Why is $\langle x|\psi\rangle$ both a number and a wavefunction?

Dìguā’s question

While studying Dirac notation, Dìguā (地瓜) came across

$$\psi(x)=\langle x|\psi\rangle,$$

and then the expansion of a quantum state

$$|\psi\rangle = \int dx\,|x\rangle\langle x|\psi\rangle.$$

There are three things here that do not feel entirely natural.

First, $\langle x|\psi\rangle$ looks like an inner product. Should an inner product not be just a real number? Why is it also called a wavefunction?

Second, even if $\langle x|\psi\rangle$ is a number, why does the ket $|x\rangle$ still appear inside the integral? An ordinary integral would seem to integrate numbers or functions, not abstract vectors.

Third, this integral looks very much like a Fourier transform. Is it performing a basis expansion, or is it performing a Fourier transform?

These confusions are not merely matters of notation. Dirac notation compresses the “abstract quantum state,” the “basis,” and the “wavefunction” into very short expressions, while also concealing some of the distributional details involved in a continuous basis.

The shortest answer

For a fixed $x$, $\langle x|\psi\rangle$ is a complex number; as $x$ varies continuously, this entire collection of complex numbers forms a complex-valued function $\psi(x)$.

$|x\rangle$ is a ket in a continuous basis, while $\psi(x)$ is the complex coefficient in front of it. The integral

$$\int dx\,\psi(x)|x\rangle$$

is the continuous version of a linear combination of vectors, and the result of the integral is still a ket.

This integral is not itself a Fourier transform. A Fourier transform appears only when the coefficients in the position representation are converted into coefficients in the momentum representation.

How can a number also be a function?

First consider a finite-dimensional vector space. Let $\{|e_i\rangle\}$ be an orthonormal basis. Then the $i$th component of the vector $|v\rangle$ is

$$v_i=\langle e_i|v\rangle.$$

For a fixed $i$, $v_i$ is only a number; when $i$ runs through all possible values, $\{v_i\}$ forms a list of coordinates.

The only difference in the position representation is that the discrete index $i$ is replaced by the continuous index $x$. Thus,

$$\psi(x)=\langle x|\psi\rangle.$$

For a fixed $x$, $\psi(x)$ is a complex number; as $x$ varies continuously, the map

$$x\longmapsto\langle x|\psi\rangle$$

is a complex-valued function. This is entirely analogous to how every term in a sequence is a number, while the sequence as a whole is not a single number.

Therefore, the following two statements are both correct:

• For a fixed $x$, $\langle x|\psi\rangle\in\mathbb C$;
• For all $x$, $\langle x|\psi\rangle$ defines the wavefunction $\psi(x)$.

Why is an inner product not necessarily real?

Only the inner product of a vector with itself,

$$\langle\psi|\psi\rangle,$$

is guaranteed to be a nonnegative real number. The inner product between two different vectors,

$$\langle\phi|\psi\rangle,$$

is generally complex. There is therefore no reason for $\langle x|\psi\rangle$ to be real.

Nor is the probability of a position measurement $\psi(x)$ itself. Rather, it is

$$|\psi(x)|^2\,dx = |\langle x|\psi\rangle|^2\,dx.$$

This represents the probability of finding the particle in the interval $[x,x+dx]$. $\psi(x)$ is a probability amplitude, not a probability.

Is $|x\rangle$ a wavefunction?

$|x\rangle$ is not itself any particular wavefunction. It is an abstract position eigenket satisfying

$$\hat x|x\rangle=x|x\rangle.$$

A ket becomes a function in a given representation only after it is projected onto a chosen basis.

To avoid mixing up two different uses of $x$, fix a position $x_0$. The wavefunction of the position eigenstate $|x_0\rangle$ in the position representation is

$$\langle x|x_0\rangle=\delta(x-x_0).$$

The wavefunction of the same ket in the momentum representation is instead

$$\langle p|x_0\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{-ipx_0/\hbar}.$$

Thus, the relationship between a ket and a wavefunction is this: the ket is an abstract vector, while the wavefunction is the coordinate representation of that vector in a chosen basis.

Why can a ket appear inside an integral?

A finite-dimensional vector can be written as

$$|v\rangle = \sum_i|e_i\rangle\langle e_i|v\rangle = \sum_i v_i|e_i\rangle.$$

Each term $v_i|e_i\rangle$ is “a complex number multiplied by a basis vector,” so every term is still a vector. Adding together all the vector components recovers the complete vector.

The position basis is labelled by the continuous parameter $x$, so the discrete sum becomes an integral:

$$|\psi\rangle = \int dx\,|x\rangle\langle x|\psi\rangle = \int dx\,\psi(x)|x\rangle.$$

Here, $\psi(x)|x\rangle$ is a ket-valued expression. The integral can be understood as a continuous linear combination, or as a vector-valued integral. Its Riemann-sum intuition is to add together many kets with different weights and then take the continuous limit.

