$\newcommand{\ket}[1]{|#1\rangle}$$\newcommand{\bra}[1]{\langle#1|}$In Principles of Quantum Mechanics (2nd edition) by Shankar, Exercise 5.1.3 asks to find the wave function of the free particle by means of applying the propagator to an wave function in $x$-space.

The propagator $U(t)$, which satisfies $\ket{\psi(t)} = U(t)\ket{\psi(0)}$ by definition, can be shown to have the form $U(t) = \exp(-iHt/\hbar)$ from Schrödinger's equation.

Now, for the free particle, the Hamiltonian is given by $H = P^2 / 2m$. However, Shankar then says the propagator for this problem is given by $$U(t) = \exp\left[\displaystyle-\frac{it}{\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}\right)\right].$$

But $H \ne -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$ !! Sure, this is its action on $\ket{\psi}$ in $x$-space, in the sense that $$\bra x H\ket{\psi(t)} = -\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}\psi(x,t)$$ but to be a pedant about the mathematics, what Shankar used for $H$ is surely not the Hamiltonian operator acting on Hilbert space.

It turns out that Shankar's "propagator" does in fact propagate the wave function in $x$-space in the sense that $\psi(x,t) = U(t)\psi(x,0)$. So it still smells like a propagator.

Now for my actual question: (that was just context)

What kind of mathematical object is Shankar's propagator (if it's not an operator on Hilbert space)? Is it an operator on a new vector space ($x$-space, perhaps)? Also, how does it relate to the actual propagator (the one that is an operator on Hilbert space)?

  • $\begingroup$ Footnote: I considered the possibility of it simply being the propagator evaluated in $x$-space in the sense where Shankar defines it as $U(x,t;x') = \langle x|U(t)|x'\rangle$, but this isn't the same thing, i.e. it acts on $\psi(x,0)$ under an integral, unlike what we have above. $\endgroup$
    – J-J
    Commented Mar 23, 2019 at 21:53
  • 2
    $\begingroup$ Why do you think the Hamiltonian is surely not $-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$? This is what is used in the Schrodinger equation. $\endgroup$
    – G. Smith
    Commented Mar 23, 2019 at 22:04
  • $\begingroup$ That is what's used in Schrodinger's equation in $x$-space, but that isn't the abstract operator $H$ which acts on $|\psi\rangle$ $\endgroup$
    – J-J
    Commented Mar 23, 2019 at 22:08
  • $\begingroup$ I think that is overly pedantic. Other than just writing $H$, do you have some way to represent the abstract Hamiltonian for a free particle that doesn’t involve $x$ or $p$? $\endgroup$
    – G. Smith
    Commented Mar 23, 2019 at 22:13
  • $\begingroup$ If you are going to ask the question you also must review both the propagator matrix elements and the inverse Weierstrass transform you are providing. $\endgroup$ Commented Mar 23, 2019 at 22:17

1 Answer 1


$\newcommand{\ket}[1]{|#1\rangle}$G. Smith's first comment got me thinking in the right direction, so I think I've figured out an answer to my question.

The propagators I'm used to are defined by $\ket{\psi(t)} = U(t)\ket{\psi(0)}$, and in order for $\ket{\psi(t)}$ to obey Schrödinger's equation, $i\hbar\frac{d}{dt}\ket{\psi(t)} = H\ket{\psi(t)}$, it follows that the propagator is given by $U(t) = \exp(-iHt/\hbar)$.

Analogously, I believe Shankar's propagator of the wave function in $x$-space must be defined by $\psi(x,t) = U(t) \psi(x,0)$. Furthermore, in order for the wave function to satisfy Schrödinger's equation in $x$-space, $i\hbar\frac{\partial}{\partial t}\psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)$, it follows that this propagator must be given by $U(t) = \exp(-\frac{it}{\hbar}(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}))$, which is what Shankar used.

These propagators must be different mathematical objects (and I'm still not sure how they're directly related except by analogy), but they do the same job and now I see why they both work when applied to the relevant vector/function, so I'm content.


To be clear, by "the wave function in $x$-space", I mean $\psi(x,t) = \langle x|\psi(t)\rangle$.

  • 1
    $\begingroup$ I don't think it is just an analogy. It is literally the position-space representation of the propagator just like $-i\hbar\frac{d}{dx}$ is the position-space representation of the momentum operator. Are they mathematically different objects? Sure. One acts on an abstract Hilbert space while the latter acts on a function space unambiguously obtained from translating the information of the Hilbert space into a specific basis. $\endgroup$
    – user87745
    Commented Mar 23, 2019 at 23:38
  • $\begingroup$ Thanks, that confirms much of what I realised. It seems obvious now but it took a long time for me to see it! $\endgroup$
    – J-J
    Commented Mar 23, 2019 at 23:45
  • $\begingroup$ Actually your notation is not entirely correct. It's ok as long as you know what you're doing. $\endgroup$
    – lcv
    Commented Mar 24, 2019 at 6:40
  • $\begingroup$ What's wrong? The ambiguity of $U(t)$? $\endgroup$
    – J-J
    Commented Mar 24, 2019 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.