Represenation of a state vector in a different basis

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Is it possible to expand a state vector in a basis where the basis vectors are not eigenvectors for some observable A? Or must it always be the case that when we expand our state vector in some basis, it will always be with respect to some observable A?
 

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  • #2
Orodruin
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Given a basis you can always construct a Hermitian operator such that the basis states are its eigenvectors and they all have different eigenvalues. This, a priori, would represent an observable. Whether this observable is easy to actually observe or holds any deeper meaning is a different issue.
 
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hilbert2
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Any state vector is an eigenvector of, for example, the identity operator ##\hat{I}##:

##\hat{I}\left|\right.\psi\left.\right> = \left|\right.\psi\left.\right>## for any ##\left|\right.\psi\left.\right>##.

This doesn't have any physical meaning, as it's an "observable that always has value 1".
 
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Does an expansion like ## \Psi = \sum_{i} a_{i}e_{i}## even have any meaning if it isn't with respect to some operator corresponding to an observable? Another question I have is, since ##\Psi## is supposed to contain all the information about the system in order to predict it's time evolution, how can we know that our representation in different basis contain this information? Sure I might have a complete set with respect to some observable, but is it complete in the sense that it captures all physical information about the system?
 
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Orodruin
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In order to have an operator, you need to define the space that operator acts on. There is no such thing as "complete set with respect to some observable". A set of basis vectors is either complete or not. Ultimately, being complete just means that you can write any state as a linear combination of your basis states.

Does an expansion like Ψ=∑iaieiΨ=∑iaiei \Psi = \sum_{i} a_{i}e_{i} even have any meaning if it isn't with respect to some operator corresponding to an observable?
Yes, it is a representation of the physical state. However, in order to relate it to observables, you need to consider how different observables act on that state.

Another question I have is, since ΨΨ\Psi is supposed to contain all the information about the system in order to predict it's time evolution, how can we know that our representation in different basis contain this information?
Because the basis you use to express something does not affect what that something is. This is basic linear algebra.
 
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Because the basis you use to express something does not affect what that something is. This is basic linear algebra.
It should, If I express the state of an electron in the momentum representation or in the +- spin representation, the two representations does not contain the same information. The first says nothing about spin and the second nothing about momentum of the particle. I don't see there being a unitary representation between those two representations. But I suppose the point is that when you express it in the spin-basis you assume the momenta is already known.
 
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Orodruin
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It should, If I express the state of an electron in the momentum representation or in the +- spin representation, the two representations does not contain the same information. The first says nothing about spin and the second nothing about momentum of the particle.
So what? You are describing two different properties of the electron here and when you look individually at each you do not need to look at the full Hilbert space - it suffices to look at the subspace for the variable you are interested in. The full state space is the product of the spin and momentum state spaces and the momentum operator on that full state space is trivially composed of the identity on the spin space and the momentum operator on the momentum space.
 
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PeroK
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It should, If I express the state of an electron in the momentum representation or in the +- spin representation, the two representations does not contain the same information. The first says nothing about spin and the second nothing about momentum of the particle. I don't see there being a unitary representation between those two representations. But I suppose the point is that when you express it in the spin-basis you assume the momenta is already known.
The state vector represents the dynamical information about the electron. When you are dealing with a spin state, you are only studying its spin. The position/momentum is not relevant. (Note: this is conceptually no different from the classical case, where you may study the spin of a rigid object without considering the motion of its centre of mass.)

And, when you are dealing with the position or momentum representation, you are only studying its "motion", and are not interested in its spin. (Again, this is conceptually analogous to what you do in classical mechanics: you don't need to consider the Earth's spin when studying its solar orbit, for example.)

To get the complete picture you would need to combine these two states into an overall state by taking the product. The state of an electron, therefore, would be of the form ##\psi(\vec{r}) \chi(\vec{s})##, which contains all the information about its "motion" and its spin.

Your course may cover this at some point, as the overall state is relevant to electron bonding, for example.
 

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