# QM Bra & Ket Linear Algebra Hermitian operator proof -- quick question

1. Aug 31, 2016

### binbagsss

1. The problem statement, all variables and given/known data

Hi,

Just watching Susskind's quantum mechanics lecture notes, I have a couple of questions from his third lecture:

2. Relevant equations

1) At 25:20 he says that
$<A|\hat{H}|A>=<A|\hat{H}|A>^*$ [1]
$<=>$
$<B|\hat{H}|A>=<A|\hat{H}|B>^*=$ [2]
where $A$ and $B$ are vectors, $H$ is a hermitian operator and $^*$ means to complex conjugate.

The proof of the bottom implying the top is trivial, setting $B=A$.

I have attempted the proof that the top implies the bottom :

3. The attempt at a solution

Let $|B>=\hat{H}|A>$

Then $<B|=<A|\hat{H^*}$

Consider $<B|\hat{H}|A>=<A|\hat{H^*}\hat{H}|A>=<A|\hat{C}|A>$

Where $\hat{H^*}\hat{H}=\hat{C}$, since $H=H^*$ is a hermitian operator, they commute and so by the theorem that operators that commute are hermitian, $\hat{C}$ is hermitian (or is this theorem made redundant by the fact that $\hat{C}$ is real and all real operators are hermitian?) , so [1] which holds for a hermitian operator is true and therefore

$<A|\hat{C}|A>=<A|\hat{C}|A>^*$ by [1]

$=<A|\hat{H^*}\hat{H}|A>=<A|\hat{H^*}|B>^*$

$=<A|\hat{H}|B>^*$ by $H=H^*$.

- Is this last line valid? I'm not sure that is the strict definition of hermitian as different notes seem to suggest different things, some define hermitian as simply $H=H^*$ others more via the inner product and [2] holding.

2. Aug 31, 2016

### andrewkirk

Unfortunately, unless the 'hat' symbol means something specific, rather than being a non-semantic aspect of the label $\hat H$, I don't think the proof is valid.
We are being asked to prove that
$$\langle A|\hat H|A\rangle=\langle A|\hat H|A\rangle^*\Leftrightarrow \langle B|\hat H|A\rangle=\langle A|\hat H|B\rangle^*$$
For the rightward direction, our starting assumption is
$$\langle A|\hat H|A\rangle=\langle A|\hat H|A\rangle^*$$
$$\langle A|\hat C|A\rangle=\langle A|\hat C|A\rangle^*$$
This is not given, so cannot be used. It could be used if what we were asked to prove were the following
$$\forall \hat H\left(\langle A|\hat H|A\rangle=\langle A|\hat H|A\rangle^*\right)\Leftrightarrow \forall \hat H\left(\langle B|\hat H|A\rangle=\langle A|\hat H|B\rangle^*\right)$$
But that is not what was written.

Also, another minor point: it is not the case that all real operators are Hermitian. They also need to be symmetric.

3. Aug 31, 2016

### binbagsss

I thought the part where I said that since $\hat C$ is Hermitian covers this, since our starting assumption holds for any Hermitian operator $\hat H$ and I argued that $\hat C$ is Hermitian by the theorem that, since $\hat{H}=\hat{H^*}$ therefore $\hat{H}$ and $\hat{H*}$ commute, operators that commute are Hermitian, and therefore the starting assumption, that holds for any Hermitian operator holds.

Thank you, I wanted this clarified, will have a think about why this is so later.

4. Aug 31, 2016

### andrewkirk

The trouble is that the nature of what is to be proved is unclear because of the lack of quantifiers ($\forall$ and $\exists$, or their verbal equivalents) in its statement.
There are several different ways to interpret what is written in the OP. One is
$$\forall \hat H\left[ isHermitian(\hat H)\Rightarrow \left(\langle A|\hat H|A\rangle=\langle A|\hat H|A\rangle^*\Leftrightarrow \langle B|\hat H|A\rangle=\langle A|\hat H|B\rangle^*\right)\right]$$
Another is
$$\forall \hat H\left(isHermitian(\hat H)\Rightarrow\langle A|\hat H|A\rangle=\langle A|\hat H|A\rangle^*\right)\Leftrightarrow \forall \hat H\left(isHermitian(\hat H)\Rightarrow\langle B|\hat H|A\rangle=\langle A|\hat H|B\rangle^*\right)$$
The first is a stronger theorem than the second, as the first implies the second, but not vice versa. The excerpt from your post seems to imply you are trying to prove the second one, not the first. Is that correct?

