I had a hard time developing an intuition for the Fourier Transform until I was introduced to Bra-ket notation in Quantum Mechanics (I come from physics). With this notation, many problems make significantly more sense now because I can fall back on the vector space intuition.
I am at a similar place with the Laplace transform as the Fourier before. I can do the transform, solve ODE's, etc., but I would like to have a similar intuition as was developed for Fourier analysis by the Bra-ket notation.
Looking at the integrals for the Laplace transform, I would naturally want to write the transform similar to the Fourier transform in the following notation,
$$ \begin{align} \mathbb{F}[f](k) &= \int dx\ \langle k | x \rangle \langle x | f \rangle \\ &= \frac{1}{\sqrt{2 \pi}} \int dx\ e^{-ikx} f(x) \end{align} $$
For the Laplace transform, I might want to write, $$ \begin{align} \mathbb{L}[f](a + i b) &= \int da\ \langle k | a + i b \rangle \langle a + i b | f \rangle \\ \end{align} $$
I don't know if this is valid for the Laplace transform. Of course I can write down whatever I want, but I am looking for a good way to think about the Laplace transform in the way I can think of the Fourier by the Dirac notation.
Could you please point me in the right direction? A paper/lecture notes/book would be really nice, or if you can easily write down here what the correct way to think about this is, I would appreciated it.
The bra-ket notation is really an adaptation of Fourier's original ideas from more than a century before Dirac. There are ways of viewing this notation in terms of cyclic/irreducible representations associated with selfadjoint operators. However, the Laplace transform is associated with a non selfadjoint operator, which is not appropriate for study using the theory of selfadjoint Quantum observables. Trying to shove this square peg into the round hold of selfadjoint Quantum observables leads to errors.
So what goes wrong? Consider $$ A=\frac{1}{i}\frac{d}{dt} $$ on a domain $\mathcal{D}(A)$ consisting of absolutely continuous functions $f$ on $[0,\infty)$ for which $f,Lf \in L^2[0,\infty)$. Define $$ e_{\lambda}(t) = e^{i\lambda t} $$ It is easily verified that $e_{\lambda} \in L^2[0,\infty)$ whenever $\Im\lambda > 0$, and $$ Ae_{\lambda} = \lambda e^{\lambda},\;\;\; \Im\lambda > 0. $$ Selfadjoint operators on separable spaces (Quantum spaces must be separable by axioms of constructibility) cannot have a countably infinite number of eigenvalues, because eigenfunctions of selfadjoint operators corresponding to different eigenvalues must be mutually orthogonal. That is not the case here: $$ (e_{\lambda},e_{\mu}) = \int_{0}^{\infty}e^{i(\lambda-\overline{\mu})t}dt = \frac{i}{\lambda-\overline{\mu}},\;\;\; \Im\lambda > 0,\,\Im\mu > 0. $$ Hence, formalism developed for selfadjoint observables (even if it were totally correct,) does not apply here. Treating this operator $A$ as if it were selfadjoint has led to wrong results on a half space. The operator $A$ is not a valid observable on the half plane, which means it cannot be a legitimate momentum operator on the half line. Details like this turn out to matter.
The operator $A$ cannot be studied using Spectral Theory for selfadjoint operators. However, It is possible to study $A$ using the theory of operators $A$ for which $(A-\lambda I)^{-1}$ exists for $\Im\lambda < 0$. Such an operator $A$ has the property that \begin{align} \Im (Af,f)& = \frac{1}{2i}\int_{0}^{\infty}\frac{1}{i}f'(t)\overline{f(t)}+\frac{1}{i}f(t)\overline{f'(t)}dt \\ & = -\frac{1}{2}\int_{0}^{\infty}\frac{d}{dt}|f(t)|^2dt \\ & = \frac{1}{2}|f(0)|^2 \ge 0. \end{align} The operators $A$ for which $\Im(Af,f) \ge 0$ are studied using $C^0$ semigroups instead. Usually, however, the study of such operators is carried out for $B=iA$ instead, which gives $\Re(Bf,f) \le 0$. An operator $B : \mathcal{D}(B)\subseteq \mathcal{H}\rightarrow\mathcal{H}$ on a complex Hilbert space $\mathcal{H}$ for which $\Re(Bf,f) \le 0$, and for which $(B-\lambda I)^{-1}$ exists for $\Re\lambda > 0$ are exactly the generators of contractive $C^0$ semigroups on $\mathcal{H}$. In other words, $B$ is as described iff there exists $T(t) : \mathcal{H}\rightarrow\mathcal{H}$ such that $$ T(0) = I,\;\;\; T(t)T(t')=T(t+t'),\;\;\; \|T(t)\| \le 1,\\ \frac{d}{dt}T(t)f = T(t)Bf,\;\;\; f \in \mathcal{D}(B). $$ The derivative equality exists in the vector sense at $t=0$ iff $f \in \mathcal{D}(B)$. So these operators $B$, while not selfadjoint, are exactly described as generators of stable time evolution systems. And, functions of $B$ may be formed using the Laplace transform $$ \mathscr{L}\{F\}|_{B} = \int_{0}^{\infty}F(t)T(t)dt. $$ In this way, $\frac{d}{dt}$ becomes the model operator for all such $B$ through the Laplace transform. This calculus is not nearly so nice as that derived for selfadjoint operators, but it shouldn't be, and cannot be, because the Quantum observable representation forces the operator to be selfadjoint.