# Linear algebra question (using braket notation).

by MathematicalPhysicist
Tags: algebra, braket, linear, notation
 P: 352 Oh no there is something wrong in your argument. The vector u is orthogonal to v, it is a specific vector. When you invoke completeness you are summing over all vectors, and not just the ones orthogonal to v. It is not true that = 0.
 P: 352 The completeness relation in this problem is $$|u> = \alpha |u> + \beta |v>$$ It is very easy to show that $$\beta =$$ And then you're left with the other term, and maybe there the completeness relation is helpful.
 Mentor P: 6,248 Linear algebra question (using braket notation). It's easy to find an example that contradicts the assertion that you're trying to prove. If fact, since $$A \left| v \right> = \left< A \right> \left| v \right> + \delta A \left| u \right>$$ is equivalent to $$A \left| v \right> - \left< A \right> \left| v \right> = \delta A \left| u \right>$$ and, from the given, $$A \left| v \right> - \left< A \right> \left| v \right> = \left( A - \left< A \right> I \right) \left| v \right> = \delta A \left| v \right>,$$ so your assertion is true iff $\delta A \left| v\right> = \delta A \left| u \right>$. Maybe you're actually trying to prove the trivial $$A \left| v \right> = \left< A \right> \left| v \right> + \delta A \left| v \right>$$.
Mentor
P: 6,248
 Quote by George Jones so your assertion is true iff $\delta A \left| v\right> = \delta A \left| u \right>$.
Oops, I see that you already know this.

Cleary,the equations $A \left| u \right> = \left| u \right>$ and $A \left| v \right> = 0$ define a unique Hermitian operator $A$. Now,

$$\left< A \right> = 0$$

$$\delta A = A$$.

Therefore,

$$\delta A \left| u \right> = A \left| u \right> = \left| u \right>$$

$$\delta A \left| v \right> = A \left| v \right> = 0.$$

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