I know that

$$\tag{1}\hat{p}~=~-i\hbar \frac{\partial}{\partial x}~.$$

How can I get

$$\tag{2}\hat{x}~=~i\hbar \frac{\partial}{\partial p}~?$$

I think this simple and I'm just over thinking it, but knowing the momentum operator (1), how can I get the position operator (2)?

  • 8
    $\begingroup$ One cannot derive $\hat{x}=ih \frac{\partial}{\partial p}$ just from the formula $\hat{p}=-ih \frac{\partial}{\partial x}$ alone. For instance $\hat{x}=\hat{c}+ih \frac{\partial}{\partial p}$ is also consistent, where $\hat{c}$ is a Casimir operator (e.g. proportional to the identity operator ${\bf 1}$). One needs to make further (conventional) assumptions in order to derive eq. (2). See e.g. this, this, and possibly this Phys.SE post. $\endgroup$
    – Qmechanic
    Commented Nov 17, 2013 at 10:03

6 Answers 6


Here's how you do it.

First, notice that for any state $|\psi\rangle$, we have \begin{align} \langle p|[\hat x, \hat p]|\psi\rangle &= \langle p|\hat x\hat p-\hat p\hat x|\psi\rangle \\ &= \langle p|\hat x\hat p|\psi\rangle - \langle p|\hat p \hat x|\psi\rangle \\ &= \langle p|\hat x \hat p|\psi\rangle - p\langle p|\hat x|\psi\rangle \end{align} Now, recall the canonical commutation relation between $\hat x$ and $\hat p$; \begin{align} [\hat x, \hat p] &= i\hbar \hat I \end{align} where $\hat I$ is the identity operator. Using this fact on the left hand side of the manipulation we just performed, and doing a little rearranging, we find that \begin{align} p\langle p|\hat x|\psi\rangle = \langle p|\hat x \hat p|\psi\rangle -i\hbar\langle p|\psi\rangle. \tag{1} \end{align} Now, focus on the first term on the right. we have \begin{align} \langle p|\hat x \hat p|\psi\rangle &= \int dx\,\langle p|\hat x|x\rangle\langle x|\hat p|\psi\rangle \\ &= \int dx\, x e^{ipx/\hbar}\left[-i\hbar \frac{d}{dx}\psi(x)\right] \\ &= i\hbar \int dx\, \frac{d}{dx}(xe^{ipx/\hbar})\psi(x) \\ &= i\hbar \int dx\, e^{ipx/\hbar}\psi(x) + i\hbar\int dx\,x\left(\frac{ip}{\hbar}\right)e^{ipx/\hbar}\psi(x) \\ &= i\hbar\psi(p) -p\int dx\,xe^{ipx/\hbar}\psi(x)\\ &= i\hbar\psi(p)+i\hbar p\frac{d}{dp}\int dx\,e^{ipx/\hbar}\psi(x) \\ &= i\hbar\psi(p)+i\hbar p \frac{d}{dp}\psi(p) \tag{2} \end{align} where we have made liberal use of integration by parts, and the following identities: \begin{align} \int dx\, |x\rangle\langle x| = \hat I, \qquad \psi(x) \overset{\mathrm{def}}{=} \langle x|\psi\rangle, \qquad \psi(p) \overset{\mathrm{def}}{=} \langle p|\psi\rangle,\qquad \langle x|\hat p|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x). \end{align} The last fact is precisely the statement that the position space representation of the momentum operator is $-i\hbar d/dx$. Now combine $(1)$ and $(2)$ to obtain \begin{align} p\langle p|\hat x|\psi\rangle = i\hbar p\frac{d}{dp}\psi(p) \tag{3} \end{align} Next, we simply note that if we denote the momentum space representation of the position operator as $D^{(p)}(\hat x)$, then by definition \begin{align} D^{(p)}(\hat x)\psi(p) = \langle p|\hat x|\psi\rangle, \end{align} Combining this with $(3)$, and noting that the resulting equation should hold for all $p$, and dividing both sides by $p$, we obtain the desired result; \begin{align} D^{(p)}(\hat x) = i\hbar \frac{d}{dp} \end{align}


