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I know that energy eigenstates are define by the equation $$ \hat{H} \psi_n(x) = E_n \psi_n(x),$$ where all the eigenstates form an orthonormal basis.

And I also know that $\hat{H}$ is hermitian, so $\hat{H} = \hat{H}^\dagger$. However, I have no intuition as to what this means. What is an easy way to understand the significance of $\hat{H}$ being hermitian. Also, is there any intuitive reasoning for why the eigenstates would form an orthonormal basis?

I am having trouble with these concepts, so any help would be appreciated.

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    $\begingroup$ looking up the spectral theorem might help $\endgroup$ Nov 18 '19 at 22:34
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    $\begingroup$ Your teacher must have evaluated $\langle \psi_m |H|\psi_n\rangle$ for you. Hermiticity dictates that is is equal to $E_n\langle \psi_m |\psi_n\rangle$ but also $E_m\langle \psi_m |\psi_n\rangle$. What do you conclude for $E_n\neq E_m$? $\endgroup$ Nov 18 '19 at 23:17
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Observables in Quantum Mechanics are chosen to be Hermitian operators because their eigenvalues, by postulate, are the outcomes of physical measurements, which always produce outcomes in terms of real numbers.

Also, an Hermitian operator has a number of eigenvalues (counted with their multiplicity) which is equal to the dimension of the vector space it acts on. It can be proved that its eigenvectors corresponding to different eigenvalues are always orthogonal, while the one corresponding to degenerate eigenvalues can be chosen orthogonal.

In both cases, the eigenvector can be normalized to 1, hence the orthonormality.

Finally, as the number of eigenvectors is the same as the dimension of the vector space, they form a basis.

What I said applies flawlessly for finite dimensional Hilbert spaces, while it needs some more care for infinite-dimensional Hilbert spaces, but for the purpose of the present discussion it might be enough.

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Are you familiar with Bra-ket notation? The inner product we associate in our (Hilbert) space of wave functions means, given two wave functions $\psi_m$ and $\psi_n$, their inner product is $\lt\psi_m|\psi_n\gt=\int\psi_m^*\psi_ndx.$

By definition $\lt\psi_m| =(|\psi_m\gt)^\dagger$. The $\dagger$ means convert the ket to a bra by taking the transpose, then take the complex conjugate.

Suppose we have $\hat{O} = \hat{O}^\dagger$. And let us suppose $\psi_n$ is an eigenstate of $\psi_n$, then $\hat{O}\psi_n=\lambda_n\psi_n$.

$\int(\hat{O}\psi_n)^*\psi_ndx=\int(\lambda_n\psi_n)^*\psi_ndx=\int\psi_n^*\lambda_n^*\psi_ndx$.

But by definition of Hermitains, we have that :

$\int(\hat{O}\psi_n)^*\psi_ndx=\int\psi_n^*\hat{O}\psi_ndx=\int \psi_n^*\lambda_n\psi_ndx.$

This in turn implies $\lambda_n=\lambda_n^*$ That is, the eigenvalues of $\hat{O}$ are real.

The above integrals can be expressed as:

$\lt\psi_n|\hat{O}\psi_n\gt=\lt\hat{O}\psi_n|\psi_n\gt$, by definition of Hermitian operator.

This means by definition of eigenfunction/eigenvalue:

$$\lt\psi_n|\lambda_n\psi_n\gt=\lt\lambda_n\psi_n|\psi_n\gt$$ $$\lt\psi_n|\lambda_n\psi_n\gt=\lt\psi_n|\lambda_n^*\psi_n\gt$$ $$0=(\lambda_n-\lambda_n^*)<\psi_n|\psi_n>$$

Since the eigenvalues are real, this follows for all wave functions.

Now, suppose:

$$\hat{H}\psi_m=E_m\psi_m$$ $$\hat{H}\psi_n=E_n\psi_n$$

Then:

$$\psi_n^*\hat{H}\psi_m=\psi_n^*E_m\psi_m$$ $$\psi_m^*\hat{H}\psi_n=\psi_m^*E_n\psi_n$$

The terms on the left are complex conjugates of each other, by Hermiticity, our energy eigenvalues are real, and our inner product is commutative. So if we subtract the equations from each other we have:

$$0=(E_m-E_n)\lt\psi_m|\psi_n\gt$$.

If our Eigenvalues are different, then the inner product of the eigenstates is zero, and therefor orthogonal.

Now, we don't have a state in the first place unless it is square integrable. If it is square integrable, it is normalizable, so we can always divide by some constant to guarantee its norm is 1. So orthogonal, normalizable states gives us an orthonormal basis.

Now the question is, is the basis complete? I'm not sure how to prove this in general, but there are theorems for the particular families of solutions depending on the potential. For example, Fourier Analysis guarantees completeness in the case of the infinite potential well. Analysis of the Hermite Polynomials addresses completeness of solutions for the Quantum Harmonic Oscillator. Legendre' polynomials answers this question for Spherical Harmonics. Laguerre Polynomials address the matter for a hydrogen atom.

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The linear operator representing an observable should have eigenvalues that are real. This isn't because the results of a measurement must logically be real -- e.g., we can measure complex impedances. But in any real-world application of the complex number system, we must always choose some arbitrary phase conventions, such as that an inductor has a positive imaginary impedance to represent the fact that the voltage leads the current by 90 degrees. (Such phase conventions are always arbitrary because we define $i$ as $\sqrt{-1}$, but this doesn't distinguish $i$ from $-i$.) These phase conventions are all independent of one another, and the classical ones are independent of the convention used for wavefunctions in quantum mechanics, which is that a state with positive energy twirls clockwise in the complex plane. Hermitian operators have real eigenvalues, and this is the reason we like Hermitian operators. (Actually Hermiticity is a little too strict.)

Also, is there any intuitive reasoning for why the eigenstates would form an orthonormal basis?

This is related to the completeness axiom of quantum mechanics, which states that for any system of interest, there exists a set of compatible observables, called a complete set, such that any state of the system can be expressed as a sum of eigenstates of these observables. In general, energy alone isn't enough; we would need to add other observables. It wouldn't make sense if we couldn't find an orthonormal basis for our complete set of observables. That would mean either that the states weren't normalizable (which we need in order to make sense of probabilities) or that there were states that were not distinguishable from one another (not orthogonal) and yet could be distinguished based on the observables (which wouldn't make sense).

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The first important property of a Hermitian operator is that it only has real eigenvalues. Try show it yourself starting from the eigenvalue equation: $$ \hat H \psi_n(x) = E_n\psi_n(x)$$ taking adjoint of the equation, and show that $E^* = E$, where * means complex conjugation.

Physically, it is significant that eigenvalues of the Hamiltonian operator are real, as energy is a physical observable, and one can only measure a real value, not a complex value.

Personally, I don't have an intuitive reason or picture for why eigenstates of Hermitian operator form a complete orthogonal basis, maybe other users can illuminate me on this.

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  • $\begingroup$ Impedance is a physical observable, but it's not real-valued. $\endgroup$
    – user4552
    Nov 19 '19 at 20:42
  • $\begingroup$ @BenCrowell I would argue that you do not directly measure a complex-valued impedance. Experimentally, one would measure the amplitude and phase (both are real quantities) of the impedance. Also, It is always possible to calculate circuit problems in terms of real-valued sinusoidal functions instead of complex exponential. it is just out of mathematical convenience that one "package" the amplitude and phase of impedance into a complex exponential. $\endgroup$
    – Leo L.
    Nov 20 '19 at 2:11

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