- #1
- 19,790
- 10,746
Definition/Summary
The Hermitian transpose or Hermitian conjugate (or conjugate transpose) M^{\dagger} of a matrix M is the complex conjugate of its transpose M^T.
A matrix is Hermitian if it is its own Hermitian transpose: M^{\dagger}\ =\ M.
An operator A is Hermitian (or self-adjoint) if it is its own adjoint: \langle Ax|y\rangle\ =\ \langle x|Ay\rangle (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 http://en.wikipedia.org/wiki/Hermitian.
Equations
\int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx
\langle b |\hat{O} |c \rangle = \langle c |\hat{O} |b \rangle ^*
Extended explanation
A matrix M is hermitian if:
M^{\dagger} = (M^T)^* = M ,
where \dagger 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
\hat{x},
momentum operator
-i\hbar \dfrac{\partial}{\partial x},
(with discrete spectrum)
z-component of angular momentum operator
\hat{L}_z .
In terms of wave functions, an operator \hat{O} is hermitian if:
\int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx
In terms of bra-ket:
\langle b |\hat{O} |c \rangle = \langle c |\hat{O} |b \rangle
Now, using the wave function formalism, some valuable identities will be presented:
Let us consider two hermitian operators \hat{A} and \hat{B}.
The expectation value:
<\hat{A}> = \int \psi ^* (\hat{A} \psi ) dx = \int (\hat{A}\psi )^* \psi dx,
is real, proof:
<\hat{A}>^* = \int ((\hat{A}\psi)^*\psi)^* dx = \int (\hat{A}\psi)\psi^* dx = \int \psi^* (\hat{A}\psi) dx = <\hat{A}>
Since \hat{A} was said to be hermitian, and \psi _1 = \psi _2 when we do expectation values.
Expectation value of \hat{A}^2:
<\hat{A}^2> = \int \psi ^* (\hat{A}(\hat{A}\psi)) dx = \int \psi^* (\hat{A}\tilde{\psi})dx =
(\tilde{\psi} = \hat{A}\psi \: \text{ is a new wavefunction} )
\int (\hat{A}\psi)^*\tilde{\psi}dx = \int (\hat{A}\psi)^*(\hat{A}\psi) dx
Now we can show another useful result:
\int \psi^* (\hat{A}(\hat{B}\psi))dx = \int(\psi^*(\hat{B}(\hat{A}\psi)))^*dx ,
prove this as an exercise.
Two more useful things:
I = \int \psi^*(\hat{A}\hat{B}+\hat{B}\hat{A})\psi = I^*
is real, show this as an exercise.
The operators always to the right if not indicated otherwise. Thus:
I = \int \psi^*(\hat{A}(\hat{B}\psi))dx + \int \psi^* (\hat{B}(\hat{A}\psi)) dx
J = \int \psi^*(\hat{A}\hat{B}-\hat{B}\hat{A})\psi = -J^*
is imaginary, show this as an exercise.
These identities are needed to prove the uncertainty relations of quantum mechanics.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
The Hermitian transpose or Hermitian conjugate (or conjugate transpose) M^{\dagger} of a matrix M is the complex conjugate of its transpose M^T.
A matrix is Hermitian if it is its own Hermitian transpose: M^{\dagger}\ =\ M.
An operator A is Hermitian (or self-adjoint) if it is its own adjoint: \langle Ax|y\rangle\ =\ \langle x|Ay\rangle (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 http://en.wikipedia.org/wiki/Hermitian.
Equations
\int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx
\langle b |\hat{O} |c \rangle = \langle c |\hat{O} |b \rangle ^*
Extended explanation
A matrix M is hermitian if:
M^{\dagger} = (M^T)^* = M ,
where \dagger 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
\hat{x},
momentum operator
-i\hbar \dfrac{\partial}{\partial x},
(with discrete spectrum)
z-component of angular momentum operator
\hat{L}_z .
In terms of wave functions, an operator \hat{O} is hermitian if:
\int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx
In terms of bra-ket:
\langle b |\hat{O} |c \rangle = \langle c |\hat{O} |b \rangle
Now, using the wave function formalism, some valuable identities will be presented:
Let us consider two hermitian operators \hat{A} and \hat{B}.
The expectation value:
<\hat{A}> = \int \psi ^* (\hat{A} \psi ) dx = \int (\hat{A}\psi )^* \psi dx,
is real, proof:
<\hat{A}>^* = \int ((\hat{A}\psi)^*\psi)^* dx = \int (\hat{A}\psi)\psi^* dx = \int \psi^* (\hat{A}\psi) dx = <\hat{A}>
Since \hat{A} was said to be hermitian, and \psi _1 = \psi _2 when we do expectation values.
Expectation value of \hat{A}^2:
<\hat{A}^2> = \int \psi ^* (\hat{A}(\hat{A}\psi)) dx = \int \psi^* (\hat{A}\tilde{\psi})dx =
(\tilde{\psi} = \hat{A}\psi \: \text{ is a new wavefunction} )
\int (\hat{A}\psi)^*\tilde{\psi}dx = \int (\hat{A}\psi)^*(\hat{A}\psi) dx
Now we can show another useful result:
\int \psi^* (\hat{A}(\hat{B}\psi))dx = \int(\psi^*(\hat{B}(\hat{A}\psi)))^*dx ,
prove this as an exercise.
Two more useful things:
I = \int \psi^*(\hat{A}\hat{B}+\hat{B}\hat{A})\psi = I^*
is real, show this as an exercise.
The operators always to the right if not indicated otherwise. Thus:
I = \int \psi^*(\hat{A}(\hat{B}\psi))dx + \int \psi^* (\hat{B}(\hat{A}\psi)) dx
J = \int \psi^*(\hat{A}\hat{B}-\hat{B}\hat{A})\psi = -J^*
is imaginary, show this as an exercise.
These identities are needed to prove the uncertainty relations of quantum mechanics.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