Assume one dimensional and $\hbar=1$

  1. By \begin{array} \hat \hat{p} |p \rangle &= p | p \rangle \\ \langle x | \hat{p} |p \rangle &= p \langle x | p \rangle \\ -i \frac{\partial}{\partial x} \langle x| p \rangle &= p \langle x| p \rangle \,\,\,\, (1) \end{array}

Because you already knew $\hat{p}=-i\frac{\partial}{\partial x}$, hence $\langle x |\hat{p}| \psi \rangle =-i \frac{\partial}{\partial x} \langle x | \psi \rangle$. Take $|\psi\rangle = | p \rangle$, we obtain (1).

Solving the differential equation (1) for $\langle x| p \rangle$, we have $$\langle x | p \rangle= \frac{1}{\sqrt{2\pi}} e^{ipx} $$, $\frac{1}{\sqrt{2\pi}} $ is a normalization constant.

2. \begin{array} \langle \langle p | \hat{x} | p' \rangle &= \int \!\! \int \langle p |x \rangle \langle x | \hat{x}| x' \rangle \langle x'|p' \rangle dxdx' \\ &= \frac{1}{2\pi} \int \!\! \int e^{-ipx+ip'x'} x' \delta (x-x') dx dx' \\ &= \frac{1}{2\pi} \int e^{-i(p-p')x} x dx \\ &= \frac{i}{2\pi} \frac{ \partial}{\partial p} \int e^{-i(p-p')x} dx \\ &= i \frac{ \partial}{\partial p} \delta(p-p') \end{array}

3. We shall show $$ \langle p | \hat{x} | \psi \rangle = i \frac{ \partial}{\partial p} \psi(p), \forall |\psi\rangle $$, here $\psi(p):=\langle p | \psi \rangle$ .

By the following procedure \begin{array} \langle \langle p | \hat{x} | \psi \rangle &= \int \langle p| \hat{x} | p' \rangle \langle p' | \psi \rangle dp' \\ &= \int i \frac{ \partial}{\partial p} \delta(p-p') \psi(p') dp' \\ &= i \frac{ \partial}{\partial p} \psi(p) \end{array}

  • $\begingroup$ I think there are two sign errors which cancel each other: if I'm not mistaken, there's a minus sign missing when introducing $\frac{\partial}{\partial p}$ in the expression for $\langle p|\hat x|p'\rangle$, which gets 'fixed' by the missing minus sign when applying the derivative of the $\delta$ function $\endgroup$
    – Christoph
    Commented Nov 18, 2013 at 8:11
  • $\begingroup$ Do you mean the fourth and fifth lines in part 2? In the fourth line, $\partial /\partial p [ e^{-i(p-p')x} ]= -i x e^{-i(p-p')x}$, a prefactor $i$ takes it back to $x$. The fifth line is just the Fourier transformation of Dirac delta function, if I am not mistaken... $\endgroup$
    – user26143
    Commented Nov 18, 2013 at 8:53
  • $\begingroup$ conceptionally, the derivative of the $\delta$ function gets applied by partial integration (see here), so there's a minus sign missing in that expression; I tried to figure out why we still get the correct result, but apparently I botched that; I'll go through the equations again and see if I can't figure it out... $\endgroup$
    – Christoph
    Commented Nov 18, 2013 at 9:23
  • $\begingroup$ In part 3, the derivative of $\delta$ function acts on $p$, the integration is over $p'$. So that should not be a problem, although i am not a mathematician... $\endgroup$
    – user26143
    Commented Nov 18, 2013 at 9:38
  • $\begingroup$ that clears up where my confusion came from - what I was aiming for was $\langle p|\hat x|p'\rangle = -i\frac{\partial}{\partial p'}\delta(p'-p)$, which has a negative sign that gets re-absorbed by partial integration; I also don't know if it's fine as-is, ie under what assumptions $\lim_{\epsilon\to0}\int\frac{\partial}{\partial p}\delta_\epsilon(p-p') \psi(p') dp' = \frac{\partial}{\partial p}\lim_{\epsilon\to0}\int\delta_\epsilon(p-p') \psi(p') dp'$ holds... $\endgroup$
    – Christoph
    Commented Nov 18, 2013 at 10:01

