Proof about hermitian operators

In summary, the conversation discusses two definitions of hermitian operators in quantum mechanics and the attempt at proving their equivalence. The first definition states that for any function f in hilbert space, <f|Qf> = <Qf|f>, while the second definition states that for any functions f and g in hilbert space, <f|Qg> = <Qf|g>. The conversation explores the possibility of proving that the first definition implies the second, with the use of a hint provided in a quantum mechanics book. The final conclusion is that the two definitions are indeed equivalent, with the condition that the operator Q must be hermitian.
  • #1
AlexChandler
283
0

Homework Statement



This is something I've been trying to prove for a bit today. My quantum mechanics book claims that the following two definitions about hermitian operators are completely equivalent

my operator here is Q (with a hat) and we have functions f,g

[tex] \langle f \mid \hat Q f \rangle = \langle Q f \mid \hat f \rangle [/tex] for any function f in hilbert space

and

[tex] \langle f \mid \hat Q g \rangle = \langle Q f \mid \hat g \rangle [/tex] for any functions f,g in hilbert space

2. Related formulas

[tex] \langle f \mid g \rangle = \int f^{\ast} g dx [/tex]

The Attempt at a Solution



Clearly the second definition implies the first, but I'm having trouble showing that the first implies the second.
My quantum mechanics book has this as an exercise and as a hint it suggests to let f=g+h
and then let f=g+ih with i being the square root of -1. I have done both of these things, expanding the inner products in terms of integrals. If i assume the first definition and let f=g+h, i can get

[tex] \langle f \mid \hat Q g \rangle + \langle g \mid \hat Q f \rangle = \langle \hat Q f \mid g \rangle + \langle \hat Q g \mid f \rangle [/tex]

doing a similar thing with f=g+ih i get the same result, and not sure where to go from here. Anybody have a better way to prove it or any ideas? thanks
 
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  • #2
I don't see any definitions in your post.
 
  • #3
Ah Sorry about that. I've just edited my post and corrected that
 
  • #4
Start with (Qf,f)=(f,Qf). If you take f=g+h and simplify, you get (g,Qh) + (h,Qh) = (Qg,h) + (Qh,g). And if you take f=g-ih and simplify, you get (g,Qh) - (h,Qg) = (Qg,h) - (Qh,g).
 
  • #5
isn't it just that

Q=Q^dagger where dagger is the transpose and complex conjugate

for a matrix to be hermitian

such that <Q| = |Q>
 

1. What is a hermitian operator?

A hermitian operator is a type of linear operator in quantum mechanics that has the property of being self-adjoint. This means that the operator is equal to its own conjugate transpose.

2. How can you prove that an operator is hermitian?

To prove that an operator is hermitian, you must show that it satisfies the condition of being equal to its own conjugate transpose. This can be done by taking the complex conjugate of the operator and comparing it to its transpose. If they are equal, then the operator is hermitian.

3. What is the significance of hermitian operators in quantum mechanics?

Hermitian operators play a crucial role in quantum mechanics as they represent physical observables, such as energy, momentum, and spin. They also have real eigenvalues, which correspond to the possible outcomes of measurements in quantum systems.

4. Can a non-hermitian operator have real eigenvalues?

No, a non-hermitian operator cannot have real eigenvalues. This is because the eigenvalues of a hermitian operator are always real, and the eigenvalues of a non-hermitian operator are complex numbers. This is one of the main differences between these two types of operators.

5. What are some examples of hermitian operators?

Some common examples of hermitian operators include the position and momentum operators, the angular momentum operator, and the Hamiltonian operator. These operators are used to describe fundamental physical quantities in quantum mechanics and play a crucial role in understanding the behavior of quantum systems.

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