Position Space and Momentum Space

We can represent a state with either \bgroup\color{black}$\psi(x)$\egroup or with \bgroup\color{black}$\phi(p)$\egroup. We can (Fourier) transform from one to the other.

We have the symmetric Fourier Transform.

\begin{eqnarray*}
f(x)&=&{1\over \sqrt{2\pi}}\int\limits_{-\infty}^\infty A(k) e...
...ver \sqrt{2\pi}}\int\limits_{-\infty}^\infty f(x) e^{-ikx} dx\\
\end{eqnarray*}


When we change variable from \bgroup\color{black}$k$\egroup to \bgroup\color{black}$p$\egroup, we get the Fourier Transforms in terms of $x$ and $p$.
\bgroup\color{black}$\displaystyle \psi(x)={1\over \sqrt{2\pi\hbar}}\int\limits_{-\infty}^\infty \phi(p) e^{ipx/\hbar} dp$\egroup
\bgroup\color{black}$\displaystyle \phi(p)={1\over \sqrt{2\pi\hbar}}\int\limits_{-\infty}^\infty \psi(x) e^{-ipx/\hbar} dx$\egroup

These formulas are worth a little study. If we define \bgroup\color{black}$u_p(x)$\egroup to be the state with definite momentum \bgroup\color{black}$p$\egroup, (in position space) our formula for it is

\begin{displaymath}\bgroup\color{black}u_p(x)={1\over\sqrt{2\pi\hbar}}e^{ipx/\hbar}.\egroup\end{displaymath}

Similarly, the state (in momentum space) with definite position \bgroup\color{black}$x$\egroup is

\begin{displaymath}\bgroup\color{black}v_x(p)={1\over\sqrt{2\pi\hbar}}e^{-ipx/\hbar}\egroup\end{displaymath}

These states cannot be normalized to 1 but they do have a normalization convention which is satisfied due to the constant shown.

Our Fourier Transform can now be read to say that we add up states of definite momentum to get \bgroup\color{black}$\psi(x)$\egroup

\begin{displaymath}\bgroup\color{black}\psi(x)=\int\limits_{-\infty}^\infty \phi(p) u_p(x) dp\egroup\end{displaymath}

and we add up states of definite position to get \bgroup\color{black}$\phi(p)$\egroup.

\begin{displaymath}\bgroup\color{black}\phi(p)=\int\limits_{-\infty}^\infty \psi(x) v_x(p) dx\egroup\end{displaymath}

There is a more abstract way to write these states. Using the notation of Dirac, the state with definite momentum \bgroup\color{black}$p_0$\egroup, \bgroup\color{black}$u_{p0}(x)={1\over\sqrt{2\pi\hbar}}e^{ip_0x/\hbar}$\egroup might be written as

\begin{displaymath}\bgroup\color{black} \vert p_0\rangle\egroup\end{displaymath}

and the state with definite position \bgroup\color{black}$x_1$\egroup, \bgroup\color{black}$v_{x1}(p)={1\over\sqrt{2\pi\hbar}}e^{-ipx_1/\hbar}$\egroup might be written

\begin{displaymath}\bgroup\color{black} \vert x_1\rangle .\egroup\end{displaymath}

The arbitrary state represented by either \bgroup\color{black}$\psi(x)$\egroup or \bgroup\color{black}$\phi(p)$\egroup, might be written simple as

\begin{displaymath}\bgroup\color{black} \vert\psi\rangle .\egroup\end{displaymath}

The actual wave function \bgroup\color{black}$\psi(x)$\egroup would be written as

\begin{displaymath}\bgroup\color{black} \psi(x)=\langle x\vert\psi\rangle .\egroup\end{displaymath}

This gives us the amplitude to be at \bgroup\color{black}$x$\egroup for any value of \bgroup\color{black}$x$\egroup.

We will find that there are other ways to represent Quantum states. This was a preview. We will spend more time on Dirac Bra-ket notation later.

Jim Branson 2013-04-22