Finding the Adjoint of an Infinite Hilbert Space Operator

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Homework Statement


Hi guys

I have the following operator on an infinite Hilbert space: Oei=ei+1, where i is a positive integer and ei are the orthonormal vectors that span the space. I have to find the adjoint of this operator.

For two arbitrary vectors x and z, I have found (Ox,z), and I have to find the O* (the adjoint) such that (Ox,z)=(x,O*z). This gives me that O*ei=ei-1.

Now I have to show that OO* = 1. How can I go about doing that?

Thanks for any input.
 
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You've done the hard work, that's trivial now! For any basis vector ei, O*ei= ei_1 so OO*ei= Oei-1= e(i-1)+1= ei.

You can show that O*O = 1 in exactly the same way.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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