Showing that general operator equations are hermitian

[UCLphysics]

Homework Statement

a)For a general operator A, show that and i(A-A⁺) are hermitian?

b) If operators A and B are hermitian, show that the operator (A+B)^n is Hermitian.

Homework Equations

The Attempt at a Solution

The first part I did,

(A+A⁺)⁺=(A⁺+A)=(A+A⁺)

i(A-A⁺)=[i(A-A⁺)]⁺=(iA)⁺-(iA⁺)⁺)=i(A-A⁺)

The second part I used binomial expansion (induction) but I was told they do not commute so this cannot be done?

Is my answer to the first part correct and what route should I take for the second? Thanks

 

[Oxvillian]

First part looks about right.

For the second part, do you really need to multiply things out? After all, if A and B are Hermitian then A+B is also Hermitian...

 

[dextercioby]

The problems seems to have no connection with <Advanced Physics> and neither with mathematics, since it's not mathematically (i.e. in agreement with mathematics) formulated and not even semantically.

 

[dextercioby]

For example, the mathematical problem for point a) should have been

"Let A be a densly defined linear operator on a complex separable Hilbert space $\mathcal{H}$.

Prove that if $D(A)\cap D\left(A^{\dagger}\right)$ is dense everywhere in $\mathcal{H}$,

then the operator $i\left(A-A^{\dagger}\right)$ is symmetric.

 

[paranormal]

Your answer to the first part is correct. As for the second part, you are correct that operators do not generally commute, so the binomial expansion method may not work. Instead, you can approach this problem by using the definition of a Hermitian operator, which states that A is Hermitian if A=A+. So, for (A+B)^n to be Hermitian, we must show that [(A+B)^n]^+=(A+B)^n.

We can start by expanding the left side using the definition of the Hermitian conjugate:

[(A+B)^n]^+=(A+B)^+*(A+B)^+*...*(A+B)^+

= (A+B)^+ * (A+B)^+ * ... * (A+B)^+

= (A^+ + B^+) * (A^+ + B^+) * ... * (A^+ + B^+)

= (A^+ * A^+ * ... * A^+) + (A^+ * B^+ * ... * B^+) + (B^+ * A^+ * ... * B^+) + (B^+ * B^+ * ... * B^+)

= A^n + A^(n-1)B + BA^(n-1) + B^n

Now, using the definition of a Hermitian operator, we know that A and B are Hermitian, so A=A+ and B=B+. Therefore, we can rewrite the above expression as:

= A^n + A^(n-1)B + AB^(n-1) + B^n

= (A^n + A^(n-1)B + AB^(n-1) + B^n)^+

= (A+B)^n

Hence, we have shown that [(A+B)^n]^+=(A+B)^n, and therefore, (A+B)^n is Hermitian.