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.


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.


@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 $x''$.

  • 1
    $\begingroup$ Have you thought about projecting the position eigenstate onto the QHO number basis? $\endgroup$ Nov 15, 2016 at 20:33
  • $\begingroup$ @alfred-centauri, that was not my original intent, but I will try and see if it helps. $\endgroup$
    – mavzolej
    Nov 15, 2016 at 22:11
  • $\begingroup$ I agree that my condition does not specify the operator uniquely. My natural suggestion is to fix it by requiring $\hat{\mathcal{O}}=\hat{\mathcal{O}}^\dagger$. $\endgroup$
    – mavzolej
    Nov 15, 2016 at 22:15
  • $\begingroup$ Good point. Thinking now of some reasonable requirement. In QFT, the states of definite position/momentum are arising as the eigenstates of the corresponding operators, but in QM we don't have an analogy of this. $\endgroup$
    – mavzolej
    Nov 15, 2016 at 22:28

1 Answer 1


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 of x and 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 entire Fock 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 coefficients c(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 awesome standard 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 $i\sqrt{2} \hat{p}=a-a^\dagger$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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