Position and Momentum Representation of States #

This article discusses the position and momentum representations of states in quantum mechanics.
The goals of this article are:

To derive $\langle x|p\rangle = e^{ipx}/\sqrt{2\pi\hbar}$
To understand the transformation between position and momentum representations

Only the one-dimensional case will be considered.

1. Summary #

Let the eigenstates of the position operator $\hat{x}$ and the momentum operator $\hat{p}$ be denoted as $|x\rangle$ and $|p\rangle$, respectively, with orthogonality defined as follows:

$$ \begin{gather} \langle x|x’\rangle = \delta(x-x’) \\ \langle p|p’\rangle = \delta(p-p’) \\ \end{gather} $$

In this context, the quantity $\langle x|p\rangle$ is given by:

$$ \begin{gather} \langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{ipx} \end{gather} $$

Defining $\psi(x) \equiv \langle x|\psi\rangle$ and $\psi(p) \equiv \langle p|\psi\rangle$, the position and momentum representations of the state $|\psi\rangle$, respectively, the two are related by the following expression:

$$ \begin{align} \left\{ \begin{aligned} \psi(p)&= \frac{1}{\sqrt{2\pi\hbar}} \int dx \psi(x) e^{-ipx/\hbar} \\ \psi(x)&= \frac{1}{\sqrt{2\pi\hbar}} \int dp \psi(p) e^{ipx/\hbar} \end{aligned} \right.\label{es2} \end{align} $$

2. Properties of Operators #

To avoid ambiguity, we clarify here that the position operator $\hat{x}$ and the momentum operator $\hat{p}$ in this article refer to the following operators in the position representation:

$$ \begin{align} \hat{x}|x\rangle &= \hat{x}|x\rangle \label{e1}\\ \hat{p}|x\rangle &= -i\hbar\frac{d}{dx} |x\rangle \label{e2} \end{align} $$

2.1. Eigenvalues and Eigenstates of the Position Operator #

The position operator is denoted as $\hat{x}$, and its eigenvalues and eigenstates are denoted as $x$ and $|x\rangle$, respectively.
The orthogonality of $|x\rangle$ and the identity operator $\hat{I}_x$ are expressed as follows:

$$ \begin{gather} \hat{x}|x\rangle = x|x\rangle \label{ex1} \end{gather} $$

$$ \begin{gather} \langle x |x’\rangle = \delta(x-x’) \label{ex2} \end{gather} $$

$$ \begin{gather} \hat{I}_x = \int dx |x\rangle\langle x|\label{ex3} \end{gather} $$

Since $\hat{x}$ is a Hermitian operator, $\hat{x}^{\dagger} = \hat{x}$, and its eigenvalues $x$ are real numbers.

2.2. Eigenvalues and Eigenstates of the Momentum Operator #

The momentum operator can be treated similarly to the position operator.

The momentum operator is denoted as $\hat{p}$, and its eigenvalues and eigenstates are denoted as $p$ and $|p\rangle$, respectively.
The orthogonality of $|p\rangle$ and the identity operator $\hat{I}_p$ are expressed as follows:

$$ \begin{gather} \hat{p}|p\rangle = p|p\rangle \label{ep1} \end{gather} $$

$$ \begin{gather} \langle p |p’\rangle = \delta(p-p’) \label{ep2} \end{gather} $$

$$ \begin{gather} \hat{I}_p = \int dp |p\rangle\langle p|\label{ep3} \end{gather} $$

Since $\hat{p}$ is a Hermitian operator, $\hat{p}^{\dagger} = \hat{p}$, and its eigenvalues $p$ are real numbers.

In particular, the position representation of the momentum operator is given by the following expression, which is assumed to be known:

$$ \begin{align} \hat{p}|x\rangle = -i\hbar\frac{d}{dx} |x\rangle \label{ep4} \end{align} $$

3. Position and Momentum Representations of States #

When expressing a state as $|\psi\rangle$, this is an abstract representation without using a specific basis.

3.1. Position Representation #

By inserting the position identity operator \eqref{ex3} before $|\psi\rangle$, we have:

$$ \begin{align} |\psi\rangle &= \hat{I}_x|\psi\rangle \label{epx1}\\ &= \int dx \langle x| \psi\rangle |x\rangle \label{epx2} \end{align} $$

Further applying $\langle x’|$ from the left, we obtain:

$$ \begin{align} \langle x’|\psi\rangle &= \int dx \langle x| \psi\rangle \langle x’|x\rangle \label{epx3}\\ &=\int dx \langle x| \psi\rangle \delta(x’-x) \label{epx4} \end{align} $$

Examining the form of equation \eqref{epx4}, the term $\langle x’|\psi\rangle$ matches the following representation of a function $f(x)$ using a delta function:

$$ \begin{align} f(x)= \int dx’ f(x’)\delta(x’-x) \label{epx5} \end{align} $$

Thus, we can associate:

$$ \begin{align} \langle x|\psi\rangle = \psi(x) \label{epx6} \end{align} $$

This is referred to as the position representation of the state.

