# How do you show that momentum is hermitian in Dirac notation?

I am trying to prove that momentum operator $\bf{\hat{p}}$ is hermitian. I know how to prove it in the $\bf{x}$ representation integrating by parts and using the fact that $\lim_{r \rightarrow \infty} \psi(\bf{ r } ) = 0$, but I haven't found any proof using abstract Dirac notation. Is there a way? If not, why?

• – Qmechanic Feb 13 at 18:24

## 3 Answers

As with most things, if you want to prove something about the momentum operator, then you have to start from some specification of what the momentum operator actually is.

If we give the momentum operator an explicit representation like $$\hat p = -i\hbar \frac{d}{dx}$$, then we can show directly that $$\langle \psi| \hat p \phi\rangle = \langle \hat p \psi|\phi\rangle$$ for any $$\psi,\phi$$ in the domain of $$\hat p$$. In the absence of such a representation, we need to specify the properties of $$\hat p$$ some other way.

One such definition is in terms of the translation operator $$\hat T(a)$$, which acts on some state $$\psi$$ as follows: $$\hat T(\epsilon)|\psi\rangle := \int dx \ \hat T(\epsilon)|x\rangle\langle x|\psi\rangle = \int dx |x+\epsilon\rangle\langle x|\psi\rangle = \int dx \ |x\rangle\langle x-\epsilon|\psi\rangle$$

The momentum operator can be defined as the infinitesimal generator of translations - in concrete terms,

$$\hat p := i\hbar \lim_{\epsilon\rightarrow 0} \frac{\hat T(\epsilon) - \mathbb I}{\epsilon}$$

From here,

$$|\hat p\psi\rangle = i\hbar \lim_{\epsilon\rightarrow 0}\int dx |x\rangle\frac{\langle x-\epsilon|\psi\rangle-\langle x|\psi\rangle}{\epsilon}$$

and

$$\langle\hat p \phi| \equiv |\hat p \phi\rangle^\dagger = -i\hbar \lim_{\epsilon\rightarrow 0}\int dx\frac{\langle\phi | x-\epsilon\rangle-\langle\phi|x\rangle}{\epsilon} \langle x |$$

which means that

$$\langle\phi | \hat p \psi\rangle = i\hbar \lim_{\epsilon\rightarrow 0} \int dx \frac{\langle \phi|x\rangle\langle x-\epsilon|\psi\rangle-\langle\phi | x\rangle \langle x |\psi\rangle}{\epsilon}$$

and

$$\langle \hat p \phi | \psi \rangle = -i\hbar\lim_{\epsilon\rightarrow 0} \int dx \frac{\langle \phi|x-\epsilon\rangle\langle x|\psi\rangle - \langle \phi|x \rangle \langle x | \psi\rangle}{\epsilon}$$

If we make the variable substitution $$x \rightarrow x+\epsilon$$ in the first term of the integrand and then make the substitution $$\epsilon \rightarrow -\epsilon$$, the negative signs will cancel out, and we see from this that $$\langle \hat p\phi |\psi\rangle = \langle\phi | \hat p \psi\rangle$$ which means that $$\hat p$$ is Hermitian.

• Could you explain why you only made the variable substitution $x\rightarrow x+\epsilon$ in the first term? If you only make the variable substitution in the first term, the $|x\rangle$ in the left and right term will not be the same, and the limit wouldn't be identical the first one. – TheAverageHijano Feb 13 at 20:53
• @TheAverageHijano The integral is over all values of x (more specifically, shifting the variable by a constant doesn’t change the integration bounds), so we can make substitutions in each term independently. If you prefer, you can split it into two integrals, perform the substitution in the first one, and then recombine them after you’ve finished. – J. Murray Feb 13 at 20:58

