MHB Is <L^2> always greater than or equal to 0 for a Hermitian operator?

[ognik]

I'm given an operator $\mathcal{L}$ is Hermitian, and asked to show $<\mathcal{L}^2>$ is $\ge 0$
I believe $<\mathcal{L}>$ is the expectation value, $=\int_{}^{}\Psi^* \mathcal{L} \Psi \,d\tau $

(Side issue: I am not sure what $d\tau $ is, perhaps a small region of space? And the interval?)

I can show that $<\mathcal{L}>$ is real, ie $<\mathcal{L}> = <\mathcal{L}^*>$, doesn't seem useful here ...

I tried $ <\mathcal{L}^2>=\int_{}^{}\Psi^* \mathcal{L}^2 \Psi \,d\tau $ $ =\int_{}^{}\Psi^* \mathcal{L} \mathcal{L} \Psi \,d\tau $ $ =\int_{}^{}(\mathcal{L}^*\Psi)^* \mathcal{L} \Psi \,d\tau $
${ (\Psi^* \mathcal{L}) = (\mathcal{L}^* \Psi)^*} $

I could argue that inside the integral we have $ (\mathcal{L} \Psi) $ times it's conjugate which must therefore be positive, but as you can see I have a spurious conjugate in $(\mathcal{L}^* \Psi)^* $ Any ideas?
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I am then asked to show that if <A> is real, then A must be Hermitian w.r.t. $\psi$
$ <A>=\int \psi^* A \psi \,d\tau = <\psi|A \psi>. $
But $A=A^*$ so <A>= $\int \psi^*A^*\psi d\tau =<A\psi|\psi> $ $ \therefore <\psi|A \psi> =<A\psi|\psi> $ and A is Hermitian

Is that good enough?

 

[Euge]

Doesn't $\mathcal{L}^* = \mathcal{L}$ since $\mathcal{L}$ is Hermitian? So $\mathcal{L}^2 = \mathcal{L}^*\mathcal{L}$, and thus

$$\langle \mathcal{L^2}\rangle = \int \Psi^*(\tau) \mathcal{L}^*(\tau) \mathcal{L}(\tau)\Psi(\tau)\, d\tau = \int (\mathcal{L}\Psi)^*(\tau) (\mathcal{L}\Psi)(\tau)\, d\tau = \|\mathcal{L}\Psi\|^2 \ge 0.$$

 

[ognik]

You're right of course Euge (I should have seen that immediately!). I saw what I did with $L^2$ somewhere while researching expectation value; it's obviously wrong - just shows how valuable this forum is, my gratitude is endless :-)

I think my 2nd part is right?

 

[Euge]

Your 2nd part is wrong. You used circular logic, namely, you assumed $A$ is Hermitian in order to prove $A$ is Hermitian.

You want to show that $\langle A\rangle = \langle A^*\rangle$. Try proving that.

 

[ognik]

Your 2nd part is wrong. You used circular logic, namely, you assumed $A$ is Hermitian in order to prove $A$ is Hermitian.

I used A = to it's complex conjugate, IE A is real. Then I showed that $ <ψ|Aψ> =<Aψ|ψ> $
My book states that 'An operator $L$ is Hermitian if $<ψ_1 |Lψ_2>=<Lψ_1 | ψ_2>$'

I would just like to understand why this is wrong, it still doesn't seem circular?

Note that in my solution, I said $But A=A^∗, so <A>= ...$, that was abbreviated, I could have said But $A=A^∗ so <A^*>= ..$ and then continued to the same conclusion, namely $ <ψ|Aψ> =<Aψ|ψ> $

You want to show that $\langle A\rangle = \langle A^*\rangle$. Try proving that.

I definitely will, want to absorb Deveno's post on terminology first ...

 

[Euge]

I used A = to it's complex conjugate, IE A is real.

How do you know that $A$ is real? The condition is that $\langle A\rangle$ is real, not that $A$ is real. So you have $\langle A\rangle = \overline{\langle A\rangle}$. Now $\langle A\rangle = \langle \Psi|A\Psi\rangle$ and so $\overline{\langle A\rangle} = \langle A\Psi|\Psi\rangle$. Hence $\langle A\rangle = \overline{\langle A\rangle}$ if and only if $\langle \Psi|A\Psi\rangle = \langle A\Psi|\Psi\rangle$.

 

[ognik]

OK - easy if one reads the question correctly <blush>

So I also tried it using the integral form and have another of my curious questions (I prefer * for conjugate):

$ <A>^*=\int (\Psi^* A \Psi)^* = \int \Psi A^*\Psi^* =<\Psi|A\Psi>^* $, but
$ <A>= \int \Psi^*A\Psi =<\Psi|A\Psi>$,

I completely understand why using the 'dirac form' gives A Hermitian, just wondering what I did wrong with the integral form approach?

