States and Group: Eigenvectors Represent One Dimension

[kent davidge]

Suppose a set of basis vectors are eigenvectors of some operator. So they will provide a one dimensional representation of that operator in the vector space?

 

[vanhees71]

If the operator $\hat{A}$ is self-adjoint, then it defines a unitary representation of a one-parameter Lie group via
$$\hat{U}=\exp(-\mathrm{i} \lambda \hat{A}).$$
It's not clear to me what you mean by "representation of that operator in vector space". A representation of an operator is usually the realization of the operator in a concrete realization of the (up to isomorphy unique) separable Hilbert space. E.g., if you choose the position representation, you work with position-wave functions, $\psi(\vec{x})$ and the representation of the components of the momentum are the derivatives $\hat{p}_j=-\mathrm{i} \hbar \partial_{j}$.

If you work in the "$A$ representation", i.e., with the wave functions which are "components" wrt. to the complete orthonormal set of eigenvectors (or generalized eigenvectors if the operator has continuous parts in its spectrum or if the spectrum is even entirely continuous), $\psi(a)=\langle a|\psi \rangle$. Then the representation of $\hat{A}$ is simply multiplication by $a$ since
$$\hat{A} \psi(a)=\langle a |\hat{A} \psi \rangle = \langle \hat{A} a |\psi \rangle=a \langle a|\psi \rangle.$$
That's of course also the case in the position representation. The components of the position vector are represented by multiplication of the wave function with the component, $\hat{x}_j \psi(\vec{x})=x_j \psi(\vec{x})$.

 

[A. Neumaier]

Suppose a set of basis vectors are eigenvectors of some operator. So they will provide a one dimensional representation of that operator in the vector space?

Then on each 1-dimensional eigenspace one has a 1-dimensional representation.

 

[kent davidge]

If the operator $\hat{A}$ is self-adjoint, then it defines a unitary representation of a one-parameter Lie group via
$$\hat{U}=\exp(-\mathrm{i} \lambda \hat{A}).$$

isn't that the case only if the group is additive?

 

[vanhees71]

Isn't this implied by the definition of "one-parameter Lie group"?

 

[kent davidge]

Isn't this implied by the definition of "one-parameter Lie group"?

oh yea, sorry

 

[strangerep]

Suppose a set of basis vectors are eigenvectors of some operator. So they will provide a one dimensional representation of that operator in the vector space?

I'm guessing your underlying question is whether the operator can be represented in terms of the eigenvectors. For a self-adjoint operator $A$ on a Hilbert space, with eigenvectors $|a\rangle$, where $a$ is a eigenvalue of $A$, one can represent $$A ~=~ \sum_a a\, |a\rangle\langle a| ~.$$ This is a simple version of "the spectral theorem". More general versions are available for more general operators, i.e., operators that are not self-adjoint, but satisfy certain other property(ies). I could sketch more detail if indeed that was what you were really asking. (?)

 

[vanhees71]

You mean
$$\hat{A}=\sum_a a |a \rangle \langle a|.$$
Your expression is, due to the completeness of the eigenkets
$$\sum_a |a \rangle \langle a|=\hat{1}.$$

 

[kent davidge]

@strangerep that's not exactly what I was meaning to ask, but I will come up with a better description of my doubts in a further thread.

 

[strangerep]

You mean [...]

Oops! Yes -- thank you.

(Post corrected.)

 

[A. Neumaier]

I'm guessing your underlying question is whether the operator can be represented in terms of the eigenvectors. For a self-adjoint operator $A$ on a Hilbert space, with eigenvectors $|a\rangle$, where $a$ is a eigenvalue of $A$, one can represent $$A ~=~ \sum_a a\, |a\rangle\langle a| ~.$$ This is a simple version of "the spectral theorem".

This is valid for self-adjoint operators with a purely discrete spectrum only. For Hermitian operators with a continuous spectrum there are no eigenvectors. Or you need to embed the Hilbert space into the dual of a nuclear space and get an analogous representation involving distribution-valued bras and kets and integrals in place of the sum. In the mixed spectrum case you need a combination of sums and integrals.

 

[strangerep]

This is valid for self-adjoint operators with a purely discrete spectrum only. [...]

Yes, I know. The intent of my (incomplete) post was merely to try and zero in on what the OP was really asking about.