Confusion about ket states and bra with position

Question

I am very confused about the bra-ket notation of states and the fact that $$\psi(x) = ⟨x|\psi⟩$$ and $$⟨x|x'⟩ = \delta(x-x')$$ are true. What does this mean? What is the ket $|x⟩$, is it just some kind of identity vector? And what even is a state $|\Psi⟩$. What does it look like, is it an infinite vector where $\Psi_n$ is just that wavefunction with the principle quantum number being $n$?.

E.g. the solution to the infinte square well is given by the wave function $$\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n \pi}{L} x\right),$$ if I am not mistaken. Then, can one say that $$ \psi_1(x) =\sqrt{\frac{2}{L}}\sin\left(\frac{1\pi}{L} x\right) $$ $$\psi_2(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{2 \pi}{L} x\right)$$ and etc.? And if so, how is that related to $⟨x|\psi⟩ = \psi(x)$?

Answer 1

As you say, $\psi(x)$ is a wavefunction, and that's a function.

Functions can be regarded as vectors, but not arrows, but rather "things that satisfy all requirements of a vecor space", i.e. they can be added up and scaled by complex numbers in the samee way arrow vectors do.

So wavefunctions, with the sum of functions and multiplication by complex scalars form a vector space.

There is also another operation called scalar product of two functions, defined as

$$f\cdot g = \int_{-\infty}^{+\infty} f^*(x)g(x) dx $$

And it can be proven that this integral-type operation satisfies all conditions for a scalar product.

So we could keep dealing with all this forever. But scientist discovered that they could make the most of all theorems they knew for vector spaces. So they decided to stop seeing wavefunctions as functions and start seeing them as vectors.

A scalar function like $\psi(\vec{r})$ is a function $\mathbb{R}^3 \ \rightarrow \mathbb{C}$, that assigns a complex number (or two real numbers) to each point in space.

We can regard them as "a collection of infinite complex numbers", each of them corresponds to one point in space.

So, take such collection and, instead of seeing it as a funcition, see it as a table of values. Each row corresponds to the $\psi$ at one point in space.

So we've got infinite rows with values. That's like a vector column. It has an infinite number of coordinates (infinite dimension), but that's not a problem.

So that's $\psi\rangle$, a vector whose infinite components are the values of $\psi(x)$.

Such infinite-dimensional vector space with scalar product is called "Hilbert's space", $\mathcal{H}$

Now, we define an operator of the dual space. This operator acts on vector and assigns them a complex number. $\mathcal{H}\rightarrow \mathbb{C}$.

One operator like this is the delta-function. We call it with a bra $\langle x|$. The scalar product of two vectors is the projection of one on another. In this case, $\langle x|$ projects $|\psi\rangle$ onto its "component", onto its "corresponding point". In terms of functions, it is

$\langle x_0|\psi\rangle = \int_{-\infty}^{+\infty} \delta(x-x_0)\cdot\psi(x) dx = \psi(x_0)$

So $\langle x|\psi\rangle=\psi(x)$, it gives the fnuction in x representation.

You can define other representations. Momentum representaton gives the components of the fourier-transform of $\psi$.

Answer 2

So the ket:

$$ |x\rangle $$

is an eigenstate of the position operator $\hat x$, so that:

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

which is confusing because of the proliferation of x's. To make it a little clearer, lets say $x=a$...no, how about $x=3$, very definite:

$$ \hat x|3\rangle = 3|3\rangle $$

What does that even mean? Well, we know what it means because we've solved the plane wave problem, and the wave function is, in the momentum representation:

$$ \langle p|3\rangle =\frac 1 {\sqrt{2\pi}} e^{-ip(3)} $$

(It might be clearer to think of momentum eigenstates in the position representation, but you asked about $|x\rangle$, which can be written analytically in the momentum representation--in an entirely complementary fashion with respect to momentum eigenstates, i.e. plane waves, as functions of $x$).

Recall:

$$ \hat x = -i\frac d{dp} $$

so that:

$$ \hat x|3\rangle = -i\frac d{dp}\frac 1 {\sqrt{2\pi}} e^{-ip(3)} = 3\frac 1 {\sqrt{2\pi}} e^{-ip(3)}=3|3\rangle$$

Now consider two kets in the $x$-basis, let's call them $|a\rangle$ and $|b\rangle$. Then:

$$\langle a|b\rangle=\frac 1{2\pi}\int{e^{ipa}e^{-ipb}dp} = \frac 1{2\pi}\int{e^{ip(a-b)}dp} = \delta(a-b) $$

I know, it seems a bit circular at first, so consider it a useful notation. Once you've solved the problem, you can label the solutions with bras and kets, which are a handy way to compute thing, and have a formal mathematical basis backing them up.

