I Question about the Propagator (as defined in Ballentine)

[jbergman]

TL;DR

What is the propagator?

I am a little bit confused about the definition of the propagator.

We start with the evolution equation for our state vector.

$$ \ket{\Psi(t)} = U(t,t_0)\ket{\Psi(t_0)} $$

Now, I would expect

$$ \Psi(x, t) = \bra{x}U(t,t_0)\ket{\Psi(t_0)} = \int \delta(x'-x) U(t,t_0) \Psi(x',t_0) dx' $$

Instead, Ballentine defines it as

$$ \Psi(x, t) = \int G(x,t;x',t_0) \Psi(x',t_0) dx'$$
with the propagator,
$$G(x,t;x',t_0)=\bra{x}U(t,t_0)\ket{x'}$$
Which can be thought of as the amplitude of a state initially in state $\ket{x'}$ and at time $t$ in state $\bra{x}$.

What I don't see is how the integral above is equivalent to the first one.

 

[dextercioby]

What picture is involved here? If Schrödinger, then it is obvious, U „gets out of the sandwitch”.

 

[jbergman]

What picture is involved here?

Schrödinger.

 

[jbergman]

What picture is involved here? If Schrödinger, then it is obvious, U „gets out of the sandwitch”.

Can you explain bit more why we can do that?

 

[Haborix]

As written, I don't see how your expected expression makes sense mathematically. The first use of $U(t,t_0)$ is as an operator on vectors in a Hilbert space. Your use isn't consistent with that. The question of picture is not going to help you move the time-evolution operator outside of an inner product.

 

[jbergman]

As written, I don't see how your expected expression makes sense mathematically. The first use of $U(t,t_0)$ is as an operator on vectors in a Hilbert space. Your use isn't consistent with that. The question of picture is not going to help you move the time-evolution operator outside of an inner product.

Can you be more specific as to what line is inconsistent? This is almost verbatim from Ballentine's book. I am guessing you are referring to the following.
$$G(x,t;x',t_0)=\bra{x}U(t,t_0)\ket{x'}$$

The way I understand it is that we can consider $\ket{x'}$ as a state that is highly localized around $x'$.

 

[PeterDonis]

We start with the evolution equation for our state vector.

$$ \ket{\Psi(t)} = U(t,t_0)\ket{\Psi(t_0)} $$

This is not correct. $\Psi$ is a function of both $x$ and $t$, not just $t$. So you have to take the $x$ dependence into account in the evolution equation. That is what Ballentine is doing.

Basically, Ballentine's evolution equation says that the amplitude at position $x$ at time $t$ is the sum of all the possible evolutions from $t_0$ to $t$ that end up at position $x$. And since there is a nonzero amplitude for the position to change to $x$ at time $t$ from any position $x'$ at time $t_0$, you have to include that in the evolution equation, and the way to do that is Ballentine's $G$ factor. Your delta function factor doesn't do that; a delta function factor says that the only way to end up at position $x$ at time $t$ is to be at position $x$ at time $t_0$, which is wrong.

 

[Haborix]

You gave an expression that you expected to find. In that expression you use $U(t,t_0)$ in a way which is inconsistent with the defining use:

$\ket{\Psi(t)}=U(t,t_0)\ket{\Psi(t_0)}$

 

[Haborix]

This is not correct. $\Psi$ is a function of both $x$ and $t$, not just $t$. So you have to take the $x$ dependence into account in the evolution equation. That is what Ballentine is doing.

What are you saying is not correct? The equation you cite is correct. However, I think when things are taken into the position basis it might be better practice to use a different variable than upper-case Psi. There would be a lot more clarity here if it were made clear what is an abstract vector in Hilbert space, what is the wavefunction (i.e. the state vector basis elements in the position basis), and similarly for the operators.

 

[jbergman]

This is not correct. $\Psi$ is a function of both $x$ and $t$, not just $t$. So you have to take the $x$ dependence into account in the evolution equation. That is what Ballentine is doing.

