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The way I have proved the Eigen values of Hermitian matrices are real like this:

I considered $H$ is a hermitian matrix.

Operator applied in ket space: $\left<\psi|H | \psi\right>= \lambda$

Operator applied in bra space: $\left<\psi|H | \psi\right> = \lambda^\star \ $

There is no state in the right hand side because of the normalization.

If we subtract the two equations we get $\lambda = \lambda ^*$ Therefore the eigenvalues are real

Have I done in the right way?

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closed as off-topic by ACuriousMind, user36790, JamalS, Cosmas Zachos, Wolpertinger Oct 22 '16 at 20:40

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    $\begingroup$ Why don't you look it up in your textbook? Also, there seems to be an error in your second equation. $\endgroup$ – freecharly Oct 6 '16 at 22:30
  • $\begingroup$ how can I rewrite the second equation? $<H \psi \ \ | \ \psi >= \lambda^*$ is it right now? $\endgroup$ – Numerical Person Oct 7 '16 at 0:15
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    $\begingroup$ $<H\psi|=<\psi|H $? It does not make sense with the used notation. Here $|\psi> $ is a unique object exactly as $<\psi|$ is. The only thing the OP may declare is that, in the second line, $H $ acts on the left bra instead of on the right ket. This is an evident example of how is awkward the bra ket notation sometimes. $\endgroup$ – Valter Moretti Oct 7 '16 at 5:35
  • $\begingroup$ @Valter amoretti, Sorry for the typo. What I have written is that, the Hermitian operator applied in the ket space in the first line and applied in the bra space in the second line. Since I have applied the H operator in the left bra space I got the complex conjugate o the eigenvalue $\lambda$. Am i right? $\endgroup$ – Numerical Person Oct 7 '16 at 17:46
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    $\begingroup$ Yes, you are right. $\endgroup$ – Valter Moretti Oct 7 '16 at 18:29
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Note that the question has nothing to do with physics. The correct proof without using Bra-Ket notation can be written as follows:

Let $A\in\mathbb{C}^{n\times n}$ and $A=A^*$. (That is to say: Let A be a hermitian matrix). If $\lambda$ is an eigenvalue of $A$ then $\lambda\in\mathbb{R}$.

proof. Suppose $v\neq 0$ and $Av=\lambda v$. Then $$\overline{\lambda}\langle v,v\rangle=\langle \lambda v,v \rangle=\langle Av,v \rangle=\langle v,Av \rangle=\langle v,\lambda v \rangle=\lambda\langle v,v \rangle .$$ Since $v\neq 0$ one can divide by $\langle v,v \rangle$ to get $\lambda=\overline{\lambda}$. $\blacksquare$

What you want would be the stronger result:

Let $\mathcal{H}$ be a $\mathbb{C}$-Hilbertspace and $H$ a possibly unbounded self-adjoint operator on $\mathcal{H}$. Then the spectrum of $H$ is real.

This is obviously more difficult to prove and going over it would only make sense if you know some functional analysis.

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