Trouble with Hermitian operators?

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SUMMARY

The discussion centers on the properties of Hermitian operators, specifically regarding the relationship between the products of two Hermitian operators, A and B. It is established that the complex conjugate of the product of two Hermitian operators follows the rule (AB)† = B†A†, leading to the conclusion that ⟨AB⟩ and ⟨BA⟩ are complex conjugates. The confusion arises from the expected value notation, where the equality |⟨AB⟩| = |⟨BA⟩| holds true, but the user mistakenly attempts to equate |⟨AB - BA⟩| with |2⟨AB⟩|, which is incorrect and should be treated as an inequality.

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  • Understanding of Hermitian operators and their properties
  • Familiarity with complex conjugation in quantum mechanics
  • Basic knowledge of expected values in quantum mechanics
  • Ability to manipulate operator algebra
NEXT STEPS
  • Study the derivation of the Heisenberg Uncertainty Principle in detail
  • Learn about the properties of Hermitian conjugation and its implications
  • Explore the concept of expected values in quantum mechanics
  • Investigate inequalities in quantum mechanics, particularly in relation to uncertainty
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Students in high school or early college studying quantum mechanics, particularly those interested in operator algebra and the foundational concepts of Hermitian operators.

21joanna12
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I am looking at the derivation of the Heisenberg Uncertainty Principle presented here: http://socrates.berkeley.edu/~jemoore/p137a/uncertaintynotes.pdf

and am confused about line (21)...

I do not understand why AB and BA are complex conjugates of each other... (I'm still in high school so I don't really have much of a background in the algebra of operators).

If I assume what is written in line 21, I think I get the following lines- if AB and BA are complex conjugates, then |\langle AB-BA \rangle |^2 = |2\langle AB\rangle |^2 = 4|\langle AB\rangle |^2 I think... The expected value signs are what is confusing me most...

Thank you for any help!
 
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21joanna12 said:
I do not understand why AB and BA are complex conjugates of each other... (I'm still in high school so I don't really have much of a background in the algebra of operators).
A hermitian operator has the property ##A^\dagger = A##, where the ##\dagger## symbol refers to hermitian conjugation, which is the extension to operators (or matrices) of complex conjugation. Also, when applying hermitian conjugation to a product of operators, you have to invert their order:
$$
(A B)^\dagger = B^\dagger A^\dagger
$$
So, if ##A## and ##B## are hermitian, you get
$$
(A B)^\dagger = B^\dagger A^\dagger = BA
$$
The complex conjugate of a bracket is given by
$$
\left( \langle \psi | A | \phi \rangle \right)^* = \langle \phi | A^\dagger | \psi \rangle
$$
When you combine these things together, you find
$$
\begin{align*}
\left( \langle AB \rangle \right)^* &= \left( \langle \psi | AB | \psi \rangle \right)^* \\
&= \langle \psi | (A B)^\dagger | \psi \rangle \\
&= \langle \psi | BA | \psi \rangle \\
&= \langle B A \rangle
\end{align*}
$$
So, as for all complex numbers ##|z| = |z^*|##, you have that ##|\langle AB \rangle| = |\langle BA \rangle|##.

21joanna12 said:
If I assume what is written in line 21, I think I get the following lines- if AB and BA are complex conjugates, then |\langle AB-BA \rangle |^2 = |2\langle AB\rangle |^2 = 4|\langle AB\rangle |^2 I think...
You can only write an inequality here, not equalities.
 
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