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alisa
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show that if A and B are both Hermitian, AB is Hermitian only if [A,B]=0. where or how do io start?
alisa said:show that if A and B are both Hermitian, AB is Hermitian only if [A,B]=0. where or how do io start?
dextercioby said:Here's how i'd do it. Consider the scalar product
[tex] \langle x, AB y\rangle [/tex] (1)
for "x" unspecified yet and [itex] y\in D(AB), \overline{D(AB)}=\mathcal{H} [/itex] an arbitrary element.
[tex] \langle x, AB y\rangle = \langle x, A^{\dagger}B y\rangle [/tex] (2)
,where [itex] D(A^{\dagger}B)=D(AB) [/itex], since [itex] D(A)\subseteq D\left(A^{\dagger}\right) [/itex]
[tex] \langle x, A^{\dagger}B y\rangle =\langle x, A^{\dagger}B^{\dagger} y\rangle [/tex] (3)
as [itex] B\subseteq B^{\dagger} [/itex]. Therefore [itex] y\in D\left(A^{\dagger}B^{\dagger}\right) [/itex] and [itex] D(AB)\subseteq D\left(A^{\dagger}B^{\dagger}\right) [/itex].
[tex] \langle x, A^{\dagger}B^{\dagger} y\rangle =\langle Ax, B^{\dagger} y\rangle [/tex] (4),
if [itex] x\in D(A) [/itex].
[tex] \langle Ax, B^{\dagger} y\rangle =\langle BAx, y\rangle [/tex] (5),
if [itex] x\in D(BA)\subseteq D(A)[/itex].
[tex] \langle BAx, y\rangle=\langle ABx, y\rangle [/tex] (6),
since, by hypothesis [itex] AB=BA [/itex].
Finally
[tex] \langle ABx, y\rangle =\langle x, \left(AB)^{\dagger}y\rangle [/tex](7)
by the definition of the adjoint. Therefore [itex] y\in D\left((AB)^{\dagger}\right) [/itex] and
[tex] ABy=(AB)^{\dagger} y {} \wedge D(AB)\subseteq D\left((AB)^{\dagger}\right) [/tex] (8),
which means the operator AB is symmetric/hermitean.
QED.
alisa said:show that if A and B are both Hermitian, AB is Hermitian only if [A,B]=0. where or how do io start?
quetzalcoatl9 said:in which case the proof would reduce to about 4 lines.
quetzalcoatl9 said:in other words, when would the domain of A and its hermitian conjugate NOT be the same?
wm said:Dear alisa, Does it go like this?
Let |n> be an eigenstate of A and B. Then:
AB|n> = Abn|n> = bn.an|n>.
Therefore AB is hermitean because bn.an is real.
In a similar expansion you will find that: [A, B]|n> = 0.
So, given |n>, AB is hermitean and [A, B] = 0.
This should point you in the right direction; check it out, wm
dextercioby said:The spectrum of a hermitean operator is not necessarily real. Actually only the eigenvalues (points in the pure point spectrum) are real, however, one cannot guarantee that [itex] |n\rangle [/itex] is an element of the Hilbert space. Therefore one cannot guarantee that "a_{n}b_{n}" is real. And even if it was real, there are examples of nonhermitean operators with real spectral values.
dextercioby said:I don't know Alisa's level of education. And in this case i'd rather make no assumption, unlike you.
A Hermitian operator is a mathematical concept used in quantum mechanics to describe observable properties of a physical system. It is a linear operator that is equal to its own adjoint, meaning that the operator and its conjugate transpose are the same.
An operator is Hermitian if it satisfies the condition [A,B]=0, meaning that its commutator with any other operator is equal to zero. This implies that the operator has real eigenvalues and its eigenvectors form an orthonormal basis.
In quantum mechanics, Hermitian operators are used to represent physical observables, such as position, momentum, and energy. The Hermitian property ensures that the measurement of these observables will always yield real values and that the operators have a complete set of eigenvectors.
No, not all operators in quantum mechanics are Hermitian. However, any observable quantity in quantum mechanics must be represented by a Hermitian operator, as this ensures that the measured values are real and that the operator has a complete set of eigenvectors.
The commutator [A,B]=0 indicates that the operators A and B can be measured simultaneously without affecting each other's values. This is known as the principle of compatibility and is essential in quantum mechanics for making simultaneous measurements of observables.