I'm curious whether I can find the overlap $\langle q | p \rangle$ knowing only the following:

  1. $|q\rangle$ is an eigenvector of an operator $Q$ with eigenvalue $q$.

  2. $|p\rangle$ is an eigenvector of an operator $P$ with eigenvalue $p$.

  3. $Q$ and $P$ are Hermitian.

  4. $[Q,P] = i \hbar$.

I'm asking because books and references I've looked in tend to assume that $Q$ is a differential operator when viewed in the $P$-basis, then show that it satisfies the commutation relation. (I haven't read one yet that proves that the form given for $Q$ is the only possible one.) I've heard, though, that we can work purely from the hypothesis of the commutation relation and prove the properties of $Q$ and $P$ from it.

  • 3
    $\begingroup$ Maybe this helps a bit? $\endgroup$
    – Fabian
    Oct 28, 2012 at 10:45

2 Answers 2


I) Here we will work in the Heisenberg picture with time-dependent self-adjoint operators $\hat{Q}(t)$ and $\hat{P}(t)$ that satisfy the canonical equal-time commutation relation

$$ [\hat{Q}(t),\hat{P}(t)]~=~ i\hbar~{\bf 1}, \tag{1}$$

and two complete sets of instantaneous eigenstates$^1$ $\mid q,t \rangle $ and $\mid p,t \rangle $, which satisfy

$$\begin{align} \hat{Q}(t)\mid q,t \rangle ~=~&q\mid q,t \rangle, \cr \hat{P}(t)\mid p,t \rangle ~=~&p\mid p,t \rangle ,\end{align}\tag{2} $$

$$ \begin{align}\langle q,t \mid q^{\prime},t \rangle~=~&\delta(q-q^{\prime}), \cr \langle p,t \mid p^{\prime},t \rangle~=~&\delta(p-p^{\prime}),\end{align}\tag{3}$$

$$\begin{align}\int_{\mathbb{R}} \!dq~ \mid q,t \rangle\langle q,t \mid~=~&{\bf 1}, \cr \int_{\mathbb{R}} \!dp~ \mid p,t \rangle\langle p,t \mid~=~&{\bf 1},\end{align} \tag{4}$$

II) We would like to give an argument$^2$ that the sought-for overlap is of the form

$$ \langle p,t \mid q,t \rangle~=~\frac{f(q,t)g(p,t)}{\sqrt{2\pi\hbar}}\exp\left(\frac{pq}{i\hbar}\right), \tag{5}$$ where $f$ and $g$ are two phase factors $|f|=|g|=1$.

In other words, we can define new instantaneous eigenstates

$$\begin{align} \mid q,t )~:=~&\frac{1}{f(q,t)} \mid q,t \rangle \cr\text{and}& \cr \mid p,t )~:=~&g(p,t) \mid p,t \rangle , \end{align}\tag{6}$$

such that the overlap takes a standard form

$$ ( p,t \mid q,t )~=~\frac{1}{\sqrt{2\pi\hbar}}\exp\left(\frac{pq}{i\hbar}\right). \tag{7}$$

The square roots in eqs. (5) and (7) are due to standard normalization factors in Fourier analysis.

III) Sketched proof: Since everything here will refer the same instant $t$, let us not write $t$ explicitly from now on. Define for convenience an anti-selfadjoint operator

$$ \hat{D}~:=~\frac{\hat{P}}{i\hbar}, \tag{8}$$

so that eq. (1) becomes

$$ [\hat{Q},\hat{D}]~=~ {\bf 1} , \tag{9} $$


$$ [\hat{Q},e^{a\hat{D}}]~=~ae^{a\hat{D}} , \tag{10} $$

where $a$ is a real number. It follows from (10) that the state $e^{a\hat{D}}\mid q \rangle $ must be proportional to $\mid q+a \rangle$, i.e. there exists a function $f(q,q+a)$ of two arguments such that

$$ e^{a\hat{D}}\mid q \rangle ~=~f(q,q+a)\mid q+a \rangle.\tag{11}$$

Since $e^{a\hat{D}}$ is an unitary operator, the function $f(q,q+a)$ must be a phase factor $|f(q,q+a)|=1$. Or one can use that

$$\begin{align}\delta(q-q^{\prime})~\stackrel{(3)}{=}~& \langle q \mid e^{-a\hat{D}}e^{a\hat{D}}\mid q^{\prime} \rangle \cr ~\stackrel{(11)}{=}~&\overline{f(q,q+a)}f(q^{\prime},q^{\prime}+a) \langle q+a \mid q^{\prime}+a \rangle \cr ~\stackrel{(3)}{=}~& |f(q,q+a)|^2\delta(q-q^{\prime}). \end{align}\tag{12}$$

