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GR0177 #28
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Quantum Mechanics$\Rightarrow$}Orthogonality

States or orthogonal when their dot-product (bra-ket) is 0. Since the eigenvectors $|i\rangle$ are given as orthonormal, $\langle i | i \rangle=1$.

$\langle \psi_i | \psi_2 \rangle =5-15+2x=0$ Solve the equation to find that $x=-10$, as in choice (E).

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 kristen2013-09-23 15:37:00 it should be 5+15+2x=0 (instead of 5-15+2x=0)Reply to this comment walczyk2011-04-07 00:14:00 this site has got some disrepair and neglect, but i still think its fantastic.Reply to this comment Phobos2010-09-13 11:37:44 As wrote jax: "You mean 5 + 15 + 2x = 0..." Maybe if I post it as a "typo" someone will notice. Reply to this comment Phobos2010-09-13 11:36:55 As wrote jax: "You mean 5 + 15 + 2x = 0..." Maybe if I post it as a "typo" someone will notice.Reply to this comment Jefe2006-10-29 20:11:11 jax is correctReply to this comment jax2005-12-07 19:19:52 You mean 5 + 15 + 2x = 0...Reply to this comment

You are replying to:
As wrote jax: "You mean 5 + 15 + 2x = 0..." Maybe if I post it as a "typo" someone will notice.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$