Exploring Quantum Mechanics: Propagators and Green's Functions

[Niles]

Hi

I am, at the moment, reading about propagators and Green's functions in QM. An example of a Green's function in some γ-basis is

$$G(\gamma, t, t') = - i\left\langle {c_\gamma \left( t \right),c_\gamma ^ \dagger\left( {t'} \right)} \right\rangle$$

Now, if I expand this in terms of eigenstates $\left| n \right\rangle$of the Hamiltonian, we obtain

$$G(\gamma, t, t') \propto \sum\limits_{n,n'} {\left\langle n \right|c_\gamma ^{} \left| {n'} \right\rangle \left\langle {n'} \right|c_\gamma ^\dag \left| {n'} \right\rangle f\left( {E_n ,E_{n'} ,t,t'} \right)}$$

where f is some function (the exact form of the Green's function is not that important). So this is the expression for the Green's function in some γ-basis written in terms of the Hamiltonians basis states (i.e., the Hamiltonian representation).

My question is: I cannot see what we mean when we distinguish between the basis and the representation of an operator (or state). I think the basis refers to the quantum numbers and the representation refers to the basis states (i.e., the functional form of the expression). But if this is correct, then any arbitrary state, say ${\left| \varepsilon \right\rangle }$, is written in ε-basis and the x-representation (e.g.) is given by ${\left\langle x \right.\left| \varepsilon \right\rangle }$?Niles.

 

[Matterwave]

It seems you are getting some terminology confused. A ket, is a geometrical object, and does not necessarily have a "basis".

A "basis" is a set of (usually orthnonormal) complete "axes", if you will, with which we can express the components of a ket.

So, the ket |psi> isn't in some "psi-basis", it's just a geometrical object that holds all the information that we could hope to obtain about the system. If we want to express |psi> in the x-basis, we take the inner product <x|psi>. x, here is the basis, and <x|psi> is the x-representation of the ket |psi>. We could equally well take <p|psi>, where now p is the basis, and <p|psi> is the p-representation of the ket |psi>.

Think in terms of a vector (in some kind of Hilbert space). The vector itself is a just an object that is independent of any kind of coordinate axes. We can represent the coordinates of that vector in any set of axes that we choose.

 

[Niles]

Thanks. I see, so in the case with the Green's function in my original post, then

$$G(\gamma, t, t') = - i\left\langle {c_\gamma \left( t \right),c_\gamma ^ \dagger\left( {t'} \right)} \right\rangle$$

is written in γ-representation. But then I (or rather, the book) go on and write the γ-representation in terms of different eigenstates, specifically the eigenstates of the Hamiltonian. Can one talk about writing a state in some representation, but expressing it in terms of other basis functions?Niles.

 

[Matterwave]

You can go from one basis to another by means of the identity operator:

$$I=\sum_n |\alpha_n><\alpha_n|$$

This means that you are now expressing whatever ket you had before, in terms of the alpha basis.

Oh...I see where this can get confusing haha. So the <x|psi> is the x-representation of the ket |psi>. But you can express |psi> in different bases, e.g. |psi>=a|a>+b|b>. And then <x|psi>=a<x|a>+b<x|b> is still the x-representation of |psi>.

Perhaps it was wrong of me to call x, and p a basis then. Perhaps it's better to call |x> and |p> the basis.

X and P are a little bit special, because we all the inner products <x|psi> and <p|psi> "representations". Whereas, for other bases, like |phi>, we just call <phi|psi> the amplitude. We could also call <x|psi> and <p|psi> the amplitudes though.

 

[Niles]

Ahh, I see. Thanks!