I'm wondering if somebody could help me to finish a simple calculation. Let me first provide motivation for the question below: I would like to write a QM amplitude in the 'QFT-style', as

$$\langle \phi_2 (t_2) \phi_1(t_1) \rangle = \langle 0 | \hat{\phi}_2(0) \operatorname{e}^{-i H (t_2 – t_1)} \hat{\phi}_1 | 0 \rangle $$

Here I assumed that $\hat{\phi}_{1,2}$ are Hermitean, which in general does not necessary have to be true.

OK, now the question: I would like to know which operator creates the position eigenstates in QM. Here's my attempt to find its matrix elements in the position representation:

\begin{gather}

\hat{\mathcal{O}} |0\rangle = |x\rangle \qquad \text{(by definition)}\\

\langle x'| \hat{\mathcal{O}} |0\rangle = \langle x'|x\rangle = \delta(x-x')\\

\int \operatorname{d}x''\langle x'| \hat{\mathcal{O}} |x''\rangle \langle x''|0\rangle =\delta(x-x')\\

\end{gather}

Here comes the dependency on Hamiltonian – we can proceed no further without specifying the particular form of $H$, and so of $\langle x'' | 0 \rangle$. Of course, for simplicity we begin with the harmonic oscillator.

$$\langle x | 0 \rangle = \pi^{-1/4} \operatorname{e}^{-x^2/2}$$

Where I chose $\dfrac{m \omega}{\hbar} = 1$ for convenience. Hence, one gets:

$$\pi^{-1/4}\int \operatorname{d}x''\langle x'| \hat{\mathcal{O}} |x''\rangle \operatorname{e}^{-(x'')^2/2} =\delta(x-x')\\

$$

Can anyone help me to find $\langle x'| \hat{\mathcal{O}} |x''\rangle $? For sure, I don't expect $\hat{\mathcal{O}}$ to be a 'nice' operator, since it does not define a properly normalised state. I expect it to be some kind of distribution, whose kernel I would like to know.

**UPDATE 1**

Few clarifications about the statement of the question:

- The question is about a $1$-d QM problem, with a single degree of freedom. In principle, it is NOT related to QFT.
- Starting from the line where I chose the particular form of Hamiltonian (QHO) and of the ground state, the definition of vacuum is $\hat{a} |0\rangle = 0$, where $\hat{a}$ is the usual QHO annihilation operator.

**UPDATE 2**

@AccidentalFourierTransform has suggested the following:

$$\langle x' |\hat{\mathcal{O}} | x''\rangle= \pi^{1/4} \operatorname{e}^{+(x'')^2/2} \operatorname{e}^{i x'' (x-x')} / (2\pi) $$

Being substituted into the previous line, this clearly leads to the correct result. Now my question is whether such definition can indeed define a (nice at least in certain sense) operator. What confuses me is the factor $\operatorname{e}^{+(x'')^2/2}$ together with the fact that we are integrating over $x''$.

## Best Answer

Well, you asked for a start, not a nice operator—but I suspect some coherent state/displacement operator review or paper has it nicely. I hasten to frame the issue, and you might choose to carry it out more efficiently.

You are seeking a representation-changing operator $\hat{\mathcal{O}}(x)$ taking you from Fock states (eigenstates of the number/energy operator) to eigenstates of the position operator $\hat{X}$ with eigenvalue

x. The operator will be a function ofxand creation operators.I use the conventions of Sakurai & Napolitano, QM, (2.3.21), $$ |n\rangle= \frac{a^{\dagger ~n}}{\sqrt{n!}} |0\rangle $$ and, significantly, since you are using the

entireFock space,(2.3.32), $$ \langle n| x \rangle= \frac{1}{\pi^{1/4}\sqrt{2^n~n!}}~ (x-\partial_x)^n~ e^{-x^2/2} ~. $$You then posit your basis change, $$ |x\rangle= \hat{\mathcal{O}}(x) |0\rangle= \sum_0^\infty c_n (x) |n\rangle, $$ and use the above relation, as you did for

n=0 to determine the coefficientsc(x).The good news is that the answer, $$ |x\rangle= \sum_0^\infty \frac{1}{\pi^{1/4}} ~ \frac {(x-\partial_x)^n a^{\dagger ~n}}{2^{n/2} n!} e^{-x^2/2 }|0\rangle $$ exponentiates, as often, $$ \hat{\mathcal{O}}(x)= \frac{1}{\pi^{1/4}} e^{(x-\partial_x)~ a^\dagger /\sqrt{2} } ~ e^{-x^2/2 }, $$ and the exponential of operators is straightforward to split by use of the degenerate CBH identity, since $[x,-\partial] =1$, so $\exp(-a^{\dagger ~2}/4) \exp(xa^\dagger/\sqrt{2})\exp(-a^\dagger \partial_x /\sqrt{2})$. And hence you get the exponential of the gradient ∂ on the right, thus a translation operator: it translates the exponent of the

x-Gaussian to $-(x-a^\dagger/\sqrt{2})^2/2$.So, finally, $$ \bbox[yellow,5px]{\hat{\mathcal{O}}(x)= \frac{1}{\pi^{1/4}} e^{-x^2/2} e^{\sqrt{2} xa^\dagger} e^{-a^{\dagger ~2}/2} =\frac{e^{x^2/2}}{\pi^{1/4}} e^{-(a^\dagger-\sqrt{2} x)^2/2} } . $$ As a lark, you may check that, since $a$ acts as a derivation w.r.t. $a^\dagger$, $$ \frac{(a+a^\dagger)}{\sqrt 2} ~\hat{\mathcal{O}}(x) |0\rangle= x~\hat{\mathcal{O}}(x) |0\rangle ~, $$ so an eigenstate of $\hat{X}$, indeed.

I have not checked the delta-orthogonality of $\langle y|x\rangle$ states here.

Edit: I found out today that this is but Prob. 14.4.a) of M Schwartz's book on QFT. In any case, working out the <x|p> in 312004 yields the plane wave, and inserting complete momentum states and integrating over them trivially produces the δ-function normalization $\langle y| x\rangle\propto \delta(x-y)$ sought.Edit II: This is, in fact, reducible to the celebrated Segal-Bargmann transform, Def 2 & Corollary 1, should you wish to pursue it more formally and stick labels on it.Edit III: I am being repeatedly asked about the connection of this oscillator vacuum $|0\rangle$ to the translationally invariant vacuum of Dirac's book, the awesomestandard ket, $~\lim_{p\to 0} |p\rangle\equiv |\varpi\rangle ~$, for which $~\hat{p} |\varpi \rangle =0~$ and $~\langle x | \varpi\rangle=1/\sqrt{2\pi \hbar}~$, as well as $~\langle x | \hat{x}|\varpi\rangle=x/\sqrt{2\pi \hbar}$.The relation is actually $$|\varpi\rangle= \exp (a^\dagger a^\dagger /2) |0\rangle {1\over (\pi \hbar)^{1/4}}~,$$ implicit in the above, but evident in this obscurely placed answer. By inspection, this state is in the kernel of $\sqrt{2} \hat{p}=a-a^\dagger$.