# Switch from the position representation to the momentum representation

If we use Fourier Transform, we can switch from the position representation to the momentum representation, like the following formula here comes the problem, if we use dirac notation we can see it is the inverse Fourier Transform we use to switch into the momentum representation This is used in one published paper. title:" A finite-dimensional quantum model for the stock market" author:Liviu-Adrian Cotfas, page 7, formula 32

also how can I get the transformation of operator in position and mentum representation via inverse Fourier Transform and Fourier Transform? \hat{p}=F^{-1}\hat{x}F many thanks!

• Can you clarify your primary question please – anon01 Oct 26 '15 at 4:48
• If you're wondering why the transform and its inverse are identical, its because there is no agreed upon convention - though once you define one, the other is obviously set as well. – anon01 Oct 26 '15 at 4:51
• first thanks for your answeing.how about my second question? – Bob zhang Oct 27 '15 at 14:16

## 1 Answer

No contrast occurs using the dirac notation.

$\phi(p)=\langle p|\Psi\rangle \\ \psi(x)=\langle x|\Psi\rangle$

So $\phi(p)=\langle p|\Psi\rangle=\int dx\langle p|x\rangle \langle x|\Psi\rangle=\frac{1}{\sqrt{2\pi\hbar}}\int dx\, e^{-ipx/\hbar}\psi(x)$ giving your first equation.

About your second question. In order to transform an operator A from say the position representation to the momentum representation you should transform the matrix elements:

$\langle p_t|A|p_l\rangle = \Sigma_{i,j}\langle p_t|x_i\rangle\langle x_i|A|x_j\rangle \langle x_j|p_l\rangle$

• but what's wrong with my derivation？ I just derive that momentum vector is obtained by inverse Fourier Transform. – Bob zhang Oct 27 '15 at 14:10
• also for the second question, next step in your derivation, you obtain that \hat{p}=F\hat{x}F^{-1} , just the opposite of mine. really thanks for your answering. @ workaholic – Bob zhang Oct 27 '15 at 14:14