# Question about Eigenvalues of Hermetian Operators Being Real Numbers

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?

• Think of conjugates, only real numbers are their own conjugate s
– user108787
Jun 5, 2016 at 12:25
• $$a+ \mathrm i b~= ~a-\mathrm i b\,,$$
– user36790
Jun 5, 2016 at 12:30
• What do you think of $b\;?$
– user36790
Jun 5, 2016 at 12:30
• 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.
– OON
Jun 5, 2016 at 14:05
• 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.
– user108787
Jun 5, 2016 at 14:27

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.