# Mathematical understanding of Quantum Mechanics

Assuming that $\phi(r) = F (\psi(r))$ for some operator $F$ in Quantum Mechanics. Then,

in our lecture today, we said that $$\phi(r) = \langle r|F |\psi\rangle = \int_{\mathbb{R}} \langle r |F| r' \rangle \langle r' | \psi\rangle dr' = \int_{\mathbb{R}} \langle r |F| r' \rangle \psi(r') dr'.$$

I think this result is somehow interesting, cause it tells us that any operator can be represented as an integral operator and also it is similar to $$y_i=\sum_{k=1}^n A_{ik}x_k$$ from Linear Algebra.

Unfortunately, I am not so sure what $\langle r |F| r' \rangle$ is. From comparison, it seems that $\langle r |F| r' \rangle = \delta(r-r') F ,$ but I am not sure how to see this, if we just look at $\langle r |F| r' \rangle$ and not at the equations above(!).

So, $|r\rangle$ are delta distributions in position space, but in order to evaluate this expectation value ($\langle r |F| r' \rangle$) , I would need to know what the dual of a distribution is, which I don't know. Could anybody explain to me how I could evaluate this expectation value?

Thus, there is really one thing that I do not consider as an anser: Claim that $\langle r |F| r' \rangle = \delta(r-r') F ,$ holds from the definitions above.

• It is to be noted that it is not surprising that QM operators are a generalisation of the linear operators from linear algebra, since infinte-dimensional Hilbert spaces generalize the notion of a normed vector space from LA. – ACuriousMind Oct 16 '14 at 11:16
• Unluckily, it is not true that any operator in a Hilbert space can be written as an integral operator. A one to one correspondence can be established (at a rigorous mathematical level) only for Hilbert Schmidt operators (between HS operators and $L^2$ kernels in both variables). – yuggib Oct 16 '14 at 11:26
• What you said is true, but the representation via integral operators is not restricted to HS operators, but the integral kernel will not be $L^2$ in that case. – Mateus Sampaio Oct 16 '14 at 12:40
• @MateusSampaio It may be possible to write a non HS operator as an integral operator. But I am quite confident it is not possible always, i.e. there are operators that cannot be written in integral form (even with "singular" integral kernels). – yuggib Oct 16 '14 at 13:40
• @yuggib I think you are right, too. I edited my answer showing a counterexample. – Mateus Sampaio Oct 16 '14 at 13:54

In a certain sense, what you said that every operator might be represented in position representation as a integral operator may be true for many operators only if you allow distributions to be used, and even though that's not always the case.

You are kind of confusing things when you say about the dual of distributions. What is a distribution is the representation of $|r\rangle$, which is not a vector of the Hilbert space $\mathcal{H}$, that is $\langle x |r\rangle=\delta(r-x)$. The problem about using this notation is that is that this lacks a lot of rigor and sometimes if you do not pay attention you may get into trouble.

What may be hard in general is to find the position representation of an operator. For example, take the free propagator $U(t)=e^{-itH}$, with $H=-\Delta$ being the free hamiltonian. It's possible to show that $$U\psi(x)=(e^{-itH}\psi)(x)=\frac{1}{\sqrt {4\pi it}}\int_\Bbb{R} e^{i\frac{(x-y)^2}{4t}}\psi(y)dy,$$ which shows that $U(t)$ may be represented as an integral operator in position representation with kernel $K(x)=\dfrac{1}{\sqrt {4\pi it}}e^{i\frac{x^2}{4t}}$. One way to find the representation of $F$ is to change the basis where the operator is a multiplication operator. This is always possible if $F=f(T)$, where $f$ is a measurable function and $T$ is a self-adjoint operator. In that case, the spectral theorem states that $$F=\int_\Bbb{R} f(t)dP^T(t),$$ where $P^T$ is the resolution of the identity of $T$. In practice it's possible to write $F=U^{-1}fU$, where $U$ is the unitary map that maps the position representation to the basis where $T$ is a diagonal operator. For example, if $T=P$, the momentum operator, then $U=\mathcal{F}$ is the Fourier transform. So if you want to find the position representation of a function $F=f(P)$ you may proceed like: $$(F\psi)(x)=(\mathcal{F}^{-1}f(p)\mathcal{F}\psi)(x)=\frac{1}{2\pi}\int e^{ipx}\left(f(p)\int e^{-ipy}\psi(y)dy\right)dp=\int K(x,y)\psi(y)dy,$$ with $$K(x,y)=\frac{1}{2\pi}\int e^{-ip(x-y)}f(p)dp$$ The last step is formal, because for us to change the order of integration, the integrals must exists, and that's not always the case, and you might get distributions, like is the case if $f(p)\in \mathcal{S}'(\Bbb{R})$, that is $f$ is a tempered distribution. E.g, taking $F=P$, that is $f(p)=p$, we find $K(x,y)=i\delta'(x-y)$. But if you take for example $f(p)=e^p$, the kernel $K(x,y)$ is not defined even in the sense of tempered distributions, although the operator $F=\exp(P)$ is well defined for a dense domain.