You don't actually need $\hat{p}$ to do this. You can start from the fact that $\hat{x}$, when applied to a position eigenfunction, has to produce the corresponding position eigenvalue ($x_0$) times that same function:

$$\hat{x}\phi_{x_0}(p) = x_0\phi_{x_0}(p)$$

In the momentum representation, position eigenfunctions take the form $\phi_{x_0}(p) = e^{-ipx_0/\hbar}$. (That comes from the Fourier transform of the delta function.) So what operator can you apply to this that pulls out a factor of $x_0$? It's not hard to figure out by some combination of intuition, and trial and error, that a derivative with respect to $p$ will do the trick, specifically

$$i\hbar\frac{\partial}{\partial p}e^{-ipx_0/\hbar} = x_0e^{-ipx_0/\hbar}$$

That's enough to identify

$$\hat{x} = i\hbar\frac{\partial}{\partial p}$$

Even if you have the definition of $\hat{p}$, this is still probably the easiest way to find $\hat{x}$ in the momentum representation. If you want to come up with a proof that explicitly involves the definition of $\hat{p}$, you have to figure out what other part of the definition of the position representation you want to discard to make it necessary to start from $\hat{p}$.

As a rough argument, you could say that the position and momentum representations are related by complex conjugation. Notice that the transformation that gets you from the position representation to the momentum representation is this:

$$\phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\int\psi(x)e^{-ipx/\hbar}\mathrm{d}x$$

and the one that gets you back is this:

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}}\int\phi(p)e^{ipx/\hbar}\mathrm{d}p$$

If you exchange $\psi(x)\leftrightarrow\phi(p)$, $x\leftrightarrow p$, and $i\leftrightarrow -i$, it leaves these transformations unchanged. This implies that you can "switch" position space and momentum space in any formula as long as you also take the complex conjugate.

Applying this reasoning to $\hat{p} = -i\hbar\frac{\partial}{\partial x}$, you get $\hat{x} = i\hbar\frac{\partial}{\partial p}$.


Dirac argues from symmetry in his Principles of QM:

In a 1-D system, $\hat{q}$ and $\hat{p}$ are both observables, with eigenvalues extending from $-\infty$ to $+\infty$, and are connected by the commutation relation $[\hat{q},\hat{p}]=i \hbar$.

Since one can interchange $\hat{q}$ and $\hat{p}$ in these equations if $i$ is replaced by $-i$, it follows that, if there is a representation in which $\hat{q}$ is diagonal and

$$\hat{p} = -i \hbar \frac{d}{dq} $$

there must also be a representation in which $\hat{p}$ is diagonal and

$$\hat{q} = +i \hbar \frac{d}{dp} $$


Actually, Dirac first does a bit of work to demonstrate that $\hat{p}$ is indeed an observable with an infinite range of eigenvalues, like $\hat{x}$.

I think the ambiguity noted by Qmechanic does not appear here because:

  • Dirac constructs the eigenstates of $\hat{q}$ specifically to give the simple, single-term expression for $\hat{p}$
  • In Qmechanic's analysis the eigenstates are specified by "the enemy" and are not adjustable.

