There is the standard argument, using the definition of the inner product; that $\langle f|A|g\rangle =\langle g|A|f\rangle ^{*}$ for a Hermitian operator $A$, given any wave vectors $|f\rangle,~ |g\rangle$.

Also consider the following:

Consider the infinite dimensional one dimensional position space, with a column vector of values of the wave function at discretized points along the $x$-axis.

In the limit of infinitesimal differences between position values, we have $\frac{\mathrm df}{\mathrm dx}\approx \frac{1}{2h} (f(x+h)-f(x-h))$, where $(x+h)$ and $(x-h)$ are the discrete position values just preceding and succeeding the position value $x$, and $h$ is sufficiently small.

Then we might talk of an infinite dimensional matrix representation of ${\rm d/d}x$, where only the two off diagonal "diagonals" adjacent to the actual diagonal has $1/2h$ and $-\:1/2h$ entries. This matrix is skew symmetric. Everywhere else we have $0$ entries.

If we multiplied this matrix with $'\mathrm i'$, this skew symmetry becomes Hermitian, which makes ${\rm i~ d/d}x$ hermitian.

Edit: As pointed out by tparker in an answer below, we get $\langle x|\partial|x^\prime\rangle =\frac{\partial}{\partial x}\langle x|x^\prime \rangle =\frac{\partial}{\partial x}\delta(x-x^\prime)$.

Since we are dealing with a discrete set of points in space here, we must have the normalization given by the Kronecker delta $\delta_{x,x^\prime}$.

Informally, we then have $\partial(\delta_{x,x^\prime})|_{x~=~(x^\prime-h)}\approx\frac{1}{2h}(\delta_{x^\prime,x^\prime}-\delta_{x^\prime-2h,x^\prime})=\frac{1}{2h}$, and also $\partial(\delta_{x,x^\prime})|_{x~=~(x^\prime+h)}\approx-\frac{1}{2h}, ~~\partial(\delta_{x,x^\prime})|_{x~=~x^\prime}\approx 0$, which again gives us a skew-symmetric matrix of the form obtained before. Here we scanned the matrix vertically in a given column, whereas in the previous calculation, we scanned a fixed row horizontally.

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    $\begingroup$ The mathematicians will complain, but this is how I personally think about it, and I think it's a great way to conceptualize it. $\endgroup$
    – knzhou
    Nov 13, 2016 at 4:16
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    $\begingroup$ @knzhou I'm not sure that the mathematicians would complain that much. Often "they", as much as do physicists, seek intuition to motivate why something should be true in order to sketch a more rigorous proof. $\endgroup$ Nov 13, 2016 at 4:43
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    $\begingroup$ @J.G. Why is that? $\endgroup$ Nov 14, 2016 at 15:03

5 Answers 5


Short answer: because for differential operators the transpose is integration by parts, and integration by parts flips signs. So, by itself, the derivative operator is antisymmetric (it's transpose is the negative of the operator). The factor of $i$ adds another sign flip when taking the Hermitian conjugate (complex conjugate and transpose) that counter-balances the sign flip from the transpose.

Slightly longer answer: you can examine linear operators in terms of their integration kernel. For the derivative operator, the kernel is: $$K(x,y) = -\delta'(x-y),$$ that is $f'(x) = -\int \delta'(x-y) f(y) \operatorname{d}y$ for any reasonably smooth function $f$. Examining a concrete Gaussian representation of the delta function at finite width gives: $$K_\sigma(x,y) = \frac{[x-y]}{\sigma^3\sqrt{2\pi}} \operatorname{e}^{-\frac{[x-y]^2}{2\sigma^2}}.$$ The advantage of looking at the integration kernal is that the transpose is easy to define in a way analogous to the way it works for matrices. For a matrix $\left[M^T\right]_{ij} = M_{ji}$, so the transpose of an operator in terms of an integration kernel is: $$\left[K^T\right](x,y) = K(y,x).$$ Inspection of $K_\sigma(x,y)$ above is sufficient to show that it satisfies: $$K_\sigma(x,y) = -K_\sigma(y,x);$$ that is, $K_\sigma(x,y)$ is antisymmetric.

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    $\begingroup$ Hey I really love that first paragraph. It's kind of dense, but excellent mnemonic value, and it also sketches how you prove it. I like the second explanation too. $\endgroup$ Nov 16, 2016 at 7:54
  • $\begingroup$ But... symmetry/ anti-symmetry of the derivative operator alone does not establish Hermiticity or non-Hermiticity. $\endgroup$ Nov 17, 2016 at 18:45

This is exactly the right intuition and indeed the way physicists think about it. Many books give the continuum version of the same argument by pointing out that the derivative of the Dirac delta function $\delta'(x - x')$ (which is basically just the "matrix element" for the derivative operator) is odd under the interchange $x \leftrightarrow x'$, i.e. the corresponding "matrix" would be skew-symmetric.

