# Hermitian operators

We know that operators can be represented by matrices.
Every operator in finite-dimensional space can be represented by a matrix in a given basis in this space.
If the transpose conjugate of the matrix representation of an operator in a given basis is the same of the original matrix representation of this operator in this basis, then this operator is hermitian.

But what confuses me is that if this applies to a given basis what guarantees that it will apply to any basis in the space?

Even in infinite dimensional spaces,for example the derivative operator D in the position basis is : <x|D|x'>=δ'(x-x'), where δ'(x-x') is the derivative of the dirac delta function, this will yield that the operator D is not hermitian. But what if the operator D is represented in another basis,will it also be non-hermitian in the other basis?
How can the operator be hermitian in general ?
I hope my question is clear.

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DrClaude
Mentor
The matrix that performs a basis transformation is unitary, and a unitary transform preserves hermicity.

The matrix that performs a basis transformation is unitary, and a unitary transform preserves hermicity.
Ok,I'm convinced in the case of finite-dimensional spaces.
But how can we prove that these operators are unitary in infinite-dimensional space?for example the derivative operator I mentioned above.Is it trivial?

blue_leaf77
Homework Helper
Well, I think the conservation of hermicity by change of basis must also apply in infinite vector spaces, otherwise, Hamiltonians in most cases in quantum mechanics (like H atom, infinite square well, and harmonic oscillator), which have infinite number of bases will not necessarily be Hermitian. Suppose you have a Hermitian operator ##H##, the matrix element of this operator in some (infinite) basis ##{|a\rangle}## is
$$\langle a | H |a' \rangle \hspace{3cm} (1)$$
Since, ##H## is Hermitian, the matrix element in that basis will satisfy ##\langle a | H |a' \rangle = \langle a' | H |a \rangle ^\dagger##. You want to prove if this relation still holds in another basis, say ##{|b\rangle}##. By utilizing completeness relation in ##{|b\rangle}## basis, one can rewrite equation (1) as
$$\langle a | H |a' \rangle = \sum_{b'}\sum_b \langle a|b'\rangle \langle b' | H |b \rangle \langle b|a'\rangle \hspace{3cm} (2)$$
Next consider ##\langle a' | H |a \rangle ## (obtained by transposing ##\langle a | H |a' \rangle##). Using the same method as above with the completeness relation, this matrix element is written as ##\langle a' | H |a \rangle = \sum_{b'}\sum_b \langle a'|b'\rangle \langle b' | H |b \rangle \langle b|a\rangle ##. The complex conjugate of this is
\begin{aligned} \langle a' | H |a \rangle^\dagger &= \sum_{b'}\sum_b \langle b'|a'\rangle \langle b' | H |b \rangle^\dagger \langle a|b\rangle \\ &= \sum_{b'}\sum_b \langle a|b\rangle \langle b' | H |b \rangle^\dagger \langle b'|a'\rangle \\ &= \sum_{b'}\sum_b \langle a|b'\rangle \langle b | H |b' \rangle^\dagger \langle b|a'\rangle \hspace{3cm} (3)\\ \end{aligned}
where the last expression is obtained using the fact that ##b## and ##b'## are dummy indices and therefore can be freely interchanged. Due to the hermicity of ##H## in ##{|a\rangle}## basis, the LHS of (2) must be equal to the LHS of (3), so are the RHS. Upon comparing one then get ##\langle b | H |b' \rangle^\dagger = \langle b' | H |b \rangle##.

By the way, I have the impression that you mix the term "infinite bases" and "non-denumerable basis". Eigenstates of harmonic oscillater are denumerable and infinite, on the other hand the position bases, which you used as example, are non-denumerable and infinite. As for the definition of ##\langle x |D| x' \rangle##, see the first linked pdf document in this recent post https://www.physicsforums.com/threads/how-is-this-equation-for-x-p-x-derived.853217/#post-5352266.

