# Orthogonality of eigenstates and hermitian statement

1. Dec 14, 2014

### rwooduk

if I derive a hermitian relation

use:

[1] $$\left \langle \Psi _{m} | H |\Psi _{n}\right \rangle =E_{n}\left \langle \Psi _{m} |\Psi _{n}\right \rangle$$

and

[2] $$\left \langle \Psi _{n} | H |\Psi _{m}\right \rangle =E_{m}\left \langle \Psi _{n} |\Psi _{m}\right \rangle$$

if i take the complex conjugate of [2]

[3] $$\left \langle \Psi _{m} | H^{*} |\Psi _{n}\right \rangle =E_{m}\left \langle \Psi _{m} |\Psi _{n}\right\rangle$$

then [3] - [1] i get

$$H_{mn}^{*} - H_{mn} = 0$$

therefore

$$H_{mn}^{*} = H_{mn}$$

BUT in my notes its given as

$$H_{nm}^{*} = H_{mn}$$

so does

$$H_{nm}^{*} = H_{mn}^{*}$$ ? and when i took the complex conjugate was that result correct, i'm getting a little confused with notation.

thanks in advance for any guidance.

2. Dec 14, 2014

### vanhees71

Defining
$$H_{mn}=\langle \Psi_m| \hat{H} \Psi_n \rangle$$
you get
$$H_{mn}^*=\langle n |\hat{H}^{\dagger} \Psi_m \rangle=\langle n|\hat{H} \Psi_m \rangle=H_{nm}.$$
Loosely speaking, an operator in Hilbert space is self-adjoint if and only if it's matrix elements make up an hermitean (usually infinite-dimensional) matrix.

If you work with the energy-eigenbasis you also get
$$H_{mn}=E_n \langle \Psi_m | \Psi_n \rangle=E_n \delta_{mn}$$
since the eigenvectors of a self-adjoint operator are perpendicular to each other (in the case of a degenerate energy spectrum you can always choose the basis to be orthonormal). From the latter equation, and the "hermitecity" of the $H_{mn}$'s you get
$$H_{mn}^*=E_n^* \delta_{mn}=H_{nm}=E_m \delta_{mn}=E_n \delta_{mn} \; \Rightarrow \; E_n \in \mathbb{R},$$
i.e., the eigenvalues of a self-adjoint operator are always real.

3. Dec 14, 2014

### Fredrik

Staff Emeritus
If you take [3]-[1], the right-hand side isn't zero, it's $(E_m-E_n)\langle\Psi_m|\Psi_n\rangle$. (This is assuming that you have already proved that the eigenvalues are real. If not, then the $E_m$ in [3] should be $E_m^*$).

4. Dec 14, 2014

### rwooduk

thanks for this! will study it through.

good point! will make the adjustment

thanks for all the replies!!

5. Dec 14, 2014

### kith

I wouldn't write $H^{*}$ here because $H$ is an abstract operator while complex conjugation refers to numbers. Complex conjugation does make sense if you chose a basis, because then your operator $H$ gets represented by a matrix which is a collection of numbers $H_{ij}$. To avoid confusion, one can use the hat notation $\hat{H}$ for the abstract operators because unfortunately, the usual linear algebra notation suggests that big letters like $H$ denote matrices.

What you already know is how to complex conjugate inner products (namely by flipping the vectors). Since the eigenstates of self-adjoint operators form a basis, you can write $\hat{H} = \sum_i E_i |\psi_i\rangle \langle \psi_i|$ which leaves you with an expression which you probably know how to handle. Or you can do what vanhees did and look at resulting state vectors like $\hat{H} |\psi_n \rangle = |\hat{H} \psi_n \rangle$. This way, you may need to reflect a bit about hermitian conjugation. Unlike complex conjugation, hermitian conjugation is defined using the abstract state vectors and operators, so $\hat{H}^{\dagger}$ is a well-defined object.

Having said that, I'm not sure if your notation actually runs into difficulties somewhere or if it can be used as a sloppy way of writing things. In either case, I think it is good to keep the distinction between the general abstract objects and the column vectors and matrices you use to actually calculate things in the back of your head.

6. Dec 14, 2014

### rwooduk

Thanks will look work through this post also! Just one thing that relates to your post, we were given a question in our notes (that I havent come onto revising yet) that reads:

So in this instance how would you represent the Hamiltonian?

7. Dec 14, 2014

### vanhees71

The usual derivation for a self-adjoint operator (btw Hermitean is not sufficient for the usual manipulations made by physicists, but that's a more subtle mathematicle issue) goes as follows
$$\hat{H} |\Psi_n \rangle=E_n |\Psi_n \rangle.$$
This implies
$$\langle \Psi_m|\hat{H} \Psi_n \rangle=E_n \langle \Psi_m|\Psi_n \rangle.$$
On the other hand, because of self-adjointness you have
$$langle \Psi_m|\hat{H} \Psi_n \rangle=\langle \hat{H} \Psi_m|\Psi_n \rangle=E_n^* \langle \Psi_m|\Psi_n \rangle.$$
Subtracting both equations from each other yields
$$(E_n-E_m^*) \langle \Psi_m|\Psi_n \rangle=0.$$
Let's assume that the Hamiltonian is non-degenerate, i.e., that each eigenvalue has a one-dimensional eigenspace. Then for $n =m$ it follows
$$E_n-E_n^*=0 \; \Rightarrow \; E_n \in \mathbb{R},$$
and for $n \neq m$ you have $E_n \neq E_m^*$, and thus
$$\langle \Psi_m|\Psi_n \rangle=0 \quad \text{for} \quad m \neq n.$$
Normalizing the eigenvectors to 1 you have an orthonormal system of eigenvectors, which is complete for a self-adjoint operator.

