Property of the adjoint operator

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SUMMARY

The adjoint operator A* is defined by the relationship <\phi|A\psi> = . In the discussion, participants explore how to demonstrate that (cA)* = c*P* using properties of inner products such as skew-symmetry, positive semi-definiteness, and linearity. The conversation emphasizes the importance of scalar multiplication with operators and the properties of inner products to derive the adjoint of a scalar multiplied operator. Additionally, the property (PQ)^{\dag} = Q^{\dag}P^{\dag} is mentioned as a more complex topic requiring further exploration.

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  • Understanding of adjoint operators in linear algebra
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  • Knowledge of scalar multiplication with operators
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Mathematicians, physicists, and students studying linear algebra or quantum mechanics who seek to deepen their understanding of adjoint operators and their properties.

ehrenfest
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The adjoint of an operator A is defined as an operator A* s.t.

&lt;\phi|A\psi&gt; = &lt;A^{*}\phi|\psi&gt;.

How would you use the properties of inner products (skew-symmetry, positive semi-definiteness, and linearity in ket) to show that (cA)* = c*P*


Note that I am using the conjugate and the adjoint symbol interchangeably. If anyone knows how to get a real adjoint symbol in LaTeX let me know.
 
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what does \dag do? I assume that A and P are supposed to be the same letter... and then it is trivial. What have you attempted?
 
Yes sorry. A and P are supposed to be the same letter.

You could use the skew-symmetry property to show that:

&lt;\phi|cA\psi&gt; = &lt;cA\psi|\phi&gt;^{*}

and that

&lt;cA^{\dag}\phi|\psi&gt; = &lt;\psi|cA^{\dag}\phi&gt;^{*}

but I do not see how that helps.
 
((cA)x | y) = ...

Unless I'm mistaken, you have to use the definition of scalar multiplication with operators, and two peoperties of the inner product. In three (four) steps, you can show what (cA)* equals.
 
I see. So, &lt;(cA)x|y&gt; = c*&lt;Ax|y&gt; = c*&lt;x|A^{\dag}y&gt; = &lt; x|c*A^{\dag}y&gt;.
 
What about the property (PQ)^{\dag} = Q^{\dag}P^{\dag}? This one seems a bit more difficult.
 
ehrenfest said:
What about the property (PQ)^{\dag} = Q^{\dag}P^{\dag}? This one seems a bit more difficult.

I wouldn't really call it difficult. Again, ( (PQ)x | y ) = ...
 

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