# Propagation amplitude and time-evolution operator

1. Jan 27, 2016

### spaghetti3451

I know that the time-evolution operator in quantum mechanics is $e^{-iHt}$.

Is this also called the Schrodinger time-evolution operator?

Also, can you guys explain why the amplitude $U(x_{a},x_{b};T)$ for a particle to travel from one point $(x_{a})$ to another $(x_{b})$ in a given time $(T)$ is the $\textit{position representation}$ of the Schrodinger time-evolution operator?

2. Jan 27, 2016

### blue_leaf77

Directly translating that statement into quantum mechanical languange, you start with a particle being in a position operator eigenfunction $|x_a \rangle$. Then this state evolve in time according to the time-independent Hamiltonian the system subjects to. This means operating the time evolution operator $U = \exp(-iHt)$ to the initial state, thus $\exp(-iHt)|x_a \rangle$. Finally after time $t=T$, you measure the particle's position in another place $x=x_b$, which translates into the projection of the last state into $|x_b \rangle$. In the end, you have $\langle x_b |\exp(-iHT)|x_a \rangle = U(x_a,x_b;T)$. The last expression may be interpreted as the matrix element of the operator $U$ in position basis.

3. Jan 27, 2016

### spaghetti3451

Thank you, this is a great answer. But, let me ask one counter-question.

What's you've mentioned is in the canonical Hamiltonian formalism. In that formalism, the propagation amplitude is in the position rep of the time-evolution operator.

But, in the path-integral formalism,

$U(x_a,x_b;T) = \int \mathcal{D}x(t)\ e^{iS[x(t)]/\hbar}$.

How is this in the position rep of the time evolution operator?

4. Jan 27, 2016

### blue_leaf77

The derivation to show that the propagation kernel in Feynman path integral and that in Schrödinger wave mechanics are identical is rather long, but it can indeed shown that they actually coincide. For this purpose you can check Sakurai's Modern Quantum Mechanics in section 2.5. The author really addresses the answer to your question.

5. Jan 27, 2016

### vanhees71

You can also find this in my QFT lecture notes (the first chapter is entirely on non-relativistic quantum mechanics in the path-integral formalism; so you don't need QFT to understand it):

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

6. Jan 27, 2016

### spaghetti3451

Thanks for the references.

So, what you're essentially saying is that it only makes sense to talk about the position representation of the time-evolution operator in Schrodinger wave mechanics?

And furthermore that, it does not make sense to talk about position (or momentum representation) in Feynman's path integral formalism?

