0

I am solving problem 14.4. (a) of Schwartz's Quantum Field Theory and the Standard Model. It is related to the simple harmonic oscillator in quantum mechanics. It asks the eigenstate of the position operator $\hat x$ in terms of creation operators acting on the vacuum. In other words, I need to find $f_x(a^\dagger)$ such that $\hat x( f_x(a^\dagger)|0\rangle)=x(a^\dagger)|0\rangle)$. My question is not related to how to solve it.

The above problem asks how to describe the position state in terms of the linear combination of energy eigenstates. As I understand, the position eigenstates do not form a basis of Hilbert space in the usual sense, since they are not an element of Hilbert space. (Although I don't know in detail, there is the Rigged Hilbert space that describes the concept of position eigenstates.) Instead, energy eigenstates form the basis of Hilbert space. All states in the Hilbert space can be described as the linear combination of the energy eigenstates. Then, can the position eigenstate can be described by a linear combination of energy eigenstates although they are not the element of Hilbert space? Since the position eigenstates are not an element of Hilbert space, I am not sure whether it can be described as the linear combination of energy eigenstates. If not, is what I need to find the best approximation of position eigenstate in terms of energy eigenstates?

Summary: Can we describe a position eigenstate as the linear combination of energy eigenstates although it is not an element of Hilbert space? If not, are we practically computing the best approximation?

Jaeok Yi
  • 194

1 Answers1

2

The equation $$|x\rangle = \sum\limits_{n=0}^\infty |n\rangle \underbrace{\langle n |x\rangle}_{\phi_n(x)^\ast} $$ holds for any orthononormal basis $|0\rangle, \, |1\rangle, \, |2\rangle,\ldots$ as a consequence of the completeness relation $$\mathbb{I}= \sum\limits_{n=0}^\infty |n \rangle \langle n |.$$

In particular, $$\delta(y-x)=\langle y | x \rangle =\sum\limits_{n=0}^\infty \langle y |n \rangle \langle n |x\rangle=\sum\limits_{n=0}^\infty \phi_n(y) \phi_n(x)^\ast, $$ i.e. the sequence of functions $\delta_N (y,x)=\sum\limits_{n=0}^N \phi_n(y) \phi_n(x)^\ast$ converges to $\delta(y-x)$ in the sense of distributions in the limit $N\to \infty$.

So the position operator "eigenstate" localized at $x_0$ (a ket obeying the relation $\hat{x} |x_0\rangle = x_0 |x_0\rangle$ in distributional sense) can be expressed as

$$ |x_0\rangle = \sum_{n=0}^\infty \phi_{n}(x_0)^{*}|n\rangle. $$

Ján Lalinský
  • 44,745
Hyperon
  • 10,007