Ordinary mathematics also allows vectors to be integrated. For example,

$$\int dt \begin{pmatrix} \cos t\\ \sin t \end{pmatrix}$$

still produces a vector. Integrating kets uses the same idea.

Formally, $\psi(x)|x\rangle\,dx$ can be viewed as an infinitesimal ket component contributed to the complete quantum state by the position interval $[x,x+dx]$.

How can we confirm that this integral really reconstructs $|\psi\rangle$?

The continuous position basis satisfies the completeness relation

$$\int dx\,|x\rangle\langle x|=I.$$

To check this, multiply from the left by an arbitrary position bra $\langle y|$:

$$\begin{aligned} \left\langle y\middle| \int dx\,|x\rangle\langle x|\psi\rangle \right. &= \int dx\,\langle y|x\rangle\langle x|\psi\rangle\\ &= \int dx\,\delta(y-x)\psi(x)\\ &= \psi(y)\\ &= \langle y|\psi\rangle. \end{aligned}$$

The ket produced by the integral and the original state $|\psi\rangle$ have the same component at every element of the position basis, so they are the same state.

Why does this notation still feel slightly “improper”?

There is a mathematical reason for this feeling. Strictly speaking, an exact position eigenstate $|x\rangle$ is not an ordinary, normalisable Hilbert-space vector, because

$$\langle x|x'\rangle=\delta(x-x').$$

$|x\rangle$ is a generalised eigenvector with distributional properties. Therefore,

$$\int dx\,|x\rangle\langle x|=I$$

is a generalised completeness relation. A more rigorous treatment requires a rigged Hilbert space or the spectral theorem.

Likewise, the individual $|x\rangle\langle x|$ is not a well-behaved ordinary operator in the way a projector in a discrete basis is. The proper expression for projection onto a position interval $\Delta$ is

$$P_\Delta = \int_\Delta dx\,|x\rangle\langle x|.$$

Thus, Dìguā’s feeling that “having a ket inside an integral is strange” does not mean that he has failed to understand the notation. Here, Dirac notation really does conceal technical details involving distributions.

Is this integral a Fourier transform?

The expression

$$|\psi\rangle=\int dx\,\psi(x)|x\rangle$$

is not itself a Fourier transform. It reconstructs the abstract quantum state in the position basis.

If we project from the left onto the momentum basis, we obtain

$$\begin{aligned} \widetilde\psi(p) &=\langle p|\psi\rangle\\ &=\int dx\,\langle p|x\rangle\psi(x)\\ &=\frac{1}{\sqrt{2\pi\hbar}} \int dx\,e^{-ipx/\hbar}\psi(x). \end{aligned}$$

This is the Fourier transform from the position wavefunction to the momentum wavefunction, with $\langle p|x\rangle$ as the transformation kernel.

The distinction between the two operations is therefore:

• $\int dx\,\psi(x)|x\rangle$ reconstructs the abstract ket from its position components;
• $\int dx\,\langle p|x\rangle\psi(x)$ converts position components into momentum components.

Wavefunctions under unitary transformations

In Dirac notation, the position wavefunction is written as

$$\psi(x)=\langle x|\psi\rangle.$$

If the same unitary transformation $U$ acts simultaneously on both the quantum state and its corresponding basis,

$$|\psi'\rangle=U|\psi\rangle, \qquad |x'\rangle=U|x\rangle,$$

then

$$\begin{aligned} \psi'(x') &=\langle x'|\psi'\rangle\\ &=\langle x|U^\dagger U|\psi\rangle\\ &=\langle x|\psi\rangle\\ &=\psi(x). \end{aligned}$$

The point is not that the graph of the wavefunction has not moved, but that a unitary transformation preserves inner products: the probability amplitude is the same at corresponding points before and after the translation.

If the original position basis $|x\rangle$ is held fixed and only the quantum state is translated, then a translation to the right by $a$ gives

$$\psi'(x)=\psi(x-a).$$

The following two comparisons are therefore not contradictory:

• $\psi'(x)=\psi(x-a)$ compares the same coordinate label $x$;
• $\psi'(x')=\psi(x)$ compares corresponding points before and after the translation, where $x'=x+a$.

The final understanding

Dìguā can compress the whole matter into three sentences:

1. For a fixed $x$, $\langle x|\psi\rangle$ is a complex number; as $x$ varies, it is the position wavefunction.
2. $|x\rangle$ is a continuous-basis ket, and $\psi(x)$ is its complex weight; integrating $\psi(x)|x\rangle$ continuously adds together vector components.
3. A ket is an abstract quantum state; a wavefunction is the coordinate representation of that ket in a chosen basis.