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I'm still slogging through Quantum Mechanics: The Theoretical Minimum and I've reached another area that baffles me.

Susskind uses the following to show that the eigenvalues of Hermitian operators are real numbers:

Given $L$ as a Hermetian operator, $\lambda$ as its eigenvalue and $|\lambda\rangle$ as its eigenvector

$$L|\lambda \rangle = \lambda |\lambda \rangle$$

$$\langle \lambda | L^\dagger = \langle \lambda | \lambda^*$$

since $L$ is Hermetian, $$L = L^\dagger$$ and

$$\langle \lambda | L = \langle \lambda | \lambda^*$$

multiply $$\langle \lambda |$$ to the first equation and $$|\lambda \rangle$$ to the second and you have

$$\langle \lambda |L|\lambda \rangle = \lambda \langle \lambda |\lambda \rangle$$

and

$$\langle \lambda | L |\lambda \rangle = \lambda^* \langle \lambda |\lambda \rangle$$

which means

$$\lambda = \lambda^*$$ and the eigenvalues are real numbers

Question Why does $\lambda = \lambda^*$mean that the eigenvalues are real numbers?

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    $\begingroup$ Think of conjugates, only real numbers are their own conjugate s $\endgroup$
    – user108787
    Jun 5, 2016 at 12:25
  • $\begingroup$ $$a+ \mathrm i b~= ~a-\mathrm i b\,,$$ $\endgroup$
    – user36790
    Jun 5, 2016 at 12:30
  • $\begingroup$ What do you think of $b\;?$ $\endgroup$
    – user36790
    Jun 5, 2016 at 12:30
  • $\begingroup$ I strongly advice you to learn some math (necessarily doing some exercises). Those are very basic facts and going further everything will only get harder for you. $\endgroup$
    – OON
    Jun 5, 2016 at 14:05
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    $\begingroup$ I self study completely, so been there, done that. At the top of each page of my notes, I write: what are my assumptions? That's where I always get stuck, thinking I know something I actually don't. Best of luck with it and stick with it. $\endgroup$
    – user108787
    Jun 5, 2016 at 14:27

1 Answer 1

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Let us write the eigenvalues as follows: $$\lambda=a+ib$$ Where $a$ and $b$ are real. By definition we must therefore have: $$\lambda^*=a-ib$$ Equating these gives us: $$a+ib=a-ib$$ $$2ib=0$$ $$b=0$$ and therefore: $$\lambda=a$$ Which is a real number.

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    $\begingroup$ …and, of course, I forgot this as well…which I'm almost positive was covered previously in Susskind. $\endgroup$ Jun 5, 2016 at 16:06

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