Proof of the Addition of Hermitian Adjoints

Question

I have been trying to work out a proof for the following statement using two linear operators A and B: $$(A + B)^\dagger = A^\dagger + B^\dagger$$ using the following definition of a hermitian adjoint of an operator $$\langle \psi_1|A^\dagger|\psi_2\rangle = (\langle\psi_1|A|\psi_2\rangle)^*$$ where * denotes the complex conjugate, and $$\dagger$$ denotes the adjoint of the operator. I have worked out the property concerning the product of the above operators but their addition doesn't allow me to use the properties of the inner product as nicely?

Answer 1

By assumption:

$$\langle\psi_1|(A+B)|\psi_2\rangle=\langle\psi_1|A|\psi_2\rangle+\langle\psi_1|B|\psi_2\rangle\implies(\langle\psi_1|(A+B)|\psi_2\rangle)^*=(\langle\psi_1|A|\psi_2\rangle+\langle\psi_1|B|\psi_2\rangle)^*$$

By definition:

$$(\langle\psi_1|A|\psi_2\rangle+\langle\psi_1|B|\psi_2\rangle)^*=(\langle\psi_1|A|\psi_2\rangle)^*+(\langle\psi_1|B|\psi_2\rangle)^*$$

By transitivity of equality:

$$(\langle\psi_1|(A+B)|\psi_2\rangle)^*=(\langle\psi_1|A|\psi_2\rangle)^*+(\langle\psi_1|B|\psi_2\rangle)^*$$

Therefore:

$$\langle\psi_1|(A+B)^\dagger|\psi_2\rangle=\langle\psi_1|A^\dagger|\psi_2\rangle+\langle\psi_1|B^\dagger|\psi_2\rangle$$

Whence:

$$(A+B)^\dagger=A^\dagger+B^\dagger$$