Say that we are handed a 1D quantum mechanical system, which satisfies the canonical commutation relation

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

and handed some choice of eigenstates $|q\rangle$ and $|p\rangle$ for every value of $q,p\in\mathbb{R}$. The eigenstates satisfy

$$\tag{2} \hat{Q} \mid q \rangle ~=~q\mid q \rangle, \qquad \hat{P}\mid p \rangle ~=~p\mid p \rangle , $$

$$\tag{3} \langle q \mid q^{\prime} \rangle~=~\delta(q-q^{\prime}), \qquad \langle p \mid p^{\prime} \rangle~=~\delta(p-p^{\prime}),$$

$$\tag{4}\int_{\mathbb{R}} \!dq~ \mid q \rangle\langle q \mid~=~{\bf 1}, \qquad \int_{\mathbb{R}} \!dp~ \mid p \rangle\langle p \mid~=~{\bf 1}. $$

Eigenstates (2) are not uniquely defined. One could in principle re-define them with phase factors

$$\tag{5} |q \rangle ~\longrightarrow~ |q \rangle^{\prime}~:=~f(q)|q \rangle, \qquad |p \rangle ~\longrightarrow~ |p \rangle^{\prime}~:=~g(p)|p \rangle, $$

where $f$ and $g$ are two phase factors $|f|=|g|=1$. The new basis would also satisfy eqs. (2-4). Thus if the enemy has chosen the eigenstates (2) for us, we should (as a minimum) expect that the position and momentum Schrödinger representation could contain non-trivial phase factors

$$\tag{6} \hat{Q}~=~q, \qquad \frac{i}{\hbar}\hat{P}~=~ \frac{1}{f(q)}\frac{\partial}{\partial q} f(q)~=~ \frac{\partial}{\partial q} +\frac{f^{\prime}(q)}{f(q)}, $$


$$\tag{7} \hat{P}~=~p, \qquad \frac{1}{i\hbar}\hat{Q}~=~ \frac{1}{g(p)}\frac{\partial}{\partial p} g(p)~=~ \frac{\partial}{\partial p} +\frac{g^{\prime}(p)}{g(p)}, $$

respectively. Note that the two Schrödinger representations (6-7) are consistent with the CCR (1). [On the other hand, it turns out that eqs. (6-7) constitute the most general position and momentum Schrödinger representations, respectively. This may e.g. be established from knowing the most general form of the $\langle p \mid q \rangle$ overlap, see e.g. this Phys.SE answer.]

Now let us return to OP's question. Assume that $\hat{P}=-i\hbar\frac{\partial}{\partial q}$, i.e. that the position phase factor $f(q)$ is a constant (independent of $q$) in eq. (6). This does not imply that $\hat{Q}=i\hbar\frac{\partial}{\partial p}$, i.e. that the momentum phase factor $g(p)$ is a constant as well in eq. (7).

We stress that the two phase factors in eqs. (6-7) are not just an academic exercise. Often, one would of course try to chose eigenstates (2) such that phase factors $f$ and $g$ vanish, as is usually assumed in elementary textbooks. However, in actual quantum systems, this may not be possible for various reasons, e.g. dependence of external parameters, or topological obstructions. In such situations the phase factors $f$ and $g$ may have important physical consequences.


If \begin{equation} q\,p\boldsymbol{-}p\,q\boldsymbol{=} i\hbar \quad \boldsymbol{\Longrightarrow} \quad p\boldsymbol{=}\boldsymbol{-} i\hbar\dfrac{\mathrm d}{\mathrm dq} \tag{a}\label{a} \end{equation} then \begin{equation} (\boldsymbol{-}p)\,q\boldsymbol{-}q\,(\boldsymbol{-}p)\boldsymbol{=} i\hbar \quad \boldsymbol{\Longrightarrow}\quad q\boldsymbol{=}\boldsymbol{-}i \hbar\dfrac{\mathrm d}{\mathrm d(\boldsymbol{-}p)}\boldsymbol{=+}i\hbar\dfrac{\mathrm d}{\mathrm dp} \tag{b}\label{b} \end{equation}

Related Link : Hermiticity of Momentum Operator (matrix) Represented in Position Basis


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