How can I prove that $i\hbar\frac{\partial}{\partial \mathbf{p}}$ is the operator of $\mathbf{x}$ in momentum space?

  • 1
    $\begingroup$ canonical commutation relation... $\endgroup$
    – mcodesmart
    Commented Oct 11, 2013 at 6:54
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Commented Feb 9, 2015 at 9:29

2 Answers 2


Actually your result doesn't quite follow. Something slightly more general does.

Everything begins and ends with the canonical commutation relationship (CCR):

$$[\hat{\mathbf{X}},\,\hat{\mathbf{P}}]=\hat{\mathbf{X}}\,\hat{\mathbf{P}} - \hat{\mathbf{P}}\,\hat{\mathbf{X}} = i\,\hbar\,\mathbf{I}\quad\quad\quad\quad(1)$$

where $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ are respectively the position and momentum observables. Suppose our quantum state space is labeled by co-ordinates wherein $\hat{\mathbf{X}}$ is the multiplication operator $f(x) \mapsto x\,f(x)$. Then the CCR doesn't quite imply that the momentum observable $\hat{\mathbf{P}}$ is $-i\,\hbar\,\nabla$, it implies it is of the form $-i\hbar\,\mathbf{Q}^{-1} \nabla \mathbf{Q}$ where $\mathbf{Q}$ is a diagonal $3\times 3$ matrix multiplication operator. But then of course we can always choose co-ordinates wherein $\hat{\mathbf{X}}$ is a multiplication operator and $\hat{\mathbf{P}} = -i\,\hbar\,\nabla$. I'll prove this result below.

Once we have this result, we simply swap the roles of $\hat{\mathbf{P}}$ and $\hat{\mathbf{X}}$, noting that we have to switch the sign of $\hbar$ to compensate for the reversal of the Lie bracket. So, the result above can be applied with $\hbar\mapsto -\hbar$ to show that we can always find co-ordinates wherein $\hat{\mathbf{P}} f(p) = p f(p)$ and $\hat{\mathbf{X}} f(p) = i\,\hbar\,\nabla\,f(p)$. QED

Afternote: Once you have the expressions $\hat{\mathbf{X}} f(x) = x\,f(x);\;\hat{\mathbf{P}} f(x) = -i\,\hbar\,\mathrm{d}_x f(x)$ it follows that the co-ordinates wherein $\mathbf{P}$ is the simple multiplication operator $\hat{\mathbf{P}} f(p) = p\,f(p)$ and the co-ordinates wherein $\hat{\mathbf{X}} f(p) = p\,f(p)$ must be related by a Fourier transform. This is because the eigenfunctions of $-i\,\hbar\,\mathrm{d}_x$ with real eigenvalues (i.e. so that $-i\,\hbar\,\mathrm{d}_x$ is Hermitian) are the functions $\exp(i\,k\,x)$ for real $k$. So this observation brings us to a second method of deriving your result, which I also give at the end.

Derivation of Basic Result

Let's go back to what we can actually derive from the CCR. Let's simplify this to a one-dimensional particle - it should be easy to see that this generalizes with 1D multiplication operators replaced by $3\times 3$ diagonal matrix multiplication operators. It can be shown that, given the CCR, $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ must have continuous spectrums - see my answer here and especially the link therein for more information. The eigenvalues of $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ can be any real values, so let's assume firstly that our co-ordinates in quantum state space have been chosen so that $\hat{\mathbf{X}}$ is a simple multiplication operator $\hat{\mathbf{X}} f(x) = x\,f(x)$ and we write the quantum state as a wavefunction $\psi(x)\in\mathbf{L}^2(\mathbb{R})$ of the position observable's eigenvalue $x$, so that now $|\psi(x)|^2$ is the probability density of the position eigenvalue given the quantum state $\psi(x)$. So now let's think of $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ as operators belonging to some linear space $\mathcal{L}$ of suitably well behaved operators on $\mathbf{L}^2(\mathbb{R})$. We need to assume that $\mathcal{L}$ is a space of diagonalizable (i.e. spectrally factorisable) operators with continuous spectrums. Then we can think of the Lie bracket $\hat{\mathbf{P}}\mapsto[\hat{\mathbf{X}},\,\hat{\mathbf{P}}]$ as linear operator $\operatorname{ad}_{\hat{\mathbf{X}}}:\mathcal{L}\to\mathcal{L}$ on the space $\mathcal{L}$ of our operators. This mapping's kernel is the linear space of operators mapped to nought by $\operatorname{ad}_{\hat{\mathbf{X}}}$; these are precisely the linear space of generalised multiplication operators $f(x)\mapsto g(x) f(x)$ for some fixed $g(x) \in \mathbf{L}^2(\mathbb{R})$ defining each such operator. The reason we know this is the whole kernel is that operators commute if and only if those operators have the same eigenvectors and those commuting with $\hat{\mathbf{X}}$ are the kernel members we seek. So a kernel member must be a multiplication operator. Now a particular solution to the CCR can readily be verified to be $f(x)\mapsto -i\,\hbar\,\mathrm{d}_x\,f(x)$, and the coset of all operators $\mathbf{P}$ fulfilling the CCR is the kernel displaced by any particular solution, so the most general $\hat{\mathbf{P}}$ we can have in this particular co-ordinate system is:

