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In the Principles of Quantum Mechanics, Dirac states that all linear operators $\alpha$ over our vector field (over the complex numbers) can be expressed as the sum of a real and an imaginary part $\alpha=\beta + i\gamma$ where $\beta$ is real (that is $\beta=\bar{\beta}$ (where $\bar{\alpha}$ is defined s.t. if $|A\rangle=\alpha|B\rangle$, $\langle A|=\langle B|\bar{\alpha}$)) and $\gamma$ is imaginary ($\gamma=-\bar{\gamma}$). Can somebody help me prove this?

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Observe that $$ A = \frac{1}{2}(A + A^\dagger) + \frac{1}{2}(A - A^\dagger) $$ where the summands are Hermitian and anti-Hermitian as desired.

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  • $\begingroup$ I am sorry, what is $A^{\dagger}$ and hermitian and anti-hermitian? $\endgroup$ – Iván Mauricio Burbano Jun 7 '15 at 15:48
  • $\begingroup$ @Ivan: Oh, in your notation, $A^\dagger = \bar A$ and Hermitian/anti-Hermitian are the properties $A^\dagger = A$ and $A^\dagger = -A$. $\endgroup$ – ACuriousMind Jun 7 '15 at 15:50
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By definition $$\alpha^*=\bar{\alpha}$$ and is called a complex conjugate so for the real part, the complex conjugate is itself while the imaginary part takes on the opposite sign.

To prove that $$\langle A|=\langle B|\bar{\alpha}$$ first define this equation to be the complex transpose of $$|A\rangle=\alpha|B\rangle$$

For the complex number alpha, the complex transpose is simply the complex conjugate. The complex transpose is just the complex conjugate of the individual components of A and B as well as turning the row vectors to column vectors or the column vectors to row vectors, also called the transpose of the vectors. Hence the name complex transpose (Refer to wikipedia complex transpose). If we denote a complex transpose as * because I don't know how to make that t cross sign thing, then

$$\langle A|=|A\rangle^*=(\alpha|B\rangle)^*=\langle B|\bar{\alpha}$$ $$_{Q.E.D}$$ P.S

I don't know if this violates the rule on giving near complete or complete answers so if it does please tell me and I will either delete or edit this post.

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  • $\begingroup$ The question is why all linear operators can be expressed as the sum of a real and an imaginary part, not what you are doing here. $\endgroup$ – ACuriousMind Jun 7 '15 at 16:07
  • $\begingroup$ Well then it is similar to how complex numbers can be expressed as a sum of real and imaginary parts. The operators can also be split into real and complex parts. Sorry I thought he meant to help prove whatever I wrote up there. $\endgroup$ – Horus Jun 7 '15 at 16:12

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