7. Jan 27, 2016

### samalkhaiat

I will try to answer all your questions in one go. Okay, start with the representation-free Schrödinger equation:
$$i \partial_{t}|\Psi (t) \rangle = H |\Psi (t) \rangle .$$
For time-independent Hamiltonian, we have the following formal solution
$$|\Psi (t + \epsilon) \rangle = e^{- i H \epsilon} | \Psi (t) \rangle . \ \ \ \ \ \ \ \ \ (1)$$
Now, consider complete orthonormal set labelled by continuous real parameter $\alpha$
$$\int d\alpha \ |\alpha \rangle \langle \alpha | = I , \ \ \ \langle \alpha | \bar{\alpha} \rangle = \delta (\alpha - \bar{\alpha}) .$$
In this $\alpha$-representation, equation (1) becomes
$$\langle \alpha | \Psi (t + \epsilon) \rangle = \int d \bar{\alpha} \ \langle \alpha | e^{- i H \epsilon} | \bar{\alpha}\rangle \langle \bar{\alpha}| \Psi (t) \rangle .$$
In wave-function language, this is written as
$$\Psi (\alpha , t + \epsilon) = \int d \bar{\alpha} \ K(\alpha , \epsilon ; \bar{\alpha}) \ \Psi (\bar{\alpha} , t) , \ \ \ \ \ \ \ (2)$$
where
$$K(\alpha , \epsilon ; \bar{\alpha}) \equiv \langle \alpha | e^{- i H \epsilon} | \bar{\alpha}\rangle ,$$
is the propagator in the $\alpha$-representation.
So, in the coordinate representation, (2) becomes
$$\psi ( x , t + \epsilon) = \int d \bar{x} \ K( x , \epsilon ; \bar{x}) \ \psi (\bar{x} , t) , \ \ \ \ \ \ \ (3)$$
and, in the momentum representation it becomes
$$\phi ( p , t + \epsilon) = \int d \bar{p} \ \Gamma( p , \epsilon ; \bar{p}) \ \phi (\bar{p} , t) . \ \ \ \ \ \ \ (4)$$
Now, using the fact that $\psi$ and $\phi$ are the Fourier transform of each other:
$$\psi (\bar{x} , t) = \frac{1}{\sqrt{2\pi}} \int d \bar{p} \ e^{i \bar{p} \bar{x}} \ \phi (\bar{p} , t) ,$$
$$\phi (p , t + \epsilon) = \frac{1}{\sqrt{2\pi}} \int dx \ e^{- i px} \ \psi (x , t + \epsilon) ,$$
you can easily find the following relation between the p-space propagator $\Gamma$ and the x-space one $K$:
$$\Gamma (p , \epsilon ; \bar{p}) = \frac{1}{2\pi} \int dx d\bar{x} \ e^{- i px} \ K(x , \epsilon ; \bar{x}) \ e^{i \bar{p}\bar{x}} .$$
In the path-integral formalism, the propagator is given by the classical action integral:
$$K(x,\epsilon;\bar{x}) = C \ e^{- i \int dt L} = C \ e^{- \frac{(\bar{x} - x )^{2}}{2 i \epsilon}} \ e^{- i \epsilon V(\bar{x})} . \ \ \ (5)$$
We will now show (as Feynman did in his original paper) that, for a certain value for the constant $C$, the Schrödinger equation in the coordinate representation follows from Eq(3) in the limit $\epsilon \to 0$.
Substituting (5) in (3) and changing the variable of integration as
$$\bar{x} - x = y ,$$
we get
$$\psi(x , t + \epsilon) = C \int dy \ e^{ - \frac{y^{2}}{2 i \epsilon}} \ e^{- i \epsilon V(x+y)} \ \psi(x + y , t) . \ \ \ \ (6)$$
The next steps involve a small head ach of Taylor expansions and Gaussian integrals. So, let us expand $V(x+y)$ and $\psi(x+y,t)$ about $y = 0$
$$V(x + y) = V(x) + y V^{’}(x) + \cdots ,$$
$$\psi(x + y , t) = \psi(x,t) + y \psi^{’}(x,t) + \frac{y^{2}}{2} \psi^{’’}(x,t) + \cdots .$$
Notice that $\exp\left(-i\epsilon V(x)\right)$ does not depend on $y$, so it can be factored out from the integral. And, since we are considering an infinitesimal time-step, i.e., the $\epsilon \to 0$ limit of the integral (6), we may write
$$e^{-i\epsilon V(x)} = 1 - i \epsilon V(x) .$$
We can also drop the factor $\exp \left(- i \epsilon y V^{’}(x) \right)$: When the integration over $y$ is performed, the contribution of this factor will be of order $\epsilon^{3/2}$. Putting all of this in Eq(6), and using
$$\int_{-\infty}^{\infty} dy \ y \ e^{- y^{2}/ 2i \epsilon} = 0 ,$$
we find
$$\psi(x , t + \epsilon ) = C \psi(x,t) \int dy \ e^{- y^{2} / 2i \epsilon} + \frac{C}{2} \psi^{’’}(x,t) \int dy y^{2} e^{- y^{2} / 2i \epsilon} - i \epsilon C V(x) \psi(x,t) \int dy e^{- y^{2} / 2i \epsilon} .$$
Now, if we make use of the following Gaussian integrals
$$\int_{-\infty}^{+\infty} dy \ e^{- y^{2} / 2i \epsilon} = \sqrt{2i \epsilon \pi} ,$$
$$\int_{-\infty}^{+\infty} dy \ y^{2} e^{- y^{2} / 2i \epsilon} = i \epsilon \ \sqrt{2i \epsilon \pi} ,$$
and take
$$C = \frac{1}{\sqrt{2i \epsilon \pi}} ,$$
we find
$$- i \frac{\psi(x , t + \epsilon) - \psi(x , t)}{\epsilon} = \frac{1}{2} \psi^{’’}(x,t) - V(x) \psi(x,t) .$$
Thus, by letting $\epsilon \to 0$, we obtain the coordinate space Schrödinger equation
$$- i \frac{\partial}{\partial t} \psi(x,t) = \frac{1}{2} \frac{\partial^{2}}{\partial x^{2}} \psi(x,t) - V(x)\psi(x,t) .$$
Okay, if you are brave enough, try to redo the above method using the momentum-space propagator $\Gamma (p , \epsilon ; \bar{p})$ in Eq(4), and obtain the momentum-space Schrödinger equation
$$i \frac{\partial}{\partial t} \phi(p,t) = \frac{p^{2}}{2}\phi(p,t) + \int d\bar{p} \ V(p - \bar{p}) \phi ( \bar{p} , t) .$$
Probably, you will be the first one to do it

Last edited: Jan 28, 2016
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