Position Space and Momentum Space

We can represent a state with either $\psi(x)$ or with $\phi(p)$. 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 $k$ to $p$, we get the Fourier Transforms in terms of $x$ and $p$.

$$\psi(x)={1\over \sqrt{2\pi\hbar}}\int\limits_{-\infty}^\infty \phi(p) e^{ipx/\hbar} dp$$

$$\phi(p)={1\over \sqrt{2\pi\hbar}}\int\limits_{-\infty}^\infty \psi(x) e^{-ipx/\hbar} dx$$

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

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

Similarly, the state (in momentum space) with definite position $x$ is

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

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 $\psi(x)$

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

and we add up states of definite position to get $\phi(p)$.

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

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

$$\bgroup\color{black} \vert p_0\rangle\egroup$$

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

$$\bgroup\color{black} \vert x_1\rangle .\egroup$$

The arbitrary state represented by either $\psi(x)$ or $\phi(p)$, might be written simple as

$$\bgroup\color{black} \vert\psi\rangle .\egroup$$

The actual wave function $\psi(x)$ would be written as

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

This gives us the amplitude to be at $x$ for any value of $x$.

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.