  • $\begingroup$ The Dirac delta in this case is like a basis element for the wave function. So only one row of the "matrix" acts on the Dirac delta element, right? Do we see the skew symmetry from there? $\endgroup$ Nov 13, 2016 at 4:38
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    $\begingroup$ @AritroPathak If the operator $\hat{D}$ represents a derivative with respect to $x$, then $\langle x | \hat{D} | x' \rangle = \delta'(x - x')$. In order to calculate the (symmetric) derivative, you need to know the vector of the vector at three consecutive components $x - h$, $x$, and $x + h$, but in the continuum limit these three locations become arbitrarily close together, so you only need to know the value of the function in an infinitesimal neighborhood of $x$. $\endgroup$
    – tparker
    Nov 14, 2016 at 5:33
  • $\begingroup$ Thank you for your answer. I only found Shankar which points out that the derivative of the Dirac delta is antisymmetric. Are there other standard texts that do the same thing? $\endgroup$ Nov 16, 2016 at 18:55

In a concrete example, one could also look at a scalar product in $C^\infty$, with $\langle \psi | \phi\rangle = \displaystyle \int \psi^* \phi \ \mathrm{d}x$.

You would have $\langle \mathrm \psi ~| ( \mathrm i ~\partial_x | \phi \rangle ) = \displaystyle \int \psi^* (\mathrm i ~\partial_x \phi)~ \mathrm{d}x = \displaystyle\int (-~\mathrm i \partial_x \psi^*) \phi~ \mathrm{d}x= \langle \psi ~| (\mathrm i~ \partial_x)^\dagger | \phi \rangle$, having partially integrated once assuming compact support.

Hence $\mathrm{d}/\mathrm{d}x$ alone would not be hermitian.


What operators are hermitian depends on what type of inner product you have in your Hilbert space and what the boundary conditions are.

If your inner product is defined as $\langle\phi|\psi\rangle:=\int_{-\infty}^\infty dx\phi^*\psi$ then $\int_{-\infty}^\infty dx\phi^*(i\frac{d}{dx}\psi)=\int_{-\infty}^\infty dx(-i\frac{d}{dx}\phi^*)\psi=\int_{-\infty}^\infty dx(i\frac{d}{dx}\phi)^*\psi$, where one had to integrate by parts and assume that boundary terms vanish. If this is not the case then one would have to add corrections to the derivate to "re-hermitize" the operator.

The simplest example is a scalar product $\langle\phi|\psi\rangle:=\int_{a}^b dx\phi^*\psi$ with again a boundary condition where the fields vanish at $a$ and $b$. This then causes that one has to add additional delta distribution terms to the derivative to re-hermitize. These delta distribution terms correspond to boundary conditions one would have to implement for a particle in a box with infinitely high walls.

Therefore what is hermitian depends on boundary conditions and inner product of the Hilbert space.


The Hermiticity of the derivative operator is dependent on the object/ functions upon which they act! These derivative functions alone are neither Hermitian, nor non-Hermitian; answers claiming otherwise are incomplete and or incorrect. It is the the momentum-operator that needs to be Hermitian if momentum is an observable, not the derivative. The momentum-operator is precisely "an operation performed on a function that extracts momentum, $p$. Proof: consider the function $e^{\mathrm ipx}$. The operation that extracts momentum in this case is $-~\mathrm i\partial_x$. Were the function instead $e^{px}$, the form of the momentum- operator in terms of a derivitive would have to be changed to $+\partial_x$. In each case the momentum operator is Hermitian in this context!. While the effects of the derivative between the two cases changes, QED.

In non-relativistic quantum mechanics, howerver, the bras and kets take on specific mathematical properties, so as to accurately model the physics happening. Usually here have $p|~\rangle=-~\mathrm i\partial_x|~\rangle$, and $p\langle ~|=+~\mathrm i\partial_x\langle ~|$ (Factors of $\hbar$ are being suppressed here.) If we combine this with the correlation that $p|~\rangle\leftrightarrow (p\langle~ |)^\dagger$, it is simple enough to analyze the properties of the derivatives mentioned, in this context. Simply check that $-~\mathrm i\partial_x|~\rangle =(+\mathrm i\partial\langle ~|)^\dagger=(\mathrm i\partial_x)^\dagger(\langle~|)^\dagger$ So the momentum operator in this form is indeed Hermitian in QM.

As for the real derivative $\partial_x$ within the standard non-relativistic quantum mechanics scenario, one may use the simple identity that any Hermitian operator multiplied by $\mathrm i$ is an anti-Hermitian operator. $(\mathrm i\Omega_H)^\dagger=-~\mathrm i\Omega_H$.


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