From the fact that

$$ e^{a\hat{D}} e^{b\hat{D}}~=~ e^{(a+b)\hat{D}},\tag{13}$$

we deduce from repeated use of eq. (11) that

$$ f(q,q+b)f(q+b,q+a+b)~=~f(q,q+a+b). \tag{14}$$

Setting $q=0$ in eq. (14) yields

$$\begin{align} f(b,a+b)~=~&\frac{f(0,a+b)}{f(0,b)} \cr \Updownarrow~& \cr f(q,q+a)~=~&\frac{f(0,q+a)}{f(0,q)}. \end{align}\tag{15}$$

Hence let us define

$$ \mid q ) ~:=~f(0,q) \mid q \rangle , \tag{16}$$

so that eq. (11) becomes

$$ e^{a\hat{D}}\mid q ) ~\stackrel{(11,15,16)}{=}~\mid q+a ).\tag{17}$$

In other words, we may identify$^3$ the operator $\hat{D}$ with the differentiation operator $\frac{\partial}{\partial q}$.

$$\begin{align} \exp\left(\frac{ap}{i\hbar}\right)\langle p \mid q) ~\stackrel{(8)}{=}~& \langle p \mid e^{a\hat{D}}\mid q )\cr ~\stackrel{(17)}{=}~& \langle p \mid q+a ),\end{align}\tag{18}$$

or in the limit $a\to 0$,

$$\begin{align} \frac{p}{i\hbar}\langle p \mid q) ~=~& \langle p \mid \hat{D}\mid q )\cr ~=~& \langle p \mid \frac{\partial}{\partial q} \mid q )\cr ~=~&\frac{\partial}{\partial q} \langle p \mid q ).\end{align}\tag{19}$$

From the ODE (19) in $q$, we deduce that

$$ \langle p \mid q) ~=~g(p)\exp\left(\frac{pq}{i\hbar}\right), \tag{20}$$

where $g(p)$ is an integration constant that can depend on the parameter $p$. It is not hard to see that

$$ |g(p)|~=~\frac{1}{\sqrt{2\pi\hbar}}.\tag{21}$$

Use e.g. that

$$\begin{align} \delta(p-p^{\prime})~\stackrel{(3,4)}{=}~& \int_{\mathbb{R}} \!dq~ \langle p \mid q)(q \mid p^{\prime} \rangle\cr ~\stackrel{(20)}{=}~&\overline{g(p)}g(p^{\prime}) \int_{\mathbb{R}} \!dq~\exp\left(\frac{(p-p^{\prime})q}{i\hbar}\right)\cr ~=~~&2\pi\hbar|g(p)|^2\delta(p-p^{\prime}). \end{align}\tag{22}$$


$^1$ Instantaneous eigenstates are often introduced in textbooks of quantum mechanics to derive the path-integral formalism from the operator formalism in the simplest cases, see e.g. J.J. Sakurai, Modern Quantum Mechanics, Section 2.5. Note that the instantaneous eigenstates $\mid q,t \rangle $ and $\mid p,t \rangle $ are time-independent states (as they should be in the Heisenberg picture).

$^2$ Here we will only work at the physical level of rigor, ignoring various mathematical subtleties connected with unbounded operators. Also we ignore any topological aspects of the canonical phase space, such as, e.g., if the position $q$ would be a periodic variable.

$^3$ Note that the $q$-representation of the momentum operator $\hat{P}=i\hbar\frac{\partial}{\partial q}$ on kets has the opposite sign of what it has on bras and wavefunctions, cf. e.g. my Phys.SE answer here.

