failexam said:
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 Schrödinger wave mechanics?
And furthermore that, it does not make sense to talk about position (or momentum representation) in Feynman's path integral formalism?
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
<br />
\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} .<br />
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
<br />
- i \frac{\psi(x , t + \epsilon) - \psi(x , t)}{\epsilon} = \frac{1}{2} \psi^{’’}(x,t) - V(x) \psi(x,t) .<br />
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
<br />
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) .<br />
Probably, you will be the first one to do it
