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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?

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- Thread starter 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?

- #2

quasar987

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From the property (AB)+=B+a+

+ denoting hermitian conjugation

+ denoting hermitian conjugation

- #3

cristo

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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?

Expand [A,B]=0, then use the hint above.

- #4

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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.

[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.

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- #5

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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.

wouldn't it be easier to just say that x,y are in the domain of the operators A, B (which i agree IS important to state, as is actually applying the operator to an element as opposed to treating it as some "algebraic" quantity as many books do)? in which case the proof would reduce to about 4 lines. in other words, when would the domain of A and its hermitian conjugate NOT be the same?

- #6

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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?

Dear

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,

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- #7

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in which case the proof would reduce to about 4 lines.

I don't know about that. I don't claim that my proof is unique/the shortest possible.

in other words, when would the domain of A and its hermitian conjugate NOT be the same?

By the Hellinger-Toeplitz theorem, iff the operator A is bounded. My proof accounts for the arbitrary character of the A and B operators.

- #8

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Dearalisa, 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

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.

- #9

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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.

Dear

Let

(1) A|x> =

(2) <x|A|x> = <x|A|x>*. Substituting (1) into (2):

(3) <x|

(4)

(5)

Thus, in the example that I gave: bn.an is real (because each factor is real).

Would that be satisfactory at

- #10

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I don't know Alisa's level of education. And in this case i'd rather make no assumption, unlike you.

- #11

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I don't know Alisa's level of education. And in this case i'd rather make no assumption, unlike you.

Dear

Thanks; and best regards,

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