If it is then your last line is valid because Hermitian implies self-adjoint whereas the opposite direction is not necessarily the case. In any case, we don't need to use the $Hermitian\Rightarrow selfadjoint$ theorem. Instead use the definition of $\hat H$ being Hermitian:
$$\forall u,v:\ \langle u|\hat Hv\rangle=\langle \hat Hu|v\rangle$$
Then we have
$$\langle A|\hat H^*|B\rangle^* =\langle \hat HA|B\rangle^* =\langle A|\hat HB\rangle^* =\langle A|\hat H|B\rangle^*$$
where the second '=' uses the Hermitian property.

Also, the Hermitianness of $\hat C$ follows easily from the inner-product definition of Hermitianness:
$$\langle \hat Cu|v\rangle =\langle (\hat H^*\hat H) u|v\rangle =\langle \hat H u|\hat H|v\rangle =\langle \hat H u|\hat Hv\rangle =\langle u|\hat H^*|\hat Hv\rangle =\langle u|\hat H^*\hat Hv\rangle =\langle u|\hat Cv\rangle$$

Last edited: Aug 31, 2016
5. Sep 2, 2016

### binbagsss

Oh thank you, I see. so the first is saying that these two inner products it is the same Hermitian operator and the second is saying that if this inner product holds for any Hermitian operator, the other one also holds for any Hermitian operator. Re-watching the video I believe it is the first that Susskind is refferring to, so my proof is not valid.

I understand the second equality uses the Hermitian property, just a quick question on the first equality, is what we've used here that

$|B^*>=<B|$, where $B$ is a vector works the same as for operators - on this note, are operators then part of the vector space itself?

or instead should it be that $|B>^*=<B|$, or are this and $|B^*>=<B|$ equivalent due to properties of complex conjugate? Thanks.

Also, back to the OP anyone have any hints then how to start the proof?
Thanks

6. Sep 2, 2016

### andrewkirk

I had a look at that part of the video. What he seems to be saying is that the following are equivalent properties of an operator $H$:
(1) $\forall A,B:\ \langle HA|B\rangle=\langle A|HB\rangle$ (ie $H$ is hermitian)
(2) $\forall A,B:\ \langle A|H|B\rangle=\langle B|H|A\rangle^*$
(3) $\forall A:\ \langle A|H|A\rangle=\langle A|H|A\rangle^*$
which, as you say, is closer to the first than the second of the possibilities I wrote.
It is easy to prove that $(2)\Rightarrow(3)$. Proving $(1)\Rightarrow(2)$ is also easy:
$$\langle A|H|B\rangle = \langle A|HB\rangle = \langle HA|B\rangle= \langle B|HA\rangle^*= \langle B|H|A\rangle^*$$
where property (1) is used in the second step.
What's needed to complete the equivalence is a proof that $(3)\Rightarrow(1)$.
We have three sets in play here:
(1) The vector space V (also a Hilbert space), whose elements are written as kets $|\psi\rangle$
(2) The dual space V*, which is the set of all complex linear functionals on $V$, ie linear functions from $V$ to $\mathbb C$. The elements of V* are written as bras $\langle\theta|$ and the dual of $|\psi\rangle$ is written as $\langle \psi|$ where dual means the functional that, when applied to $|\psi\rangle$ gives the square of its norm $\|\ |\psi\rangle\ \|^2$.
(3) The space $O$ of linear operators on V, which are linear functions from V to V. These are usually represented by letters without any angle brackets around them.