Basically, Ballentine's evolution equation says that the amplitude at position $x$ at time $t$ is the sum of all the possible evolutions from $t_0$ to $t$ that end up at position $x$. And since there is a nonzero amplitude for the position to change to $x$ at time $t$ from any position $x'$ at time $t_0$, you have to include that in the evolution equation, and the way to do that is Ballentine's $G$ factor. Your delta function factor doesn't do that; a delta function factor says that the only way to end up at position $x$ at time $t$ is to be at position $x$ at time $t_0$, which is wrong.

The first equation you said is wrong is verbatim from the book.

The exact text is,
"The time evolution of the quantum state vector, $\ket{\Psi(t)} = U(t, t_0)\ket{\Psi{t_0}}$, can be regarded as the propagation of an amplitude in configuration space,

$$ \Psi(x, t) = \int G(x,t;x',t_0) \Psi(x',t_0) dx',$$

where
$$G(x,t;x',t_0)=\bra{x}U(t,t_0)\ket{x'}$$
"

 

[jbergman]

You gave an expression that you expected to find. In that expression you use $U(t,t_0)$ in a way which is inconsistent with the defining use:

$\ket{\Psi(t)}=U(t,t_0)\ket{\Psi(t_0)}$

I see. Then is it fair to say the following?
$$ \Psi(x, t) = \bra{x}U(t,t_0)\ket{\Psi(t_0)} $$
I am assuming how I converted that expression on the right to an integral is the problem.

 

[Haborix]

I see. Then is it fair to say the following?
$$ \Psi(x, t) = \bra{x}U(t,t_0)\ket{\Psi(t_0)} $$
I am assuming how I converted that expression on the right to an integral is the problem.

That looks right to me. Now you just need to drop the identity operator in a useful spot to get the final result.

 

[PeterDonis]

The first equation you said is wrong is verbatim from the book.

The book appears to be using that equation as a definition of a shorthand notation, since in the very next equation (the one that the OP is asking where it came from) it has $\Psi$ as a function of both $x$ and $t$, not of $t$ alone. In other words, the ket $\ket{\Psi(t)}$ is a shorthand for $\Psi(x, t)$. Otherwise it would be a mystery where the $x$ dependence comes from.

 

[PeterDonis]

I think when things are taken into the position basis it might be better practice to use a different variable than upper-case Psi. There would be a lot more clarity here if it were made clear what is an abstract vector in Hilbert space, what is the wavefunction (i.e. the state vector basis elements in the position basis), and similarly for the operators.

I agree with this. Unfortunately, such confusing notation is ubiquitous in the literature.

It looks like @Haborix has gotten you on the right track, so feel free to ignore my previous posts.

 

[Haborix]

I agree with this. Unfortunately, such confusing notation is ubiquitous in the literature.

I try not to judge too harshly. I am inclined to take the Fourier transform of $f(x)$ and get $f(p)$ most of the time .

 

[samalkhaiat]

Summary:: What is the propagator?
G(x,t;x',t_0)=\bra{x}U(t,t_0)\ket{x'}

That does not mean that you can write G(x,t;x^{\prime},t_{0}) \equiv \langle x |U(t,t_{0})|x^{\prime}\rangle = \delta (x - x^{\prime}) U(t,t_{0}) . To go from time-evolution in the space of kets, |\Psi (t)\rangle = U(t)|\Psi (0)\rangle, to propagation in the space of wavefunctions, \Psi (x,t) = \langle x |\Psi (t)\rangle, you need to insert the identity operator 1 = \int dx^{\prime} \ |x^{\prime}\rangle \langle x^{\prime}| as follows

\langle x |\Psi (t)\rangle = \int dx^{\prime} \ \langle x | U(t) |x^{\prime}\rangle \langle x^{\prime}|\Psi (0)\rangle , which is nothing but \Psi (x,t) = \int dx^{\prime} \ G(x,t;x^{\prime},0) \Psi (x^{\prime},0) .