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I know that even if we have:

$$a |\alpha \rangle = \alpha |\alpha\rangle$$

We don't have:

$$a^{\dagger} |\alpha \rangle = \alpha^* |\alpha\rangle$$

Actually as explained in the second answer here Eigenvalue for the creation operator for a coherent state $a^{\dagger}$ doesn't have eigenvalues.

But what disturbs me is the "matrix" vision.

If I choose a basis composed of $|\alpha\rangle$ and then some other states such that I have an orthogonal basis in the end, I can write down the (infinite) matrix $a$ in this basis.

The first diagonal coefficient will be $\alpha$. Then to go to $a^{\dagger}$ I conjugate transpose the matrix which will give me $\alpha^*$ on the first diagonal element.

Then through this vision it seems that $|\alpha\rangle$ is eigenstate of $a^{\dagger}$ with $\alpha^*$ as eigenvalue.

Where is the mistake in this reasoning ? Is it because we work in infinite dimension space so the matrix vision can induce errors ?

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    $\begingroup$ It is interesting to note that while your argument here already fails in the finite-dimensional case as yu-v's answer shows, an operator and its adjoint can indeed only have different eigenvalues in the infinite-dimensional case, cf. math.stackexchange.com/q/1431546/143136 $\endgroup$
    – ACuriousMind
    Jan 3, 2020 at 14:39
  • $\begingroup$ @ACuriousMind different eigenvectors you mean ? Because in finite dimension diag(a,b) and its hermitic conjugate have different eigenvalues for $a$ and $b$ complex $\endgroup$
    – StarBucK
    Jan 3, 2020 at 14:41
  • $\begingroup$ Matrices are only well-defined for linear maps on finite-dimensional vector spaces $\endgroup$ Jan 3, 2020 at 14:42
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    $\begingroup$ I meant "different" in the sense that the eigenvalues of one are not the complex conjugates of the eigenvalues of the other. $\endgroup$
    – ACuriousMind
    Jan 3, 2020 at 14:44
  • $\begingroup$ Hint: review creation and annihilation operators in the number representation: infinite-dimensional matrices. The answer should be evident. $\endgroup$ Jan 3, 2020 at 15:04

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Your mistake - while $\alpha$ will be on the diagonal (and $\alpha^*$ on the diagonal of the hermitian conjugate) it is not guaranteed that all the other terms in the row (and the column in the Hermitian conjugate representation of $a^{\dagger}$) will be zero. Therefore it will not be an eigenstate. It will have the property $\langle \alpha | a^{\dagger} |\alpha\rangle = \alpha^*$ as this is the value on the diagonal.

Note, for example, that the coherent states do not form a normal orthogonal basis, but an over-complete basis. Such that $\langle \alpha | \beta \rangle \propto \exp(|\alpha-\beta|^2)$ if I remember correctly.

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  • $\begingroup$ Oh yeah right... bad mistake for me. Thanks $\endgroup$
    – StarBucK
    Jan 3, 2020 at 14:38
  • $\begingroup$ So actually we cannot use the $\{|\alpha\rangle \}$ as an orthonormal basis because its over complete. So I cannot have a matrix vision with this set. And on the other hand if I choose $|\alpha\rangle$ being the first vector of a basis and completing with other vectors, then doing the conjugate we will have extra term on the first column because the other vector of this basis are no longer eigenvector of $a$. Thus $|\alpha \rangle$ is no longer an eigenvector of $a^{\dagger}$. Would you agree with my reformulation ? $\endgroup$
    – StarBucK
    Jan 3, 2020 at 14:44
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    $\begingroup$ Yes, I think I agree. $\endgroup$
    – user245141
    Jan 3, 2020 at 14:52

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