So in answer to your question, No, operators are part of (3). They are not part of the vector space (1). However, when an operator is chained with a bra it becomes a bra, eg $\langle A|M$ is a linear function from V to $\mathbb C$ because M takes an input ket in V and delivers another ket in V, and $\langle A|$ then takes that and gives a number in $\mathbb C$.

The first equality you ask about is simply the definition of the adjoint operator $H^*$ (often written $H^\dagger$). It is a theorem of Hilbert space theory that for any operator $H$ there exists an operator $H^\dagger$, called its adjoint, such that, for any $A\in V$ we have $\langle HA|=\langle A|H^\dagger$

Post-Script: OK I think I've got it:
Use the linearity of inner products to expand
$$\langle A+B|H|A+B\rangle$$
Then take the imaginary part of that, which given (3), must be zero. The result (2) will readily follow.

Last edited: Sep 2, 2016
7. Sep 2, 2016

### binbagsss

nice one I see what you're doing but can't quite get it. I now have :

$<A|H|B>+<B|H|A>=0$
$-<B|H|A>=<A|H|B>=-<B|H|A>*$

8. Sep 2, 2016

### binbagsss

Oh I see, thank you that makes sense, my confusion stemmed because you said we can do this using only the Hermitian property.

Thank you for clarifying the theorem, I have a few questions on this concept/formalisation of the spaces involved.

You introduced spaces (1),(2) and (3), and from the Hilbert space adjoint operator exisiting theorem, I guess consistent with this notation that we have (4) $O^\dagger$ for the adjoint operators.

I am a little confused however because you said the theorem regarding the operators holds on Hilbert space, also, next to (1) you have (also a Hilbert space) .

My initial guess at the formalism of the spaces was that all (4) spaces lie within the Hilbert space? But after seeing (Hilbert space) next to (1) I'm not too sure.

Also reading about Hilbert space some notes say it says 'possesses strucute of inner product' , so for this to be true it must contain all spaces (1)-(4) surely?

Thanks

9. Sep 2, 2016

### andrewkirk

Very good. We do indeed, although I would label it as $O^*$ because for some reason (at least in the texts I have used), dual spaces use asterisks rather than daggers. As I presented it above, $H^\dagger$ is in $O$ rather than $O^*$, because it takes a ket as input and gives out a ket. However we can also view it as being in (4), and this introduces the beautiful symnmetry of the whole thing: we can read the expression $\langle A|M| B\rangle$ either from left to right, or from right to left. Read from left to right (the usual way) it means:
$$\langle A|\left(M\left(|B\rangle\right)\right)$$
which can also be written in function composition form as
$$\left(\langle A|\circ M\right)\left(|B\rangle\right)$$
Here we are interpreting the ket as a basic item in $V$ ((1)), the operator as in (3) and the bra as in (2). So we are applying the operator to the ket to get another ket, then applying the bra to that to get a scalar.

BUT, we can also interpret it from right to left, in which case it means
$$|B\rangle\left(M\left(\langle A|\right)\right)$$
or equivalently
$$\left(|B\rangle\circ M\right)\left(\langle A|\right)$$
Here we are interpreting the operator $M$ as in (4), the bra as our basic element, and the ket as an element of $V^{**}$, which is the space of linear functionals on $V^*$, which we could call (5).

It is a theorem of Hilbert Space theory that these two interpretations give the same scalar result, and that there are isometric isomorphisms (IIs) everywhere. In particular (5) is II to (1) and (4) is II to (3), and these two are identified in the notation. That is, it is not specified when we write a ket whether we mean an item in (1) or (5), and it is not specified for an operator whether we mean an item in (3) or (4) because whichever one we choose (as long as we do so consistently between the pairs, so that domains match ranges when composing functions) we get the same result.
Yes, there may be a piece still missing. That last line tells us that $Im\langle A|H|B\rangle=-Im\langle B|H|A\rangle$. To complete the proof we need to prove that the Real parts are equal. I'll have to come back to that later.

Last edited: Sep 3, 2016