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I know this kind of question has been asked before but I did not understand it completely. So while studying operators and eigenstates, I came across two formulas,

$\hat{A}|\psi\rangle = |\phi\rangle$ and, $\hat{A}|\psi\rangle = a|\psi\rangle.$

So according to me if $|\psi\rangle$ is an eigen vector of the operator it returns a scalar number. But if $|\psi\rangle$ is not a eigen vector of the operator we can represent it as a linear sum of eigenvectors which led me to believe that eigenvectors form a basis vectors of the operator but when I searched that if they are the same thing I found it was not the case and I thought that eigenvectors can be made a basis for the space but it's not always the case. I guess my question what's the difference between them and is eigenvector just a special case of basis vector?

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A complete set of basis vectors is a set of vectors such that any vector in the given vector space can be represented as a unique linear combination of the members of the said set. It can be intuitively imagined as well as easily proved that all sets of $n$ linearly independent vectors constitute a complete set of basis vectors for a vector space of dimensionality $n$ where the dimensionality of the vector space is defined to be the maximum cardinality that a set of linearly independent vectors can have.

As you can see, the definition of the basis vectors has nothing to do with the eigenvalue problem of any operator, as such. Moreover, eigenvectors corresponding to different eigenvalues of an operator would always be orthogonal but the members of a set of basis vectors only need to be linearly independent. This shows that not all sets of basis vectors can be seen as eigenvectors of some operator.

However, if a Hermitian operator is non-degenerate then the set of its eigenvectors corresponding to different eigenvalues forms a complete set of basis vectors on the vector space. Thus, you can use the eigenvectors of a non-degenerate Hermitian operator to form a basis. However,

  • In general, the set of all eigenvectors of a Hermitian operator need not form a basis for the trivial reason that if $\vert\lambda\rangle$ is an eigenvector then so is $a\vert\lambda\rangle $ and the two are clearly not linearly independent.
  • But, more non-trivially, a set of all eigenvectors of a Hermitian operator corresponding to different eigenvalues might also not form a basis because of degeneracy. For example, if one of the eigenvalues is two-fold degenerate and you only pick one eigenvector corresponding to one eigenvalue then the resultant set would not be complete -- you would be one vector short.
  • As you might have recognized, the solution to the problem of forming a basis on the vector space such that the basis vectors are also the eigenvectors of a given operator is the Gram--Schmidt procedure. You can use the Gram--Schmidt procedure to find a basis that, by construction, diagonalizes the given operator and now you have found a set of vectors whose members are both eigenvectors of the said operator and also forms a basis on the given vector space.

So, neither is a set of basis vectors necessarily a set of eigenvectors for some operator, and nor is a set of eigenvectors of a Hermitian operator necessarily a complete set of basis vectors. But, given a Hermitian operator, you can always construct a set of vectors such that it is a complete set of basis vectors as well as diagonalizes the given operator.

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What's the difference between an eigenbasis and an ordinary basis?

Well! No difference, in particular, I mean the action of both of them is the same as the vector space. If you have a transformation matrix that describes the $90^o$ rotation, then no matter what basis you choose it would remain the same transformation, that is $90^o$ rotation.

Though In quantum mechanics, there is a special role played by eigen basis due to the following reasons.

If the particle is in a state $|\psi\rangle $, measurement of the variable (corresponding to ) $\Omega$ will yield one of the eigenvalues $\omega $ with probability $P(\omega)\propto |\langle \omega |\psi\rangle|^2$ . The state of the system will change from $|\psi\rangle $ to $|\omega \rangle$ as a result of the measurement.

In mathematics too, If you choose your basis to be an eigen basis, the transformation matrix becomes diagonal and it's easy to work with them.


No eigenvectors are not special cases of basis. You should rather say, Eigen bases are a special case of basic in which the operator is diagonal.

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  • $\begingroup$ Oh yes I meant eigenbasis but got it mixed up with eigen vector. Thanks for pointing it out. $\endgroup$
    – aakash
    Apr 14, 2021 at 19:00

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