# Quantum state in continuous basis [duplicate]

If I have an arbitrary state $$|\psi\rangle$$ and want to represent it in a continuous basis, for example the position basis in $$x$$-direction, I will get $$|\psi\rangle = \int dx\, \langle x|\psi\rangle|x\rangle,$$ with $$\langle x|\psi\rangle$$ being the wavefunction (that is, the amplitude for the state $$|\psi\rangle$$ to be in an eigenstate of $$\hat{x}$$).

In the discrete case, this would be a sum with all the different probabilities of the state $$|\psi\rangle$$ being in one of the eigenstates of $$\hat{x}$$, which can be represented as a vector with each entry has the probability of the state $$|\psi\rangle$$ being in that eigenstate $$|x_i\rangle$$.

What is the equivalent representation with a continuous basis? How I can “visualize” a state when it is represented in continuous bases. If I represent it in a discrete base, I can “visualize” it as a vector. But how can I do it when I am in the continuous space?

• Does $L^2(\mathbf{R})$ tell you anything? The elements of this "function space" can also be interpreted as "vectors". Commented Jan 4, 2023 at 12:32
• Unfortunately not, I am an electrical engineer, just starting to investigate more quantum mechanics. But I already realized that I am lacking in general a bit of mathematical formalism. Commented Jan 4, 2023 at 12:35
• You may visualize an ever increasing number of discrete components. Sums turn to integrals. Commented Jan 4, 2023 at 12:53
• Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis.(ADDENDUM) Commented Jan 4, 2023 at 12:57
• You seem to be asking how to visualize a function of a single variable $x$. To visualize such a function you plot it. Plot the function $\psi(x)$. If $\psi(x)$ is real and $x$ is one-dimensional you can do this with two axes, just like you normally visualize functions.
– hft
Commented Jan 5, 2023 at 1:13

In the discrete case with dimension $$N$$, I have a basis $$\{ \left| x \right\rangle\}$$ that is orthonormal and complete, meaning that $$\left\langle x \middle| y \right\rangle \, = \, \delta^{\,}_{x,y} ~~~\text{and}~~~\sum\limits_{x=1}^N \, \left| x \middle\rangle \hspace{-0.4mm} \middle\langle x \right| \, = \, \mathbb{1} \, ,~$$ with $$\mathbb{1}$$ the identity and $$\delta^{\,}_{a,b}$$ the Kronecker delta function (it's 1 if $$a=b$$ and 0 if $$a \neq b$$). I can write any state using this basis as $$\left| \psi \right\rangle \, = \, \mathbb{1} \, \left| \psi \right\rangle \, = \, \sum\limits_{x=1}^N \, \left| x \middle\rangle \hspace{-0.4mm} \middle\langle x \middle| \psi \right\rangle \, = \, \sum\limits_{x=1}^N \, \psi^{\,}_x \, \left| x \right\rangle \, , ~$$ where $$\psi^{\,}_x \, \equiv \, \left\langle x \middle| \psi \right\rangle$$ is the overlap of the state $$\left| \psi \right\rangle$$ with the basis state $$\left| x \right\rangle$$.
In the continuous case with infinite dimension, I simply replace the sums above with integrals, as noted by @cosmas-zachos in a comment. I once again have a basis $$\{ \left| x \right\rangle\}$$ that is orthonormal and complete, meaning that $$\left\langle x \middle| y \right\rangle \, = \, \delta \left( x - y \right) ~~~\text{and}~~~\int \, {\rm d}x \, \left| x \middle\rangle \hspace{-0.4mm} \middle\langle x \right| \, = \, \mathbb{1} \, ,~$$ with $$\mathbb{1}$$ the identity as usual, and $$\delta (z)$$ the Dirac delta function (which is infinite at $$z=0$$ and zero elsewhere). I can write any state using this basis as $$\left| \psi \right\rangle \, = \, \mathbb{1} \, \left| \psi \right\rangle \, = \, \int \, {\rm d}x \left| x \middle\rangle \hspace{-0.4mm} \middle\langle x \middle| \psi \right\rangle \, = \, \int\, {\rm d}x \, \psi (x) \, \left| x \right\rangle \, , ~$$ where $$\psi (x) \, \equiv \, \left\langle x \middle| \psi \right\rangle$$ is the overlap of the state $$\left| \psi \right\rangle$$ with the basis state $$\left| x \right\rangle$$, known as the wavefunction when $$x$$ corresponds to position states.
Compared to the familiar inner product in finite dimensions, $$\vec{v} \cdot \vec{w} \, = \, \left\langle v \middle| w \right\rangle\, =\, \sum_{x=1}^N \, \left\langle v \middle| x \middle\rangle \hspace{-0.4mm} \middle\langle x \middle| w \right\rangle \, = \, \sum_{x=1}^N \, v^{\ast}_x \, w^{\vphantom{*}}_x \, ,~$$ in the infinite-dimensional case the inner product is instead $$\vec{v} \cdot \vec{w} \, = \, \left\langle v \middle| w \right\rangle \, = \, \int \, {\rm d}x \, \left\langle v \middle| x \middle\rangle \hspace{-0.4mm} \middle\langle x \middle| w \right\rangle \, = \, \int \, {\rm d}x \, v^{\ast} (x) \, w (x) \, .~$$
As far as "visualization", just think of it the same way as the discrete case with dimension $$N$$ and imagine that $$N \to \infty$$. See also these lecture notes.