  • 1
    $\begingroup$ Standard conventions for Schrödinger & Heisenberg pictures for a Schrödinger Hamiltonian $\hat{H}_S$ with no explicit time dependence: $\quad$ TDSE: $\quad \hat{H}_S|\psi(t)\rangle_S ~=~ i\hbar d_t|\psi(t)\rangle_S.$ $\quad |\psi(t)\rangle_S ~=~ \hat{U}(t)|\psi\rangle_H.$ $\quad \hat{U}(t)~:=~\exp\left(-\frac{i}{\hbar}\hat{H}_St\right).$ $\quad \hat{A}_H(t)~=~\hat{U}(t)^{-1} \hat{A}_S\hat{U}(t).$ $\quad$ HEOM: $\quad d_t\hat{A}_H(t)~=~\frac{i}{\hbar} [\hat{H}_H,\hat{A}_H(t)] +(\partial_t\hat{A}_S)_H.$ $\quad \hat{H}_H=\hat{H}_S.$ $\endgroup$
    – Qmechanic
    Oct 29, 2012 at 0:15
  • 1
    $\begingroup$ Generalization to a Schrödinger Hamiltonian $\hat{H}_S$ with explicit time dependence: $$\quad \hat{U}(t)~:=~\left\{\begin{array}{rcl} T\exp\left[-\frac{i}{\hbar}\int_0^t\! dt^{\prime}~\hat{H}_S(t^{\prime})\right] &\text{for}& t ~\geq~0, \cr AT\exp\left[-\frac{i}{\hbar}\int_0^t\! dt^{\prime}~\hat{H}_S(t^{\prime})\right] &\text{for}& t ~\leq~0. \end{array}\right.$$ $\quad i\hbar d_t\hat{U}(t) ~=~\hat{H}_S(t)\hat{U}(t).$ $\quad$ HEOM: $\quad d_t\hat{A}_H(t)~=~\frac{i}{\hbar}[\hat{H}_H(t),\hat{A}_H(t)] +(\partial_t\hat{A}_S)_H.$ $\endgroup$
    – Qmechanic
    Oct 29, 2012 at 0:22
  • $\begingroup$ The explicit time dependence in $(\partial_t\hat{A}_S)_H$ stems from the explicit time dependence in $\hat{A}_S$; not from the explicit time-dependence in $\hat{H}_S$. $\quad \hat{A}_H(t)~=~\hat{U}(t)^{-1} \hat{A}_S\hat{U}(t).$ So $\quad \hat{A}_H(t_2)~=~\hat{U}(t_2)^{-1}\hat{U}(t_1)\hat{A}_H(t_1)\hat{U}(t_1)^{-1}\hat{U}(t_2)$, which can generally not be written in terms of $\hat{U}(t_2,t_1)=\hat{U}(t_2)\hat{U}(t_1)^{-1}$ or, say, $\hat{U}(-t_2,-t_1).$ $\endgroup$
    – Qmechanic
    Nov 22, 2013 at 23:22
  • $\begingroup$ Heisenberg instantaneous eigenstate: $\quad \hat{A}_H(t)|A,t\rangle_H ~=~A|A,t\rangle_H.$ $\quad |A,t\rangle_H~:=~\hat{U}(t)^{-1}|A,0\rangle_H.$ $\quad$ So $\quad \hat{A}_S|A,0\rangle_H ~=~A|A,0\rangle_H.$ The notion of instantaneous eigenstate is only consistent if the corresponding operator $\hat{A}_S$ has no explicit time dependence. $\quad \psi(x,t)={}_H\langle x,t | \psi\rangle_H={}_H\langle x,0| \psi(t)\rangle_S$. $\quad {}_H\langle B,t_2 |A,t_1\rangle_H= {}_H\langle B,0|\hat{U}(t_2,t_1)|A,0\rangle_H.$ $\endgroup$
    – Qmechanic
    Feb 9, 2015 at 13:11
  • $\begingroup$ Interaction picture: $\quad \hat{H}_S~=~\hat{H}_{0,S}+\hat{V}_S.$ $\quad |\psi(t)\rangle_I ~:=~ \hat{U}_0(t)^{-1}|\psi(t)\rangle_S~=~ \hat{U}_I(t)|\psi\rangle_H.$ $\quad \hat{A}_I(t)~:=~\hat{U}_0(t)^{-1}\hat{A}_S\hat{U}_0(t) ~=~\hat{U}_I(t)\hat{A}_H(t)\hat{U}_I(t)^{-1}.$ $\quad $ $S$, $H$ & $I$ pictures agree at $t=0$. $\quad\hat{U}_I(t)~:=~\hat{U}_0(t)^{-1}\hat{U}(t)$. $\quad i\hbar d_t\hat{U}_I(t) ~=~\hat{V}_I(t)\hat{U}_I(t).$ $\quad$ IPEOM: $\quad d_t\hat{A}_I(t)~=~\frac{i}{\hbar}[\hat{H}_{0,I}(t),\hat{A}_I(t)] +(\partial_t\hat{A}_S)_I.$ $\quad$ $\endgroup$
    – Qmechanic
    Nov 8, 2018 at 10:26