PeroK
Homework Helper
Gold Member
Ok,I'm convinced in the case of finite-dimensional spaces.
But how can we prove that these operators are unitary in infinite-dimensional space?for example the derivative operator I mentioned above.Is it trivial?
In general, the definition of Hermitian does not depend on a particular basis. Every operator has a Hermitian-conjugate and if the two are equal then the operator is Hermitian.

An informal definition of the Hermitian conjugate of an operator ##T## is the operator ##T^\dagger##, such that for all vectors (or functions) ##u, v##:

##<T^\dagger u, v> = <u, Tv>##

And ##T## is Hermitian if ##T^\dagger = T##

Note that the inner product of two vectors or functions does not depend on the basis.

So, the question of whether an operator could be Hermitian in one basis and not in another simply does not arise.

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PeroK
Homework Helper
Gold Member
PS
##D^\dagger = -D## hence ##iD## is Hermitian. (In any basis!)

@blue_leaf77 but shouldn't you use ∫|x><x'|dx=I instead of ∑|x><x'| =I ?since we are dealing with an infinite basis.

@PeroK
<T†u,v>=<u,Tv><T†u,v>=<u,Tv> =

And TTT is Hermitian if T†=T
but while dealing with infinite-dimensional basis, this relation, as you said ,should be satisfied : <g|k|f>=<f|k|g>*, let here K=-id/dx.
which means that $\int_a^b\int_a^b\langle g | x\rangle\langle x | k| x'\rangle \langle x' |f\rangle\, dx\, dx' =(\int_a^b\int_a^b\langle f| x\rangle\langle x | k |x'\rangle\langle x' | g\rangle\, dx\, dx')*$
so by using integration by parts that yields that K won't be hermitian unless $\left. -ig(x)f(x)\right|_a^b =0$
so K is hermitian only if the surface term vanishes. As the inner product is independent of the chosen basis we can say that the surface term will vanish even for any basis other than position basis if K is hermitian.

PeroK
Homework Helper
Gold Member
@blue_leaf77 but shouldn't you use ∫|x><x'|dx=I instead of ∑|x><x'| =I ?since we are dealing with an infinite basis.

@PeroK

but while dealing with infinite-dimensional basis, this relation, as you said ,should be satisfied : <g|k|f>=<f|k|g>*, let here K=-id/dx.
which means that $\int_a^b\int_a^b\langle g | x\rangle\langle x | k| x'\rangle \langle x' |f\rangle\, dx\, dx' =(\int_a^b\int_a^b\langle f| x\rangle\langle x | k |x'\rangle\langle x' | g\rangle\, dx\, dx')*$
so by using integration by parts that yields that K won't be hermitian unless $\left. -ig(x)f(x)\right|_a^b =0$
so K is hermitian only if the surface term vanishes. As the inner product is independent of the chosen basis we can say that the surface term will vanish even for any basis other than position basis if K is hermitian.
##D = \frac{d}{dx}## is an unbounded operator on the space on square integrable functions. Formally, therefore, you need to restrict your function space to functions whose derivative is square integrable. Otherwise, ##Df## may not be square integrable at all and ##<Df, g>## may be undefined.

Also, the eigenfunctions of ##D## are not square integrable. Technically, therefore, the momentum operator ##-i\hbar D## is a Hermitian operator with no eigenfunctions (in the function space). It is possible to get round this in QM by considering the basis of (exponential) eigenfunctions of ##-i\hbar D## even though they are not part of the Hilbert space of square integrable functions. This leads to some mathematics where care is required! I don't pretend to understand all the implications of this, by the way. It goes under the term "rigged Hilbert Spaces". But, I haven't looked closely at that.

In short, things can get wild and woolly when using the eigenfunctions of the momentum and position operators, and you have to be careful in what you're doing.

PS In your example, the problem is not whether ##K## is Hermitian or not. The problem is whether ##K## is a well-defined (bounded) operator on your function space.