$$f(x) \mapsto (-i\,\hbar\,\mathrm{d}_x + g(x))\,f(x) = -i\,\hbar\,\mathbf{Q}^{-1} \mathbf{D} \mathbf{Q} f(x)\quad\quad\quad\quad(2)$$

where $\mathbf{D} f(x) = \mathrm{d}_x f(x)$ and $\mathbf{Q} f(x) = \exp\left(\frac{i}{\hbar} h(x) \right) f(x)$ where $h(x) = \int g(x)\mathrm{d}x$. So now we can rotate our state space co-ordinates so that $f(x) \mapsto \mathbf{Q} f(x)$ and the $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ observables transform in this space as $\hat{\mathbf{X}}\mapsto \mathbf{Q}\hat{\mathbf{X}}\mathbf{Q}^{-1} = \hat{\mathbf{X}}$ and $\hat{\mathbf{P}}\mapsto \mathbf{Q}\hat{\mathbf{P}}\mathbf{Q}^{-1} = -i\,\hbar\,\mathbf{D}$. Therefore:

The CCR alone implies there is an orthogonal co-ordinate system for the quantum state space wherein:

$$\begin{array}{lcl}\hat{\mathbf{X}} f(x) &=& x\,f(x)\\ \hat{\mathbf{P}} f(x) &=& -i\,\hbar\,\mathrm{d}_x\,f(x)\end{array}\quad\quad\quad\quad(3)$$

or, with a swap of roles of $\mathbf{X}$ and $\mathbf{P}$ together with a sign change of $\hbar$:

$$\begin{array}{lcl}\hat{\mathbf{P}} f(p) &=& p\,f(p)\\ \hat{\mathbf{X}} f(p) &=& i\,\hbar\,\mathrm{d}_p\,f(p)\end{array}\quad\quad\quad\quad(4)$$

Second Method for Deriving Your Result

We might begin with de Broglie's hypohthesis that momentum eigenstates are plane waves (i.e. with functional form $\exp(i\,k\,x)$ in position co-ordinates with momentum $\hbar k$ as our "fundamental" axiom. To factorise a state $\psi(x)$ into a superposition of such waves, we of course use the Fourier transform. De Broglie's hypothesis is then equivalent to the statement that momentum co-ordinates are the position co-ordinates Fourier transformed and rhe momentum operator, by our de Broglie hypothesis momentum formula, multiplies these by $\hbar\,k$, then we transform back to position co-ordinates. So our momentum operator in position co-ordinates must be:

$$\psi(x) \to \frac{\hbar}{2\,\pi}\int_{-\infty}^\infty\exp(i\,k\,x)\,k\,\int_{-\infty}^\infty\exp(-i\,k\,u)\psi(u)\,\mathrm{d}u\,\mathrm{d}k=\\-i\,\hbar\,\mathrm{d}_x\,\left(\frac{1}{2\,\pi}\int_{-\infty}^\infty\exp(i\,k\,x)\,\int_{-\infty}^\infty\exp(-i\,k\,u)\psi(u)\,\mathrm{d}u\,\mathrm{d}k\right) = -i\,\hbar\,\mathrm{d}_x\,\psi(x)\quad\quad\quad\quad(5)$$

So, if we now Fourier transform our co-ordinates so that:

$$\Psi(p) = \frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^\infty\exp\left(i\,\frac{p}{\hbar}\,x\right)\psi(x)\mathrm{d} x\quad\quad\quad\quad(6)$$

then we see that (e.g. by integration by parts)

$$p\,\Psi(p) = \frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^\infty\exp\left(i\,\frac{p}{\hbar}\,x\right)\left(-i\,\hbar\,\mathrm{d}_x \psi(x)\right)\,\mathrm{d} x\quad\quad\quad\quad(7)$$

and also (by simply differentiating under the integral)

$$i\,\hbar\,\mathrm{d}_p \Psi(p) = \frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^\infty\exp\left(i\,\frac{p}{\hbar}\,x\right)\, x \,\psi(x)\,\mathrm{d} x\quad\quad\quad\quad(8)$$

and so (7) and (8) are the momentum and position operators, respectively, in momentum co-ordinates, hence we have a second way to understand your result.


