# Proving Operator identities (Quantum Physics)

How would I go about showing:

$$\hat{A}^{\dagger} + \hat{B}^{\dagger} = \left( \hat{A} + \hat{B} \right) ^{\dagger}$$

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You wouldn't. Its an axiom. Or perhaps there is some deeper meaning in terms of higher-category-monoidal-bridget-fonda-F theory. – user346 Mar 27 '11 at 8:31

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It is not an axiom. First you should remind yourself of the definition of the $\dagger$ operation. – Qmechanic♦ Mar 27 '11 at 9:28

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I figured somebody would have a problem with that statement. I'm just finished fighting one battle and am too tired to start another. So please please, if you disagree by all means throw poop at my statements. I'm just too sleep-deprived to throw any back at present ;) – user346 Mar 27 '11 at 9:41

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Agree with Qmechanic. It's a property of adjoints. But I am not sure whether or not we should answer this kind (i.e. pure math) of questions here. – Marek Mar 27 '11 at 9:45
Dear @Deepak Vaid. Please be sure that my above comment was mainly aimed at user2789, not you. – Qmechanic♦ Mar 30 '11 at 17:18
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## 1 Answer

$$\left\langle \chi | \left( A + B \right)^\dagger | \psi \right\rangle = \left\langle \psi | (A +B) | \chi \right\rangle^* =$$ $$=\left\langle \psi | A | \chi \right\rangle^* + \left\langle \psi | B | \chi \right\rangle^* = \left\langle \chi | A^\dagger | \psi \right\rangle + \left\langle \chi | B^\dagger | \psi \right\rangle = \left\langle \chi | (A^\dagger +B^\dagger) | \psi \right\rangle$$ for all $|\chi\rangle$, $|\psi\rangle$.

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-1 because bra-ket notation totally confuses the matters here (it's not clear when the operators act to the left and when to the right). – Marek Mar 27 '11 at 10:04
+1, I think it is a totally legitimate answer to this trivial question. An operator is fully determined by its matrix elements with respect to any pair of vectors, here called $\chi$ and $\psi$, and the Hermitian conjugate operator's matrix elements are just given by those of the original one in the opposite order, with $*$. I just divided the long equation into two lines and added the missing complex conjugation signs $*$. The bracket notation is OK for any space with linear operators, and by convention, the primary action of operators is on the kets - but it also works on the bras! – Luboš Motl Mar 27 '11 at 10:12

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At any rate, the point of the proof is that the Hermitian conjugation is a linear operation (more precisely antilinear) acting on the space of operators, so if the coefficients in the linear combinations are real, and they're $1,1$ in this case, the terms may be conjugated one-by-one. – Luboš Motl Mar 27 '11 at 10:16

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@Luboš: the reason bra-ket notation "is OK" for any space with linear operators is because it satisfies lots of useful properties of which this one (i.e. linearity of adjoints) is one case. So the usage of bra-ket notation is kind of circular here. The correct way is to work from the $\left<x, Ay\right> = \left<A^{\dagger}x, y\right>$ where everything is crystal clear. – Marek Mar 27 '11 at 11:17
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