This is somehow beyond my level , I'm still a beginner in quantum mechanics.
But always when I struggle with concepts or postulates of QM in infinite-dimensional spaces "rigged hilbert space" comes around.It seems interesting,but of course still far from my level of grasp.I need more practice in this stuff.
Thanks for all anyway :)

blue_leaf77
Homework Helper
@blue_leaf77 but shouldn't you use ∫|x><x'|dx=I instead of ∑|x><x'| =I ?since we are dealing with an infinite basis.
The bases I used before are of infinite dimension, it's just that they "numerable". The position basis is both infinite and non-denumerable.

The bases I used before are of infinite dimension, it's just that they "numerable". The position basis is both infinite and non-denumerable.
what do you mean by" numerable" and "non-denumerable"? but if you can try to simplify it to me.

blue_leaf77
Homework Helper
what do you mean by" numerable" and "non-denumerable"? but if you can try to simplify it to me.
Numerable = discrete eigenvalues, a few examples would be the harmonic oscillator and infinite square well.
Non-denumerable = continuous eigenvalues, examples are position and momentum eigenbases as well as the ionized part of hydrogen eigenbasis.

PeroK
Homework Helper
Gold Member
but while dealing with infinite-dimensional basis, this relation, as you said ,should be satisfied : <g|k|f>=<f|k|g>*, let here K=-id/dx.
which means that $\int_a^b\int_a^b\langle g | x\rangle\langle x | k| x'\rangle \langle x' |f\rangle\, dx\, dx' =(\int_a^b\int_a^b\langle f| x\rangle\langle x | k |x'\rangle\langle x' | g\rangle\, dx\, dx')*$
so by using integration by parts that yields that K won't be hermitian unless $\left. -ig(x)f(x)\right|_a^b =0$
so K is hermitian only if the surface term vanishes. As the inner product is independent of the chosen basis we can say that the surface term will vanish even for any basis other than position basis if K is hermitian.
I looked again at this. If you have functions on a finite interval, then you have the restriction ##f(a) = f(b) = 0##. For overall continuity of your wavefunction, which must be 0 outside this interval. With this restriction, ##K## is Hermitian.

Note that if you have a larger Hilbert Space including functions that are not 0 at the end-points, then ##K## would not be Hermitian on that Hilbert Space. This is not to do with a change of basis, but a change of Hilbert Space altogether.

The "momentum" operator, therefore, is not Hermitian on all Hilbert spaces. But, it is Hermitian on Hillbert Spaces of valid wavefunctions.

Numerable = discrete eigenvalues, a few examples would be the harmonic oscillator and infinite square well.
Non-denumerable = continuous eigenvalues, examples are position and momentum eigenbases as well as the ionized part of hydrogen eigenbasis.
The "position and momentum eigenvectors" are not square integrable functions and thus are not vectors in Hilbert space. The Hilbert space in the latter examples have a numerable basis.

blue_leaf77
Homework Helper
The "position and momentum eigenvectors" are not square integrable functions and thus are not vectors in Hilbert space.
We know that already. But they are still eigenbases of their respective operators, otherwise the relations ##\hat{p}|p'\rangle = p'|p'\rangle## and ##\hat{x}|x'\rangle = x'|x'\rangle## will not be defined.
The Hilbert space in the latter examples have a numerable basis.
Which example, the ionized part of H atom? The bound state (negative energy) is indeed numerable, but the positive energy eigenstate is not.

We know that already. But they are still eigenbases of their respective operators, otherwise the relations ##\hat{p}|p'\rangle = p'|p'\rangle## and ##\hat{x}|x'\rangle = x'|x'\rangle## will not be defined.

Which example, the ionized part of H atom? The bound state (negative energy) is indeed numerable, but the positive energy eigenstate is not.
I was using the mathematical definition of basis, consisting of (square integrable) vectors in Hilbert space.

Mathematical surprises and Dirac’s formalism in quantum mechanics (PDF) discusses these issues. I have not read it all, but the introduction and conclusions are worth a look. See pages 1, 19(and footnote) and 36. (Or 3,21,38 as the PDF is numbered).

PeroK
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