Propagator in the derivation of path integrals

[Happiness]

Why isn't (5.298) the following instead?

##K(x, t_1; x', t_0) = \delta(x-x')\,e^{-\frac{i}{\hbar}H(t_1-t_0)}##

My reasoning:

Since [itex]\Psi(x, t_1) = e^{-\frac{i}{\hbar}H(t_1-t_0)}\,\Psi(x, t_0)\\
= e^{-\frac{i}{\hbar}H(t_1-t_0)}\,\int\delta(x-x')\Psi(x', t_0)\,dx'[/itex]

The exponential operator with ##H## is acting on the variable ##x##. So when it is pulled inside the integral, it should just be acting on ##\Psi(x', t_0)## so that after applying ##\delta(x-x')## and integrating, we get the same answer as the one when the exponential operator is acting on ##\Psi(x, t_0)##:

[itex]\Psi(x, t_1) =\int\delta(x-x')e^{-\frac{i}{\hbar}H(t_1-t_0)}\Psi(x', t_0)\,dx'[/itex]

Edit: I believe there is a typo at (5.304). It should be ##(x_n - x_{n-1})## instead of ##(x_n + x_{n-1})##. Correct?

 

[blue_leaf77]

The exponential operator with ##H## is acting on the variable ##x##. So when it is pulled inside the integral, it should just be acting on ##\Psi(x', t_0)## so that after applying ##\delta(x-x')## and integrating, we get the same answer as the one when the exponential operator is acting on ##\Psi(x, t_0)##:

It should be acting on any function which contains ##x## which is in this case the delta function. You can also obtain the propagator by starting to work in the braket notation. Starting from
$$|\psi,t_1\rangle = \exp(i\hat{H}(t_1-t_0))|\psi,t_0\rangle$$.
Insert ##\int dx' |x'\rangle \langle x'|## just in front of ##|\psi,t_0\rangle## and multiply from the left with ##\langle x|##. So that you get
$$\langle x|\psi,t_1\rangle =\int dx' \langle x|\exp(i\hat{H}(t_1-t_0))|x'\rangle \langle x'|\psi,t_0\rangle$$.
Then use the property ##\langle x|\hat{H}|\phi\rangle = H(x)\langle x|\phi\rangle## for an arbitrary state ##|\phi\rangle## to arrive at the same expression for the propagator.

 

[Happiness]

It should be acting on any function which contains ##x## which is in this case the delta function.

If the exponential operator turns ##f(x)## into ##g(x)## and if we allow it to act on ##x'## and hence ##f(x')##, it would turn ##f(x')## into ##g(x')##. And after applying ##\delta(x-x')## and integrating, we get back ##g(x)##.

You seem to suggest that the exponential operator can only act on ##x## and the variable that it acts on cannot be changed. I thought so too, until I saw (5.304), where it is not acting on ##x##, but on ##x_n## or ##x_{n-1}##. Also, ##V(x)## becomes ##V(x_{n-1})## in (5.304). So I guess that when ##V(x)## is pulled into the integral with respect to ##x_{n-1}##, ##V(x)## becomes ##V(x_{n-1})## and so the operator ##\hat{H}## acts on ##x_{n-1}## when it is inside the integral with respect to ##x_{n-1}##.

You can also obtain the propagator by starting to work in the braket notation. Starting from
$$|\psi,t_1\rangle = \exp(i\hat{H}(t_1-t_0))|\psi,t_0\rangle$$.
Insert ##\int dx' |x'\rangle \langle x'|## just in front of ##|\psi,t_0\rangle## and multiply from the left with ##\langle x|##. So that you get
$$\langle x|\psi,t_1\rangle =\int dx' \langle x|\exp(i\hat{H}(t_1-t_0))|x'\rangle \langle x'|\psi,t_0\rangle$$.
Then use the property ##\langle x|\hat{H}|\phi\rangle = H(x)\langle x|\phi\rangle## for an arbitrary state ##|\phi\rangle## to arrive at the same expression for the propagator.

Do we deduce the propagator ##\ K=\int dx' \exp\big(iH(x)(t_1-t_0)\big)|\ x'\rangle \langle x'|\ ## from the fact that
##\langle x|\psi,t_1\rangle =\int dx' \exp\big(iH(x)(t_1-t_0)\big)\ \langle x|x'\rangle \langle x'|\psi,t_0\rangle\ ## is true for any ##\langle x|##?

 

[blue_leaf77]

You seem to suggest that the exponential operator can only act on ##x## and the variable that it acts on cannot be changed. I thought so too, until I saw (5.304), where it is not acting on ##x##, but on ##x_n## or ##x_{n-1}##. Also, ##V(x)## becomes ##V(x_{n-1})## in (5.304). So I guess that when ##V(x)## is pulled into the integral with respect to ##x_{n-1}##, ##V(x)## becomes ##V(x_{n-1})## and so the operator ##\hat{H}## acts on ##x_{n-1}## when it is inside the integral with respect to ##x_{n-1}##.

I would rather choose to work out everything from the braket notation to know which comes from which. So, the equation ##\exp(-iH\Delta t) \delta(x_n-x_{n-1})## can be derived from the braket notation
$$
\langle x_n| \exp(-i\hat{H}\Delta t) |x_{n-1} \rangle
$$
which is approximately equal to
$$
\langle x_n| \exp(-i\frac{\hat{p}^2}{2m}\Delta t) \exp(-iV(\hat{x})\Delta t) |x_{n-1} \rangle
$$
Note that up to this point everything is still in the operator form. Then again use the completeness relation, this time for ##|k\rangle## in between the two exponential operators above. Then
$$
\langle x_n| \exp(-i\frac{\hat{p}^2}{2m}\Delta t) \exp(-i\hat{V}\Delta t) |x_{n-1} \rangle = \int dk \langle x_n| \exp(-i\frac{\hat{p}^2}{2m}\Delta t) |k\rangle \langle k|\exp(-iV(\hat{x})\Delta t) |x_{n-1} \rangle
$$
Now use ##\hat{p}^2 |k\rangle = k^2 |k\rangle## and ##V(\hat{x}) |x_{n-1} \rangle = V(x_{n-1}) |x_{n-1} \rangle## where ##k^2## and ##V(x_{n-1}) ## are just numbers (scalars) not operators anymore. The above equation will reduce to
$$
\int dk \exp(-i\frac{k^2}{2m}\Delta t) \langle x_n|k\rangle \langle k|x_{n-1} \rangle \exp(-iV(x_{n-1})\Delta t) = \frac{1}{2\pi} \int dk \exp(-i\frac{k^2}{2m}\Delta t + ik(x_n-x_{n-1})) \exp(-iV(x_{n-1})\Delta t)
$$
Note: I think there is another mistake in that book for writing ##ik(x_n+x_{n-1}) ## instead of ##ik(x_n-x_{n-1}) ##.