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$\newcommand{\DD}[2][]{\frac{d^2 #1}{d^2 #2}}\newcommand{\matrixelement}[3] \langle#1|#2|#3\rangle} \newcommand{\PP}[2][]{\frac{\partial^2 #1}{\partial #2^2}}\newcommand{\dd}[2][]{\frac{d#1}{d#2}}\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}$
A tabular comparison of coordinate and momentum representations is presented.
The coordinate representation is widely used. In the coordinate representation we work with the orthonormal basis vectors are eigenvectors of position.
In the momentum representation the orthonormal basis is chosen to be the eigenvectors of momentum operator(s) and roles of position and momentum get interchanged.
A comparison of important equations in the two representations is presented below.
| Comparison of coordinate and momentum representations. | ||
| Item | Coordinate representation | Momentum representation |
| Choice of basis Vectors | \(\ket{\vec{r}}\) are simultaneous eigen-vectors of position operators. | \(\ket{\vec{p}}\) are simultaneous eigen-vectors of momentum operators. |
| Orthogonality | \( \innerproduct{\vec{r}^{\,{'}{'}}}{\vec{r}^{\,{'}}}=\delta(\vec{r}{'}-\vec{r})\) | \( \innerproduct{\vec{p}^{\,{'}{'}}}{\vec{p}^{\,{'}}}=\delta(\vec{p}{'}-\vec{p})\) |
| Completeness formula | \(\int d^3r \ket{\vec{r}}\bra{\vec{r}}= \hat{I}\) | \(\int d^3p \ket{\vec{p}}\bra{\vec{p}}= \hat{I}\) |
| Expansion in the basis | \( \ket{\psi} = \int d^3r \ket{\vec{r}\,}\innerproduct{\vec{r}\,}{\psi}\,\) | \( \ket{\psi} = \int d^3r \ket{\vec{p}\,}\innerproduct{\vec{p}\,}{\psi}\,\) |
| State representative | Coordinate space wave function \(\tilde{\psi}(\vec{r}) \equiv \innerproduct{\vec{r}\,}{\psi}\)\ | Momentum space wave function \(\tilde{\psi}(\vec{p}) \equiv \innerproduct{\vec{p}\,}{\psi}\) |
| Norm squared :\(\Vert\psi\Vert^2\) | \(\innerproduct{\psi}{\psi}=\int|\psi(\vec{r}\,)|^2d^3r\) | \(\innerproduct{\psi}{\psi}=\int|\tilde{\psi}(\vec{p}\,)|^2 d^3p\) |
| Scalar product \(\innerproduct{\psi}{\phi}\) | \(\int \psi(\vec{r})^* \phi(\vec{r}) d^3r\) | \(\int \psi(\vec{p})^* \phi(\vec{p}) d^3r\) |
| Action of position operator \(\hat{x}\) | \(\hat{x} \psi(\vec{r})= x \psi(\vec{r})\) | \( \hat{x} \tilde{\psi}(p) = i\hbar \dfrac{\partial}{\partial p_x} \tilde{\psi}(p)\) |
| Action of momentum operator \(\hat{p}_x\) | \(\hat{p_x} \psi(\vec{r})= -i\hbar \dfrac{\partial}{\partial x} \psi(\vec{r})\) | \(\hat{p} \tilde{\psi}(\vec{p})= p_x \tilde{\psi}(\vec{p})\) |
| [Operator for \({F}(q_j,p_k)\) | \(\widehat{F}\Big(\hat{q}_j,-i\hbar\dfrac{\partial}{\partial q_k}\Big)\) | \(\widehat{F}\Big(i\hbar\dfrac{\partial}{\partial p_j},p_k\Big)\) |
It must be noted that two different can be written for the same classical dynamical variable. Classically, \(pq\) and \(qp\) are the same functions but the operators \(\hat{p}\hat{q}\) and \(\hat{q}\hat{p}\) are different. So which operator should correspond to the classical product \(pq\)? Should it be \(\hat{p}\hat{q}\) or \(\hat{q}\hat{p}\)? This is known as {\it ordering problem}. The only guideline available is that the operator corresponding to a dynamical variable must be hermitian. So, in case of \(pq\), we must select \[ pq \to \frac{1}{2}(\hat{p}\hat{q} + \hat{p}\hat{q})\] For more complicated expressions the requirement of hermiticity is not sufficient to get a unique answer.
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