3.2. Momentum Representation #

Similarly, by inserting the momentum identity operator \eqref{ep3} before $|\psi\rangle$ and applying $\langle p’|$ from the left, we have:

$$ \begin{align} \langle p’|\psi\rangle &=\int dp \langle p| \psi\rangle \delta(p’-p) \label{epx14} \end{align} $$

This results in:

$$ \begin{align} \langle p|\psi\rangle = \psi(p) \label{epx16} \end{align} $$

yielding the momentum representation of the state.

We now have two different representations of the same state $|\psi\rangle$: the position representation and the momentum representation.
However, $\psi(x)$ in equation \eqref{epx6} and $\psi(p)$ in equation \eqref{epx16} must fully represent the same state.

This naturally leads to the question:

How can the position representation $\psi(x)$ and momentum representation $\psi(p)$ be directly transformed into each other?

The following discussion addresses this question.

3.3. Transformation Between Position and Momentum Representations #

By inserting identity operators into equations \eqref{epx6} and \eqref{epx16}, we can examine the relationship between $\psi(x)$ and $\psi(p)$.

Inserting $\hat{I}_p$ into equation \eqref{epx6} gives:

$$ \begin{align} \psi(x)=\langle x|\psi\rangle &= \langle x| \left[\int dp |p\rangle\langle p|\right] |\psi\rangle \\ &= \int dp \langle x|p\rangle \langle p|\psi\rangle \\ &= \int dp \langle x|p\rangle \psi(p) \label{epx21}\\ \end{align} $$

This allows $\psi(x)$ to be expressed as an integral of $\psi(p)$ multiplied by $\langle x|p\rangle$.

Similarly, inserting $\hat{I}_x$ into equation \eqref{epx16} gives:

$$ \begin{align} \psi(p)=\langle p|\psi\rangle &= \langle p| \left[\int dx |x\rangle\langle x|\right] |\psi\rangle \\ &= \int dx \langle p|x\rangle \langle x|\psi\rangle \\ &= \int dx \langle p|x\rangle \psi(x) \label{epx22}\\ \end{align} $$

This allows $\psi(p)$ to be expressed as an integral of $\psi(x)$ multiplied by $\langle p|x\rangle$.

Summarizing these results:

$$ \begin{align} \left\{ \begin{aligned} \psi(p)= \int dx \langle p|x\rangle \psi(x) \\ \psi(x)= \int dp \langle x|p\rangle \psi(p) \end{aligned} \right.\label{epx23} \end{align} $$

Equations \eqref{epx21} and \eqref{epx22} enable the direct transformation between $\psi(x)$ and $\psi(p)$.

The remaining question is: what is $\langle x|p\rangle = (\langle p|x\rangle)^*$?
Let us derive this next.

3.4. Derivation of $\langle x|p\rangle$ #

As a heuristic approach, let us consider the quantity $\langle x|\hat{p}|p\rangle$ to determine $\langle x|p\rangle$.

This quantity can be manipulated in two different ways:

Considering $\hat{p}|p\rangle$

First, applying the operator $\hat{p}$ to its eigenstate gives the following transformation:

$$ \begin{align} \langle x|\hat{p}| p\rangle &= \langle x| \hspace{0.25em} \Bigl( \hat{p}| p\rangle\Bigr) \\ &= p \langle x| p\rangle \label{epx24} \end{align} $$

Considering $\langle x|\hat{p}$

Alternatively, we can apply $\hat{p}$ to the bra in the position representation. In this case, $\langle x|\hat{p} = \langle x|(-i)\hbar\frac{d}{dx}$.
By inserting the identity operator $\hat{I}_x$ and transforming the expression, we obtain:

$$ \begin{align} \langle x|\hat{p}| p\rangle &= \langle x| \hat{p}\hat{I}_{x’}| p\rangle\bigr) \\ &= \langle x| \left(-i\hbar\frac{d}{dx}\right)\hspace{0.2em} \left(\int dx’ |x’\rangle\langle x’|\right) | p\rangle \\ &= \int dx’ \langle x| x’\rangle \left(-i\hbar\frac{d}{dx} \right)\langle x’ | p\rangle \\ &= -i\hbar\frac{d}{dx} \langle x | p\rangle \label{epx25} \end{align} $$

Here, equation \eqref{ex2} is used.

