1. ## Quantum Probs

1.) Given A and B are both hermitian operators,

a.) Prove the product $AB$ will be hermitian only if $A$ and $B$ commute

b.) Prove $(A+B)^n$ is hermitian.

2.) Through the use of dirac notation, prove the eigenstates of a hermitian operator that belong to 2 different eigenvalues are orthogonal

2. Originally Posted by DiscreteW
1.) Given A and B are both hermitian operators,

a.) Prove the product $AB$ will be hermitian only if $A$ and $B$ commute

b.) Prove $(A+B)^n$ is hermitian.

2.) Through the use of dirac notation, prove the eigenstates of a hermitian operator that belong to 2 different eigenvalues are orthogonal
I will get to this as soon as I am able.

-Dan

3. Originally Posted by DiscreteW
1.) Given A and B are both hermitian operators,

a.) Prove the product $AB$ will be hermitian only if $A$ and $B$ commute

b.) Prove $(A+B)^n$ is hermitian.
A Hermitian operator is defined as an operator C such that $C^{\dagger} = C$.

So let's take any two Hermitian operators A and B and any two arbitrary kets $| a >, | b >$. (Please note that $| a >, | b >$ are not being taken as eigenkets of A or B.)

$< a | AB | b >$

$= < a | A^{\dagger} B^{\dagger}| b >$

$= < a | (BA)^{\dagger}| b >$

$= < a | (AB)^{\dagger}| b >$ (since [A, B] = 0)

If you don't require quite this amount of formality, then
$AB = A^{\dagger} B^{\dagger} = (BA)^{\dagger} = (AB)^{\dagger}$
will do just fine.

Part b) of this is easy. Just let C = A + B. Since A and B are Hermitian, so is C. Obviously C commutes with itself, so we know that $(A + B) ^2$ is Hermitian by the theorem in 1a). Now, $C^2$ and C obviously commute, thus etc.

-Dan

4. Originally Posted by DiscreteW
2.) Through the use of dirac notation, prove the eigenstates of a hermitian operator that belong to 2 different eigenvalues are orthogonal
So assume we are given a Hermitian operator A and two normalized eigenkets $| a >, | b >$ such that a and b are not equal.

First we show that the eigenvalues of A are real. (If you have already proven this, then just skip to the next part.)

We have
$< c | A | c > = < c | A^{\dagger} | c > ^* = < c | A | c > ^*$

but
$< c | A | c > = < c | (c) | c > = c < c | c > = c$
and
$< c | A | c > ^* = < c | (c) | c > ^* = c^* < c | c >^* = c^*$ (since $< c | c > = 1$ is real.)

Thus $c^* = c$ for all c.

So to prove the theorem consider
$< a | A | b > = < a | (b) | b > = b < a | b >$

Again:
$< a | A | b > = < b | A^{\dagger} | a > ^* = < b | A | a > ^* = < b | (a) | a > ^* = a^* < b | a > ^* = a^* < a | b >$

But the eigenvalues of a Hermitian operator are real, thus
$< a | A | b > = a < a | b >$

Thus
$< a | A | b > - < a | A | b > = b < a | b > - a < a | b >$

$0 = (b - a) < a | b >$

So either b - a = 0, which is not possible according to assumption, or $< a | b > = 0$.

-Dan

5. VERY thorough and helpful as usual. I'm grateful that this site has someone good at both physics and math to help with these! The answers will very clear. Thanks.

I have an exam next week, and I plan to study this weekend, so I'll probably have more questions if you have time. Thanks again.

6. Originally Posted by DiscreteW
VERY thorough and helpful as usual. I'm grateful that this site has someone good at both physics and math to help with these! The answers will very clear. Thanks.

I have an exam next week, and I plan to study this weekend, so I'll probably have more questions if you have time. Thanks again.
(I'm blushing!) Thank you, and feel free to post any time. I can't always promise I'll be around to get to them, but I will if I can.

-Dan