# Propagator in the derivation of path integrals

1. Jan 1, 2016

### 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 $\Psi(x, t_1) = e^{-\frac{i}{\hbar}H(t_1-t_0)}\,\Psi(x, t_0)\\ = e^{-\frac{i}{\hbar}H(t_1-t_0)}\,\int\delta(x-x')\Psi(x', t_0)\,dx'$

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 $\Psi(x, t_0)$:

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

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?

Last edited: Jan 1, 2016
2. Jan 1, 2016

### blue_leaf77

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.

3. Jan 1, 2016

### Happiness

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 $x_{n-1}$.

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 $\langle x|$?

Last edited: Jan 1, 2016
4. Jan 1, 2016

### blue_leaf77

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 $ik(x_n-x_{n-1})$.