0
$\begingroup$

For an arbitrary state vector $|\Psi\rangle$, we can have the completeness relation, $$ |\Psi\rangle=\sum_{i} |\Phi_i\rangle\langle\Phi_i |\Psi\rangle \tag{1} $$ a resolution of the identity.

$|\Phi_i\rangle\langle\Phi_i|$ represents a projection operator.

I just want to know what are the conditions to respect on ket $|\Phi_i\rangle$ to apply the relation $(1)$ ?

I have already seen that $|\Phi_i\rangle$ vectors have to be the eigenvectors of an observable.

Is this always the case ?

Is an arbitrary orthonormal basis ($|\Phi_i\rangle$) not enough for the relation (1) to apply, (without being systematically eigenvectors)?

$\endgroup$
3
$\begingroup$

The condition that the $|\Phi_i\rangle$ be an orthonormal basis for the space is a sufficient condition for the completeness relation that you quote.

There is no requirement that those vectors be eigenvectors of any given observable (although, once you have such an orthonormal basis, there always exist nontrivial operators for which the $|\Phi_i\rangle$ are eigenvectors, as you can construct those operators explicitly).


On the other hand, it is not necessary for the $|\Phi_i\rangle$ to be an orthonormal basis for the condition to hold. One simple example is to take $N$ vectors in dimension $2$, given by $$ |\Phi_k⟩ = C \left[ \cos\left(\frac{2\pi k}{N}\right) |0⟩ + \sin\left(\frac{2\pi k}{N}\right) |1⟩ \right], $$ where $\{|0⟩,|1⟩\}$ is a basis for the state, $0<C<1$ is a non-unity normalization constant, and $k=1,2,\ldots,N$. It is an instructive exercise to prove, via an explicit calculation, that these vectors satisfy a completeness relation identical to the one you quote, even though they are too many to be linearly independent. (Such a set is generally known as a vector-space frame; in a QM context, it's known as a POVM.)

$\endgroup$
1
$\begingroup$

You can derive the completeness relation in a couple of interesting places: but the completeness relation implies, $$|\Phi_k\rangle=\sum_i|\Phi_i\rangle~\langle\Phi_i|\Phi_k\rangle$$and if your vectors are linearly independent, this is sufficient to conclude that $\langle\Phi_i|\Phi_k\rangle=\delta_{ik}$ i.e. orthonormality. But you can have it for non orthonormal sets of vectors: it would just require them to be linearly dependent on each other in some way. Compare for example with the way that the coherent states resolve the identity in a similar way, albeit with an integral.

Of course, once you know that they are orthogonal, you know that they are an eigenbasis for some operator. that is less dramatic than you are thinking, because you can generate one by construction, and it doesn't have to be in anyway useful at a physical level. For example,$$ \hat K=\sum_i i~|\Phi_i\rangle\langle\Phi_i|. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.