I The most general state function

1. Apr 3, 2016

From my studying of QM, I have been learning about state functions that are supposed to describe a particular system. Now for most purposes, it seems that I have been computing the position wave function (or, through Fourier transforms, the momentum) which evolves with time. But I am just wondering: can any (Hermitian) operator act on this same wave function and yield sensible output? From my studying, I have learned that non-commuting operators act on different eigenvectors to yield direct/immutable measurements (eigenvalues), but that they can still act on different functions and just yield different results. I have attached a small piece from Griffiths that says when there is coupling between different observables, that they have to be included in the state function. But if they are not coupled, do we just separate these different observables at all times?

Why exactly do we call it a state function if only certain operators can act on it? Is there a way to generalize this wave function so that any operator (e.g. spin, energy, angular momentum, position) can act on it and it would yield the expected results from a measurement of this system we are trying to describe? What's the most general way to describe a system, or do we have to use different functions to characterize different aspects of a system (e.g. spin, position, energy)?

File size:
14.9 KB
Views:
43
2. Apr 3, 2016

bhobba

I don't know what you mean by state function. In QM we have states. QM operators can act on any state.

Exactly what a state is and what's going on with operators, eigenfunctions etc is elucidated by a very important theorem not usually discussed in intermediate texts like Griffiths - Gleason's Theorem:
http://www.kiko.fysik.su.se/en/thesis/helena-master.pdf.

Also see post 137:

Thanks
Bill

3. Apr 3, 2016

blue_leaf77

What do you mean by "not coupled" and "separate"?

4. Apr 3, 2016

It was with regards to footnote 7 in Griffiths (which is also related to the image attached):

"In the absence of coupling between spin and position, we are free to assume that the state is separable in its spin and spatial coordinates. This just says that the probability of getting spin up is independent of the location of the particle. In the presence of coupling, the general state would take the form of a linear combination: $\psi_+ (\vec{r}) \chi_+ + \psi_- (\vec{r}) \chi_-$"

5. Apr 3, 2016

blue_leaf77

Whether the spin and space is coupled or not, that depends on the state being considered. For one particle system, the space and spin are not coupled if the state in question is an eigenfunction of the spin component, e.g. $S_z$. For the case of coupling between space and spin, such as the eigenstates of relativistic hydrogen atom which takes on the form of $\psi_+ (\vec{r}) \chi_+ + \psi_- (\vec{r}) \chi_-$, this state is not an eigenfunction of $S_z$.

6. Apr 3, 2016

Okay. Just to clarify, would the $S_z$ operator applied to $\psi_+ (\vec{r}) \chi_+ + \psi_- (\vec{r}) \chi_-$ still yield possible values for measurements? And besides time, spin, and position/momentum components, are there any other coordinates typically considered when describing the overall state of a system?

I guess I'm still trying to figure out what the state of a system really means since I understand that hermitian operators corresponding to observables can be applied to them to yield possible measurements. But this question of being able to generally describe a state for any arbitrary system is escaping me.

7. Apr 4, 2016

blue_leaf77

Yes of course.
A quantum system is uniquely described by a set of quantum numbers corresponding to the maximal set of commuting observables, in the (nonrelativistic) H atom they are $n,l,m,m_s$. The spatial coordinate is just the probability of finding the electron in the specified coordinate.
A state of a quantum system can be understood as a mathematical object which determines the probability distribution of every observables, i.e. if you know the state of a system, then it is as good as you knowing which values with which probabilities a measurement of any observable will yield.

Last edited: Apr 4, 2016