Consider the matrix element $$\langle\phi|\hat{\mathbf p}|\chi\rangle$$ of the momentum operator $$\hat{\mathbf p}$$ between two arbitrary states $$|\phi\rangle, |\chi\rangle.$$ In order to calculate this matrix element, we can insert the identity decomposed in the momentum eigenvectors $$\left\{|\mathbf p\rangle\right\},$$ $$1 = \int d{\mathbf p} |{\mathbf p}\rangle\langle{\mathbf p}|,$$ at each side of the operator $$\hat{\mathbf p}$$. That is, $$\langle\phi|\hat{\mathbf p}|\chi\rangle = \int d{\mathbf p}_1 \int d{\mathbf p}_2 \langle\phi|{\mathbf p}_1\rangle\langle{\mathbf p}_1|\hat{\mathbf p}|{\mathbf p}_2\rangle\langle{\mathbf p}_2|\chi\rangle = \\ =\int d{\mathbf p}_1 \langle\phi|{\mathbf p}_1\rangle\langle{\mathbf p}_1|\chi\rangle {\mathbf p}_1.$$ Using this expression you can check that $$\langle\phi|\hat{\mathbf p}|\chi\rangle^* (= \langle\chi|\hat{\mathbf p}^\dagger|\phi\rangle) = \langle\chi|\hat{\mathbf p}|\phi\rangle.$$ As this is valid for any pair of states, we have $$\hat{\mathbf p}^\dagger=\hat{\mathbf p}$$.

• I think that is the case only if $\langle \bf{p}_1 | \hat{\bf{p}} | \bf{p}_2 \rangle \in \mathbb{R}$ – MBolin Mar 19 '18 at 18:28
• You should have applied operator $\bf{p}$ over $|\bf{p}_2\rangle$ instead of $\langle\bf{p}_1|$, we don't know if momentum is Hermitian a priori. – TheAverageHijano Feb 13 at 18:45
• @TheAverageHijano I think he did it and later used $\langle p_1 | p_2 \rangle = \delta(p_1 - p_2)$ to end up with only $p_1$ – ErickShock Feb 13 at 19:17
• The trouble with this is that there's nothing about $\hat {\mathbf p}$ which ties it to the momentum operator. Instead, you've used the fact that $\hat{\mathbf p}$ has a complete set of orthonormal eigenvectors with real eigenvalues. In other words, you've shown that those two properties imply that $\hat{\mathbf p}$ is Hermitian, but not that the momentum operator has them. – J. Murray Feb 13 at 19:40

Sakurai's Modern Quantum Mechanics first postulates that the position operator must be hermitian ($$X^\dagger = X$$). Later he introduces the infinitesimal translation operator that translates $$x$$ by $$x + dx$$: $$T(dx) |x\rangle = |x+dx\rangle.$$ He then imposes some reasonable conditions on $$T$$: (i) it must be unitary to preserve the norm of any ket: $$T^\dagger (dx) T(dx) = 1,$$ (ii) two successive translations by $$dx'$$ and $$dx''$$ must be equal to a single translation by $$dx' + dx''$$: $$T(dx'')T(dx') = T(dx''+dx'),$$ (iii) the inverse operator $$T^{-1}$$ must be equal to a translation in the opposite direction: $$T^{-1}(dx) = T(-dx),$$ (iv) when $$dx\to0$$ the translation operator must be equal to unity: $$\lim_{dx \to 0} T(dx) = 1.$$ You can check that, up to first order in $$dx$$, choosing $$T$$ as $$T(dx) = 1 - i\frac{P}{\hbar} dx$$ where $$P$$ is a hermitian operator, $$i$$ the imaginary unit, and $$\hbar$$ just a number it'll satisfy all the properties above. Now, this $$P$$ generates translations (we may call it "the generator of translations"). However, it's also possible to show that the classical momentum $$p$$ is the generator of translations in classical mechanics. If we define momentum as the generator of translations, then the quantum mechanical operator $$P$$ corresponds to the classical $$p$$, and we may call both of them momentum. The constant $$\hbar$$ is just there to make the units of both momenta match.