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Hermitian
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Definition/Summary
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The Hermitian transpose or Hermitian conjugate (or conjugate transpose) [itex]M^{\dagger}[/itex] of a matrix [itex]M[/itex] is the complex conjugate of its transpose [itex]M^T[/itex].
A matrix is Hermitian if it is its own Hermitian transpose: [itex]M^{\dagger}\ =\ M[/itex].
An operator [itex]A[/itex] is Hermitian (or self-adjoint) if it is its own adjoint: [itex]\langle Ax|y\rangle\ =\ \langle x|Ay\rangle[/itex] (in the finite-dimensional case, that means that its matrix is Hermitian).
In quantum theory, an observable must be represented by a Hermitian operator (on a Hilbert space).
For other uses of the adjective "Hermitian", see [url]http://en.wikipedia.org/wiki/Hermitian[/url]. |
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Equations
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[tex] \int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx [/tex]
[tex] \langle b |\hat{O} |c \rangle = \langle c |\hat{O} |b \rangle ^*[/tex] |
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Recent forum threads on Hermitian
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Breakdown
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Physics
> Quantum
>> Foundations
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Extended explanation
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A matrix [itex] M [/itex] is hermitian if:
[tex] M^{\dagger} = (M^T)^* = M , [/tex]
where [itex] \dagger [/itex] is called the hermitian conjugate, and is thus a combination of matrix transpose and complex conjugation of each entry in the matrix.
In quantum mechanics, observable quantities are assigned by hermitian operators. Examples of those are:
(with continuous spectrum)
position operator
[tex] \hat{x}, [/tex]
momentum operator
[tex] -i\hbar \dfrac{\partial}{\partial x}, [/tex]
(with discrete spectrum)
z-component of angular momentum operator
[tex] \hat{L}_z . [/tex]
In terms of wave functions, an operator [itex] \hat{O} [/itex] is hermitian if:
[tex] \int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx [/tex]
In terms of bra-ket:
[tex] \langle b |\hat{O} |c \rangle = \langle c |\hat{O} |b \rangle [/tex]
Now, using the wave function formalism, some valuable identities will be presented:
Let us consider two hermitian operators [itex] \hat{A} [/itex] and [itex] \hat{B} [/itex].
The expectation value:
[tex] <\hat{A}> = \int \psi ^* (\hat{A} \psi ) dx = \int (\hat{A}\psi )^* \psi dx,[/tex]
is real, proof:
[tex] <\hat{A}>^* = \int ((\hat{A}\psi)^*\psi)^* dx = \int (\hat{A}\psi)\psi^* dx = \int \psi^* (\hat{A}\psi) dx = <\hat{A}> [/tex]
Since [itex] \hat{A} [/itex] was said to be hermitian, and [itex] \psi _1 = \psi _2 [/itex] when we do expectation values.
Expectation value of [itex] \hat{A}^2 [/itex]:
[tex] <\hat{A}^2> = \int \psi ^* (\hat{A}(\hat{A}\psi)) dx = \int \psi^* (\hat{A}\tilde{\psi})dx = [/tex]
[tex] (\tilde{\psi} = \hat{A}\psi \: \text{ is a new wavefunction} )[/tex]
[tex] \int (\hat{A}\psi)^*\tilde{\psi}dx = \int (\hat{A}\psi)^*(\hat{A}\psi) dx [/tex]
Now we can show another useful result:
[tex] \int \psi^* (\hat{A}(\hat{B}\psi))dx = \int(\psi^*(\hat{B}(\hat{A}\psi)))^*dx ,[/tex]
prove this as an exercise.
Two more useful things:
[tex] I = \int \psi^*(\hat{A}\hat{B}+\hat{B}\hat{A})\psi = I^* [/tex]
is real, show this as an exercise.
The operators always to the right if not indicated otherwise. Thus:
[tex] I = \int \psi^*(\hat{A}(\hat{B}\psi))dx + \int \psi^* (\hat{B}(\hat{A}\psi)) dx [/tex]
[tex] J = \int \psi^*(\hat{A}\hat{B}-\hat{B}\hat{A})\psi = -J^* [/tex]
is imaginary, show this as an exercise.
These identities are needed to prove the uncertainty relations of quantum mechanics. |
Commentary
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