In the harmonic oscillator case, you can label the energy eigenstates with energy $E_n$ as:

$$ |n\rangle $$

and of course they satisfy:

$$ H|n\rangle = E_n|n\rangle $$

along with:

$$ \langle n|m \rangle = \delta_{nm}$$

and that in the position representation:

$$ \langle x|n\rangle=\psi_n(x)=\sqrt{\frac 2 L}\sin{\frac{n\pi}L x} $$

You can then express $H$ in the position representation, and apply to $\psi_n(x)$, and verify $H|x\rangle=E_n|x\rangle$, but it's a lot easier just to write that with kets.

Answer 3

Let me do a comparison with the finite dimensional case. A vector $v$ is an abstract entity belonging to a finite dimensional Hilbert space $\mathcal{H}$. Now in order to make actual computations with it, we usually handle with its components $v \to (v_1,v_2,\ldots)$. Once we choose an orthonormal basis the components are just the scalar products with the basis vectors $v_i = \langle v| e_i\rangle$. They are thus numbers because the scalar product provides a map $\mathcal{H}\times \mathcal{H} \to \mathbb{C}$.

In the infinite dimensional case the label $i$ of $e_i$ can assume infinitely many values. In this particular case it is actually continuous: $$ i \to x\,,\quad e_i \to |x\rangle\,. $$ And so the wave function $\psi(x)$ it simply the "$x$th" component of $|\psi\rangle$, $$ v_i \to \psi(x)\,,\quad\langle v | e_i\rangle \to \langle \psi | x\rangle\,. $$ As I said before this requires the basis $|x\rangle$ to be orthonormal and when $x$ is a continuous parameter the condition to be imposed is with the Dirac $\delta$. Moreover we must also impose that it is complete, namely $$ \int dx\,|x\rangle \langle x| = \mathbb{1}\,, $$ which is an operator equation. We can use this property to compute scalar products explicitly $$ \langle \psi|\chi\rangle = \langle \psi|\mathbb{1} |\chi\rangle = \int dx\, \langle\psi|x\rangle\langle x|\chi\rangle = \int dx \,\psi(x)\,\chi^*(x)\,. $$ In the same way as we would compute $\langle v | w\rangle = \sum_i v_i w^*_i$.

Answer 4

Its best to take Diracs notation as physicists generally do, which is with enough rigour to do physics with.

Although his notation is often identified with infinite dimensional hilbert spaces and the like, even there it is not a fully rigorous mathematical modelling of his notation; for example, the Dirac delta function, means we ought to consider distributions, and hence a rigged Hilbert space. For an example of an alternative axiomatisation, it's not generally well known, for example, that finite QM, often used in QI, is modelled by dagger compact closed categories.

I'd say concentrate on the physics rather than focusing on how to make the physics mathematically rigorous; this does not mean you should ignore mathematics, but it ought to know its place. You might also want to reflect on just how long it took to make the calculus mathematically rigorous (it's still ongoing even now!), and had we to wait until it was made physicists would not have made much progress on physics; I would expect a much longer time framework to make QFT rigorous...

Answer 5

Bras and kets correspond to states of the system and are regarded as vectors.So, a ket |x⟩ is a vector, a position vector. A ket |psi⟩ is also a vector but one which describes the state of the system. |x⟩ represents basis which we call position basis. If we now want to know what does our vector look like in the position basis we just do scalar (inner) product of this vector with the base vector |x⟩. This way we get our coefficients in this basis which should be known to you i guess from ordinary vectors (arrows). These coeficients depend on basis value x so we can write them as function psi(x). So quantum states of a system can be regarded as a vector space. In this space, functions psi(x) are just coefficients. But, these functions, as a solutions of some differential equation which is linear (eg Schrodinger) form a vector space themselves. Vector space of all of solutions of a diff. equation. Following example is easier because it is finite dimensional. Consider spin state of an electron. It can have two values when measured, +0.5 i -0.5. These two states can be considered as a basis. Now, we can prepare electrons in some combination of these states. How do we know that they are prepared in this way? We prepare them and then we measure the spin and we can get 50 % in one state and 50% in the other base state so we know that this state is prepared with fifty fifty probability. This probability is what we connect to the coeficients of the states in some basis.So we can say that spin state is actually, before measurement, in a state |s⟩= 0.7|+0.5⟩ + 0.7|-0.5⟩ where 0.7 is square root of 0.5. So, if we square the coefficients, we get probabilities or, in case of continuos variable like position, we get probability density.This machinery describes very well the nature of quantum world..