 

[Euge]

$ <A>^*=\int (\Psi^* A \Psi)^* = \int \Psi A^*\Psi^* =<\Psi|A\Psi>^* $, but
$ <A>= \int \Psi^*A\Psi =<\Psi|A\Psi>$

Wait, $(\Psi^* A\Psi)^* = (A\Psi)^*(\Psi^*)^* = (A\Psi)^*\Psi$, and so

$$\langle A\rangle^* = \int (A\Psi)^*\Psi = \langle A\Psi|\Psi\rangle.$$

Now $\langle A\rangle = \langle \Psi|A\Psi\rangle$, so then $\langle \Psi|A\Psi\rangle = \langle A\Psi|\Psi\rangle$ as $\langle A\rangle = \langle A\rangle^*$.

 

[ognik]

Wait, $(\Psi^* A\Psi)^* = (A\Psi)^*(\Psi^*)^* = (A\Psi)^*\Psi$

I thought with a product conjugated, the order stayed the same, only swaps with Transpose?

 

[Euge]

The intermediate step of the first equality is $(\Psi^*)^*(A\Psi)^*$.

 

[ognik]

The intermediate step of the first equality is $(\Psi^*)^*(A\Psi)^*$.

Yes; Don't follow how you are then able to swap the order?

 

[Euge]

What do you mean, and why are you referring to a transpose? When multiplying complex numbers, you're allowed to swap the order, because the complex numbers are commutative. For example, $(1 + i)(1 - i) = 2 = (1 - i)(1 + i)$.

 

[ognik]

Transpose - I was referring to $ (AB)^T = B^TA^T$ , ie the order swaps.
But for conjugate I understood $ (AB)^*=A^*B^*$, can't swap the order?

Good point about swapping complex numbers, but was under the impression you can't do that if one of the 'numbers' includes an operator. So I think $ (A\Psi)^* $ must be $A^*\Psi^* $ and can't be swapped?

Your approach suggests an operator acting on a complex number results in a complex number so we can treat $(A\Psi)$ as a complex number which we can swap with another complex number; is that always the case, or are we assumed to be working only with operators for which that is true?

Also $\Psi$ can be a vector, function or a state - those are also commutative.
I think I have understood something new - if this all makes sense?

 

[Euge]

Transpose - I was referring to $ (AB)^T = B^TA^T$ , ie the order swaps.
But for conjugate I understood $ (AB)^*=A^*B^*$, can't swap the order?

Yes, I know about the transpose, but I don't see how that is relevant to the problem of showing $\langle A\rangle $ is real $\implies$ $A$ is Hermitian.

Good point about swapping complex numbers, but was under the impression you can't do that if one of the 'numbers' includes an operator. So I think $ (A\Psi)^* $ must be $A^*\Psi^* $ and can't be swapped?

Realize that $A^*$, the adjoint of $A$, is not the same as the conjugate of $A$. So I don't understand what you mean here.

Your approach suggests an operator acting on a complex number results in a complex number so we can treat $(A\Psi)$ as a complex number which we can swap with another complex number; is that always the case, or are we assumed to be working only with operators for which that is true?

Really, you're supposed to tell us the conditions of $A$ and $\Psi$, giving the full definitions. However, judging from the context of your other posts, I assumed at the least that $\Psi$ is a state or wavefunction (so it is complex-valued) and $A$ is an operator acting on such functions. Then $A\Psi$ is a complex-valued function, and the equation $[\Psi^*(A\Psi)]^* = (\Psi^*)^*(A\Psi)^*$ makes sense.

 

[ognik]

Realize that $A^*$, the adjoint of $A$, is not the same as the conjugate of $A$. So I don't understand what you mean here...

I think I see the problem, my book uses * for complex conjugate and $\dagger$ for adjoint/hermitain; there seems to be no majority agreement on notation here - this is one of the main reasons I am trying to put together a simple table for me to use, specifically for this course :-)

Yes, I know about the transpose, but I don't see how that is relevant to the problem of showing $\langle A\rangle $ is real $\implies$ $A$ is Hermitian..

I was saying that only with Transpose do we swap the order, not with complex conjugate. Adjoint/Hermitian include a transpose operation so they also must swap, so your * is my $\dagger$ and both have inherent swaps. So all good now.

Really, you're supposed to tell us the conditions of $A$ and $\Psi$, giving the full definitions. However, judging from the context of your other posts, I assumed at the least that $\Psi$ is a state or wavefunction (so it is complex-valued) and $A$ is an operator acting on such functions. Then $A\Psi$ is a complex-valued function, and the equation $[\Psi^*(A\Psi)]^* = (\Psi^*)^*(A\Psi)^*$ makes sense.

Thanks for that, the problem didn't give context for $A$ and $\Psi$, but I am sure your assumption is correct. So I can treat complex functions/states with normal complex algebra. (just something I don't think I had come across explicitly, seemed 'intuitively' likely, but I wanted to check)