To work this out we shall use an eigen basis of position ${\left|x\right\rangle}$

We represent a ket-state in this basis as $\left|\psi\right\rangle$ and an linear operator ${M}$ such that $${M\left|\psi\right\rangle} = \left|M\psi\right\rangle$$

We thus have: $${\left\langle x\right.\left|\psi\right\rangle} = {\psi}{(x)}$$ $${\left\langle x\right.\left|M\psi\right\rangle} = M{\psi}{(x)}$$

Thus $M$ acts as a linear operator on the space of all functions represented by ${\psi}{(x)}$. Obviously $M{\psi}{(x)}$ returns a function. By the canonical commutation relations(CCR) we have:

$$xp-px = ih$$

We know $x$, it is just the multiplication of $x$ into the wave function. But how do we get the most general definition of $p$ as a linear operator? Here's how:

We start with a derivative operator $K$ such that:

$$\left\langle x\right|K\left|\psi\right\rangle = {\psi'}{(x)}$$

$K$ is linear and purely imaginary operator (I'll let you work that out). We also find that:

$$\left\langle x\right|Kx\left|\psi\right\rangle = x{\psi'}{(x)} + {\psi}{(x)} = \left\langle x\right|xK\left|\psi\right\rangle + \left\langle x\right.\left|\psi\right\rangle$$

We find that $Kx-xK=1$, we can plug $K$ into the CCR and get $$p=-ihK$$. Over... or is it?? Try instead $K+f(x)$, where $f(x)$ is some arbitrary function of x. Now we get

$$\left\langle x\right|(K+f)\left|\psi\right\rangle = x{\psi'}{(x)} + f(x)x{\psi}{(x)} + {\psi}{(x)} = \left\langle x\right|x(K+f)\left|\psi\right\rangle + \left\langle x\right.\left|\psi\right\rangle$$

Thus, $$p=-ih(K+f)$$ is the actually most general solution of the CCR.

You say, 'Oh my!! All hope is lost'. I say 'haha never fear, let us look more deeply into the problem.' Let us try this:

$$\left\langle x\right|Kf\left|\psi\right\rangle = f(x){\psi'}{(x)} + f'(x){\psi}{(x)} = \left\langle x\right|fK\left|\psi\right\rangle + \left\langle x\right|f'\left|\psi\right\rangle$$

We then get: $$Kf-fK=f'$$. 'How does this help?', you say. 'Be patient boy, we are not limited to just one eigenbasis.' I perform the following transformation:

$\left|x\right\rangle \to {e^{-i{\gamma}}}\left|x\right\rangle$

Naturally we can also say:

$\left\langle x\right| \to \left\langle x\right| {e^{i{\gamma}}}$

$\left|\psi\right\rangle \to {e^{-i{\gamma}}}\left|\psi\right\rangle$

where ${\gamma}$ is some random function of $x$. Here's why this transformation is acceptable (it is called a unitary transformation):

$$\left\langle x\right|\left.\psi\right\rangle \to \left\langle x\right| {e^{i{\gamma}}} {e^{-i{\gamma}}} \left|\psi\right\rangle = {\psi}{(x)}$$

So it preserves probability.

A linear operator $M$ transforms as:

$$M \to {e^{-i{\gamma}}}M{e^{i{\gamma}}}$$

So we will plug in our operator $K+f$ here (it's transform is denoted by to K^*):

$$K^* = {e^{-i{\gamma}}}(K+f){e^{i{\gamma}}} = {e^{-i{\gamma}}}K{e^{i{\gamma}}} + f$$

Using $Kg-gK=g'$, we get:

$$K^* = {e^{-i{\gamma}}}{e^{i{\gamma}}}({i{\gamma'}} + K) + f = K + {i{\gamma'}} + f$$

Choose an arbitrary phase such that ${i{\gamma'}(x)}$ is always equal to $-f(x)$.

By choosing an arbitrary phase as above we get $K^* = K$, therefore, here the most general solution of $p$ here is $-ihK$. Throughout my answer I could have represented $K$ as $d/{dx}$, but the choice of symbol really doesn't matter since I already said that $K$ yields the derivative.

Refer to Dirac's 'Principles of Quantum mechanics' for the same derivation in higher dimensions (Section 21-22). Also you will find, in may places in physics, that we have fairly liberal interpretations of operators, we select schemes based on pure convenience.


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