is a kind of basis for the Hilbert space of state vectors. Note however that "position" doesn't correspond to three dimensions of the Hilbert space, but rather all of them (infinitely many), and that the same thing can be said about any observable.