Most Hamiltonians have not only a discrete spectrum ("bound states") but also a continuous part ("scattering states") or even only a continous spectrum (e.g., for the free particle). Then the issue becomes a bit more involved, and you have to deal with "generalized eigenstates" (distributions). They are normalizable to a $\delta$ distribution but are not square-integrable functions:
$$\langle E|E' \rangle=\delta(E-E').$$
Now you should think about what happens here, if the Hamiltonian is not self-adjoint (which sometimes occurs in applications, when you consider certain approximations, where socalled "optical potentials" occur). Since this is a homework, you should send questions corresponding to this to the homework forum!

8. Dec 14, 2014

### kith

Is there more context? Being complex isn't a property of the Hamiltonian. You can always chose a basis of eigenstates which makes the resulting matrix diagonal and therefore real. So the question seems a bit strange.

As I understand the second sentence, they ask whether the specific result holds for the general hermitian case. So to me, this doesn't seem to be about non-hermitian operators.

Last edited: Dec 14, 2014
9. Dec 15, 2014

### rwooduk

Thanks that's very detailed and helpful.

The original question wasnt a homework question, it was a derivation I was trying to figure out, but when the complex conjugate of the Hermitian was mentioned it reminded me of the above question (which is not homework, I'm currently revising for exams in January). The original question I asked doesnt really relate to it. But again thanks for the help!

10. Dec 16, 2014

### dextercioby

One cannot seriously expect to understand the concept of hermiticity (of matrices/operators) while using the bra-ket notation.

Last edited: Dec 16, 2014
11. Dec 16, 2014

### vanhees71

Why not? It's just a clever notation for scalar products of vectors in a Hilbert space. What one should avoid are notations like $\langle \psi |\hat{A}|\phi \rangle$, where it is not always clear, how one thinks the operator to act. Usually it acts to the right. This can be made clear by the notation $\langle \psi | \hat{A} \phi \rangle$, which I thus always use. Of course, some care is always good. There can be false conclusions from an abuse of notation. A nice paper about this is

Gieres, F.: Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys. 63, 1893, 2000
http://arxiv.org/abs/quant-ph/9907069

12. Dec 16, 2014

### Fredrik

Staff Emeritus
The first of these notations is what I'd call bra-ket notation. In this notation, the state vectors are written as $|\psi\rangle$ and $|\phi\rangle$. Your notation is just the convention to use a vertical line instead of a comma in the inner product. In the former notation, the state vectors involved are denoted by $|\psi\rangle$ and $|\phi\rangle$. In the latter notation, they're denoted by $\psi$ and $\phi$.

Bra-ket notation makes it pretty hard to discuss things like the definition of the adjoint operation. I prefer the $\langle x,y\rangle$ notation myself, so I'll use it here. Let $\mathcal H$ be a Hilbert space, and let $\mathcal H^*$ be its dual space, i.e. the set of linear maps from $\mathcal H$ into $\mathbb C$, turned into a vector space by the standard definitions of addition and scalar multiplication. The Riesz representation theorem says that for each $\phi\in\mathcal H^*$, there's a unique $x\in\mathcal H$ such that $\phi=\langle x,\cdot\rangle$. (In other words, ...such that $\phi(y)=\langle x,y\rangle$ for all $y\in\mathcal H$). If $A$ is a bounded linear operator on $\mathcal H$, then for each $x\in\mathcal H$, the map $y\mapsto\langle x,Ay\rangle$ is bounded and linear, and therefore an element of $\mathcal H^*$. So by the theorem, there's a unique $z\in\mathcal H$ such that $\langle x,Ay\rangle=\langle z,y\rangle$. The map $x\mapsto z$ with domain $\mathcal H$ can easily be shown to be a linear operator. It's denoted by $A^*$ or $A^\dagger$, and is called the adjoint of $A$.

It would be hard to even say something like this in bra-ket notation, precisely for the reason you mentioned. $\langle\psi|A|\phi\rangle$ can be interpreted as the inner product of $|\psi\rangle$ and $A|\phi\rangle$, or as the inner product of $A^*|\psi\rangle$ and $|\phi\rangle$. (The latter interpretation can be described as "$A$ is acting to the left"). The equality in bra-ket notation that corresponds to $\langle A^*x,y\rangle =\langle x,Ay\rangle$ is $\big(\langle\psi| A\big)|\phi\rangle =\langle\psi|\big(A|\phi\rangle\big)$.

In presentations such as Sakurai's, it's assumed right at the start that every kind of "multiplication" that we might want to do is associative. The existence of the adjoint operation is very well hidden in this associativity assumption.

Last edited: Dec 16, 2014