Yes, operators in quantum mechanics can be understood basically as infinite matrices, $|r\rangle$ as basis vectors and $\psi(r)\equiv \langle\psi|r\rangle \sim \psi_r \sim "\psi_i"$ as components of the state vector numbered by a continuous index $r$. $\langle r|F|r'\rangle$ are indeed just matrix components of the operator $F$.

Generally $\langle r|F|r'\rangle \neq f \delta(r-r')$. For the case $\langle r|F|r'\rangle = f \delta(r-r')$ you would have $\phi(r) = f \psi(r)$, i.e. $F$ would be just $f$ times identity on the mentioned vector space.

When you have a hermitean operator which commutes with the position operator $R$, i.e. $[F,R]=0$ it has the same eigenvectors as $R$ and will thus will be diagonal in the $|r\rangle$ eigenvector basis. (Hermiticity ensures non-degeneracy of eigenspaces.) This is all a straightforward extension of things you should know from linear algebra. In that case you have $\langle r|F|r'\rangle = f(r) \delta(r-r')$, where $f(r_0)$ is the eigenvalue of a vector $|r_0\rangle$.

But you can have also different operators. Consider a translation operator $T_a$ which makes $T_a|r\rangle = |r + a\rangle$. Obviously $\langle r|T_a|r'\rangle = \delta(r-r'-a)$. There is no general recipe for finding $\langle r|F|r'\rangle$, because it all depends on what $F$ represents. From requiring $F$ to represent something physical such as momentum or position, you are usually able to derive how it should act on your basis through certain procedures. (I like how this is done in chapter 3 of Ballentine's Quantum Mechanics)

You can once again recall that there is no way to know matrix coefficients without at least knowing what the matrix does - and in most problems you are just given the coefficients in a certain basis as a definition of the matrix itself. In this case, you are usually given the coefficients or action of the matrix "by physics" - it is not an entirely mathematical problem.

Unfortunately, I am not so sure what ⟨r|F|r′⟩ is ... in order to evaluate this expectation value

it's not an expectation, it is a matrix element; think of it as the components of the operator $F$ on the position basis.

If the operator is 'diagonal' on the position basis then $\langle r|F|r' \rangle$ is zero except when $r = r'$.

Thus, for example, if $F = R$ is the position operator then

$$\langle r|R|r' \rangle = r'\delta(r - r')$$

and then

$$\phi(r) = r\psi(r)$$

Now, consider the case that $F$ operating on a position eigenstate returns a sum of two position eigenstates:

$$F|r'\rangle = \frac{1}{\sqrt{2}}\left(|r' + 1\rangle + |r' - 1\rangle\right)$$

Then

$$\langle r|F|r' \rangle = \frac{1}{\sqrt{2}}\left(\delta(r' - 1 - r) + \delta(r' + 1 - r)\right)$$

and then

$$\phi(r) = \frac{1}{\sqrt{2}}\left(\psi(r+1) + \psi(r-1)\right)$$

• ah, so my claim is only true if $F$ is diagonal on the position basis, right? – Solid State Physicist Oct 16 '14 at 12:26
• @TobiasHurth, it isn't clear to me what $F$ represents on the right hand side of your final equation. The left hand side is not an operator so what is $F$ on the right hand side? – Alfred Centauri Oct 16 '14 at 12:32
• The key idea here is that the result of $F|r'\rangle$ is another ket which may be expressed as a combination of position kets. We might e.g. have $F|r'\rangle = a|r''\rangle + b|r'''\rangle$, similar to what Alfred wrote but a little more general. Then the matrix elements of $F$ are zero except when $r''=r$ or $r'''=r$. (since they're given by $a\langle r | r''\rangle + b\langle r | r'''\rangle$) – Wouter Oct 16 '14 at 12:46