7
$\begingroup$

I'm struggling to see how $\langle x|\Psi\rangle =\Psi(x)$. I have read a few previous bra ket questions in here but still not clear. Any good book for understand the bra-ket notation in more rigorous way.

$\endgroup$
  • $\begingroup$ near duplicate of physics.stackexchange.com/questions/364208/… $\endgroup$ – ZeroTheHero Jul 3 at 14:00
  • 1
    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Jul 4 at 1:26
12
$\begingroup$

Consider the case of a vector space of countable dimension, with some orthonormal set of basis kets $\left\{\vert\mathbf{e}_i\rangle\right\}$. The orthonormality condition is stated as $\langle \mathbf{e}_i \vert \mathbf{e}_j \rangle = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta. We can then expand any vector in this basis, $$\vert \psi \rangle = \sum_i \psi_i \vert \mathbf{e}_i \rangle, $$ where the $\psi_i$ are the components of $\vert \psi \rangle$, i.e. they are the projections of $\vert \psi \rangle$ along the basis vectors $\vert \mathbf{e}_i \rangle$, which we can state more technically by noting that the identity matrix can be written as $$I = \sum_i \vert \mathbf{e}_i \rangle \langle \mathbf{e}_i \vert,$$ in which case $$\vert \psi \rangle = I \vert \psi \rangle = \sum_i \vert \mathbf{e}_i \rangle \langle \mathbf{e}_i \vert \psi \rangle,$$ i.e. $$\psi_i = \langle \mathbf{e}_i \vert \psi \rangle.$$

Now we generalize this to an uncountable basis. For example, we define the position basis as the set $\left\{\vert \mathbf{x} \rangle \,\vert\, \mathbf{x} \in \mathbb{R}^3 \right\}$. Now the orthonormality condition is slightly modified (for technical details you can read about "rigged" Hilbert spaces), $\langle \mathbf{x} \vert \mathbf{x}' \rangle = \delta^{3}(\mathbf{x} - \mathbf{x}')$ where $\delta^3$ is the three-dimensional Dirac delta. Then we can expand the identity operator (it is no longer a matrix when the basis is uncountable) as $$I = \int_{\mathbb{R}^3} d^3\mathbf{x}\, \vert \mathbf{x} \rangle \langle \mathbf{x} \vert.$$ Then, as before, we expand a vector $\vert \psi \rangle$ as $$\vert \psi \rangle = I\vert \psi \rangle = \int d^3\mathbf{x} \, \vert \mathbf{x} \rangle \langle \mathbf{x} \vert\psi \rangle \equiv \int d^3\mathbf{x} \, \psi(\mathbf{x})\,\vert \mathbf{x} \rangle,$$ so the wavefunction $\psi(\mathbf{x})$ is simply the components of the vector $\vert \psi \rangle$ along the basis vectors $\vert \mathbf{x} \rangle$, just like in the countable case. The only difference is that now $\mathbf{x}$ labels the basis vectors instead of the discrete index $i$. $$\psi_i \equiv \langle \mathbf{e}_i \vert \psi \rangle \leftrightarrow \psi(\mathbf{x}) \equiv \langle \mathbf{x} \vert \psi \rangle \quad\,\,$$ $$\vert\psi\rangle = \sum_i \psi_i \vert \mathbf{e}_i \rangle \leftrightarrow \vert \psi \rangle = \int d^3\mathbf{x}\, \psi(\mathbf{x}) \vert \mathbf{x} \rangle$$ For a pedagogical introduction, I recommend the notes found on this page, in particular "Block 1: Mathematical Foundations".

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

You can define $|\Psi\rangle$ as: \begin{equation} |\Psi\rangle=\int \Psi(y) |y\rangle d^3y \end{equation} With $\{|y\rangle \ \,|\,y \in \mathbb{R}^3\}$ the basis of the hilbert space $H$ of positions. Since $\langle x|$ is the linear form such that $\langle x|(|y\rangle)\equiv \langle x|y\rangle=\delta^{(3)}(x-y)$ we have : \begin{equation} \langle x|\Psi\rangle=\int \Psi(y) \delta^{(3)}(x-y) d^3y=\Psi(x) \end{equation}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.