Those things are surely not enough to find the inner product $\langle q|p\rangle$ uniquely. For example, starting with the conventional $Q,P$, you may redefine them by a canonical transformation, for example by $$ Q\to Q'=Q, \quad P\to P'= P + Q^3 $$ Then $P', Q'$ obey all the four conditions in the same way as $P,Q$. They also have eigenstates and $|p'\rangle$ states are something else than $|p\rangle$. In fact, eigenstates with a large eigenvalue $P'$ are close to $Q$ eigenstates because $Q^3$ easily dominates. The inner product – which is nothing else than the wave function of the $P'$ eigenstate written in the $Q$ representation – will end up being different. It will be a complicated solution of the equation for $\psi$ saying that it's a $P+Q^3$ eigenstate.

The conditions you wrote can only tell you that $$\langle q | PQ | p \rangle = \langle q | QP | p \rangle - i\hbar \langle q | p \rangle = (qp - i\hbar )\langle q | p \rangle $$ so they're only enough to say how $Q$ acts on the $p$ eigenstates and vice versa, in this combination. But the inner product itself isn't related to itself in any sense, so it can't be determined.

Let me mention that even if you imposed additional conditions that would say that $P$ and $Q$ are physically what they should be – e.g. that they scale properly under the scaling of $Q,P$ – the inner product would still be undetermined because at least the phase of the $|p\rangle$ and $|q\rangle$ eigenstates is arbitrary. In fact, even the "absolute value part of the" normalization is a matter of conventions that could be modified.

More generally, and I would say that this point is often overlooked, many properties of "wave functions" similar to your inner products are intermediate, convention- and representation-dependent quantities that don't have a direct physical impact (i.e. direct link with observable quantities). It's really the properties of observables such as the four conditions that you described that may be considered objective facts.

  • 1
    $\begingroup$ I don't understand then what is the extra condition required in order to determine $<p|q>$. Is it that $p$ generates position-translations? $\endgroup$ Dec 8, 2017 at 6:23
  • $\begingroup$ I obviously used all conditions. Yes, the commutator between X and P is what says that P generates translations of X. I have used that condition as well so I don't understand how a person who has read my answer could "then" misunderstand why the condition is needed to determine the overlap. $\endgroup$ Dec 23, 2017 at 13:25
  • 1
    $\begingroup$ Your first sentence says that the four assumptions given in OP's post are not enough to uniquely determine $<p|q>$. What I meant by my question is what is the extra information you need to get $<p|q> = \frac{e^{i\phi}}{\sqrt{2\pi\hbar}} e^{ipx}$, up to some phase? In your post you mention something about $Q$ and $P$ must scale in a certain way. $\endgroup$ Dec 24, 2017 at 5:20
  • $\begingroup$ The extra information you need is the commutator of P,Q and I think that my answer explains where this information is used and why. If you don't believe me that two observables P,Q with a general commutator don't have to have plane-wave eigenstates relatively to each other, I can easily find as many counterexamples as you want. $\endgroup$ Dec 26, 2017 at 17:59

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