#### [SOLVED] What exactly is $\langle x |$?

By sangstar

It's a linear functional, but what exactly does it do? It maps a wavefunction $$|\psi \rangle$$ to an element of $$\mathbb C$$, but what.. exactly does that mean? I know heuristically it maps $$\psi$$ to its position basis, but from its description, how does it do this? #### @yuggib 2019-02-11 10:28:21

The object $$\langle x\rvert$$ has several mathematical properties, however one should take some care in dealing with it.

The usual physical "definition" is the following: given a wavefunction $$\lvert \psi\rangle\in \mathscr{H}$$ describing a (non-relativistic) quantum particle, then $$\langle x\vert \psi\rangle=\psi(x)$$ is the value of the wavefunction at $$x$$, and $$x\mapsto \lvert \psi(x)\rvert^2$$ is the probability density for the particle's position.

The above "definition", however, is not mathematically accurate. There is one foremost problem: there is no set of everywhere defined functions from $$\mathbb{R}^d$$ to $$\mathbb{C}$$ forming a complete Hilbert space. If one starts with smooth functions, such as $$C_0^\infty(\mathbb{R}^d)$$ or $$\mathscr{S}(\mathbb{R}^d)$$, and puts an inner product on it, the result space is not complete with respect to the topology induced by the inner product. Taking the completion, one usually gets a Lebesgue (or Sobolev) space, with the typical exaxmple being $$L^2(\mathbb{R}^d)$$. Such space consists of equivalence classes of almost everywhere defined functions, and these objects may not be defined in some points of $$\mathbb{R}^d$$ (because, roughly speaking, a point is not measurable).

Therefore, the functional $$\langle x\rvert$$ can be defined rigorously (by the above definition) only on a subset of the Hilbert space of wavefunctions (albeit a dense one). From now on, for simplicity let us assume the space of wavefunctions to be $$L^2$$. One choice for the dense subset of definition of the functional could be $$C(\mathbb{R}^d)\cap L^2(\mathbb{R}^d)$$, however the most natural one (for many interesting reasons that would be too long to discuss here) is the Schwartz space of rapidly decreasing functions $$\mathscr{S}(\mathbb{R}^d)$$. The functional $$\langle x\rvert$$ is, with respect to the usual topology of the Schwartz space, continuous for all $$x\in\mathbb{R}^d$$, and linear as well (this is easy to check). It is therefore a so-called distribution, belonging to the space $$\mathscr{S}'(\mathbb{R}^d)$$.

Physically, the set of distributions $$\{\langle x\rvert\}_{x\in\mathbb{R}^d}$$ is thought of as a "position basis", because heuristically these distributions behave as eigenvectors for the position operator. This is, however, only true in a weak sense: given any Schwartz wavefunction $$\lvert\psi\rangle$$, and the position operator $$\hat{x}$$, then $$\hat{x}\lvert\psi\rangle$$ is still a Schwartz function, whose value at $$x$$ is given by $$\langle x\rvert\hat{x}\lvert\psi\rangle=x\psi(x)$$. Therefore, since $$\hat{x}$$ is self-adjoint, it can be seen as acting on the left on the distribution in the following way: $$\langle x\rvert\hat{x}=x\langle x\rvert$$ (a sort of eigenvalue equation). Nonetheless, let me remark that $$\langle x\rvert\notin L^2$$, and therefore it is not a true eigenvector of the position operator (only true wavefunctions can be eigenvectors). It is a so-called generalized eigenvector. There is a (bit complicated, and in my opinion not so useful for quantum mechanics) reasonably well-developed mathematical theory of generalized eigenvectors for operators with continuous spectrum, that was developed mostly by the russian school (Gel'fand, Shilov). #### @Andrew Steane 2019-02-11 10:49:05

I think, for learning QM, it's best not to worry about what $$\langle \psi |$$ is. Rather, focus on $$|\psi\rangle$$---a state vector, and $$\langle \phi | \psi\rangle$$---an innner product between one state vector and another. If you are still bugged by what $$\langle \psi |$$ is, then say to yourself "it's the Hermitian conjugate of $$| \psi\rangle$$".

The combination $$\langle B | A\rangle$$ has many useful interpretations. Rather than saying merely the true but rather abstract "it maps from a state to a complex number", you can keep in the forefront of your mind all of the following:

$$\langle B | A\rangle$$ can be thought of as "the overlap of $$|A\rangle$$ and $$| B\rangle$$" or "the degree to which $$|A\rangle$$ is like $$|B\rangle$$" or "the coefficient of $$|B\rangle$$ if you write $$|A\rangle$$ as a superposition of $$|B\rangle$$ and other orthogonal things" or "that which, when mod-squared, gives the probability that a measurement of a system in state $$|A\rangle$$ will yield $$|B\rangle$$." #### @Žarko Tomičić 2019-02-10 01:05:38

For any vector space you can define a set of linear functionals that map the elements of this vector space into numbers.Nobody, i belive, cares how. It just does that. By this definition, space of linear functionals constitutes a vector space in its own right. For this vector space we can define such functionals which, when acting on the elements of the original vector space basis give just one or zero. These are called dual basis vectors of the dual space, dual space being just this space of linear functionals. With these functionals defined in this way we can write dual vectors in this dual basis and act on vectors. We can have inner products and such things.It is mathematicaly more meaningfull to define inner or scalar product in this way.