Since equations \eqref{epx24} and \eqref{epx25} originate from the same expression, they must be equivalent. This yields the differential equation:

$$ \begin{align}\label{epx26} p \langle x| p\rangle = -i\hbar\frac{d}{dx} \langle x | p\rangle \end{align} $$

Solving this equation for $\langle x|p\rangle$, we find:

$$ \begin{align}\label{epx27} \langle x| p\rangle = C e^{ipx/\hbar} \end{align} $$

where $C$ is an arbitrary constant that must be determined from additional conditions.

Next, let us determine the constant $C$. We use the fact that applying the identity operator to the state $|p\rangle$ does not change it:

$$ \begin{align} |p\rangle &= \hat{I}_p \hat{I}_x|p\rangle \\ &=\hat{I}_p \int dx|x\rangle \langle x|p\rangle \\ &=\int dx \int dp’|p’\rangle\langle p’|x\rangle \langle x|p\rangle\\ &=|C|^2 \int dp’|p’\rangle \int dx e^{i(p-p’)x/\hbar} \\ &=|C|^2 \int dp’|p’\rangle 2\pi \delta((p-p’)/\hbar) \label{epx28}\\ &=|C|^2 2\pi \hbar |p\rangle \label{epx29} \end{align} $$

This expression remains unchanged as only the identity operator is inserted. Thus, we obtain:

$$ \begin{align}\label{epx30} |C|^2 2\pi \hbar=1 \hspace{1em}\rightarrow\hspace{1em} |C|=\frac{1}{\sqrt{2\pi\hbar}} \end{align} $$

In equation \eqref{epx29}, we use:

$$ \begin{align}\label{epx31} \delta(ax)=\frac{1}{|a|}\delta(x),\hspace{1em}(a\gt 0) \end{align} $$

The phase of $C$ is arbitrary and can be set to yield a real value. Therefore, we set $C = \frac{1}{\sqrt{2\pi\hbar}}$.
Thus, we obtain:

$$ \begin{align}\label{epx32} \langle x| p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar} \end{align} $$

This concludes the derivation of $\langle x|p\rangle$.

From the above, substituting into equation \eqref{epx23}, we derive the transformation formula between position and momentum representations:

Equation \eqref{epx33} corresponds precisely to a Fourier transform.

4. Supplementary Notes #

4.1. Alternative Normalization Methods for Position and Momentum Representations #

Let us redefine the orthogonality of the eigenstates of position and momentum as follows¹².

$$ \begin{align} \langle x|x’\rangle &= \delta(x-x’) \label{ea1}\\ \langle p|p’\rangle &= 2\pi\hbar\delta(p-p’)\label{ea2} \end{align} $$

Under this normalization, the momentum identity operator is redefined as:

$$ \begin{gather}\label{ea3} \hat{I}_p = \int \frac{dp}{2\pi\hbar} |p\rangle\langle p| \end{gather} $$

$$ \begin{align} \psi(p)&=\int dx\langle p|x\rangle\psi(x) \label{ea4}\\ \psi(x)&=\int \frac{dp}{2\pi\hbar}\langle x|p\rangle\psi(p)\label{ea5} \end{align} $$

In this case, the quantity $\langle x|p\rangle = C e^{ipx/\hbar}$ leads to the constant $C$, derived similarly to equation \eqref{epx28}, as:

$$ \begin{align} |p\rangle &= \hat{I}_p \hat{I}_x|p\rangle \\ &=\hat{I}_p \int dx|x\rangle \langle x|p\rangle \\ &=\int dx \int \frac{dp’}{2\pi\hbar}|p’\rangle\langle p’|x\rangle \langle x|p\rangle\\ &=|C|^2 \int \frac{dp’}{2\pi\hbar} |p’\rangle \int dx e^{i(p-p’)x/\hbar} \\ &=|C|^2 \int \frac{dp’}{2\pi\hbar}|p’\rangle 2\pi \delta((p-p’)/\hbar) \label{ea6}\\ &=|C|^2 |p\rangle \label{ea7} \end{align} $$

This results in:

$$ \begin{align}\label{ea8} |C|=1 \end{align} $$

Thus, we have:

$$ \begin{gather} \langle x|p\rangle = e^{ipx} \end{gather} $$

Finally, writing the position and momentum representations of the state $|\psi\rangle$, we find the following relationship between the two:

$$ \begin{align} \left\{ \begin{aligned} \psi(p)&= \int dx \psi(x) e^{-ipx/\hbar} \\ \psi(x)&= \int \frac{dp}{2\pi\hbar} \psi(p) e^{ipx/\hbar} \end{aligned} \right.\label{es2xx} \end{align} $$

The definitions $\hat{x}|x\rangle = x|x\rangle$ and $\hat{p}|p\rangle = p|p\rangle$ remain unchanged. ↩︎
運動量表示 -物理とはずがたり ↩︎