On pate 89 of Dirac's book, The Principles of Quantum Mechanics, he writes:

Let us treat the linear operator $\frac{d}{dq}$ according to the general theory of linear operators of section 7. We should then be able to apply it to a bra $\langle\phi(q)$, the product $\langle\phi\frac{d}{dq}$ being defined, according to (3) of section 7 , by

$$\{\langle\phi\frac{d}{dq}\}\psi\rangle = \langle\phi\{\frac{d}{dq}\psi\rangle\}$$

for all functions $\psi(q)$.

This is fine, I understand this. He then writes

Taking representatives, we get

$$\int\langle\phi\frac{d}{dq}|q'\rangle dq'\psi(q') = \int\phi(q')dq'\frac{d\psi(q')}{dq'}$$

It looks as if he's integrated both sides wrt to the eigenvalues of $q$ and simplified it in a way I don't understand.


I believe the difficulty stems from the archaic notation. What he is trying to do is show that $-\mathrm{i}\mathrm{D}$ is Hermitian, where $\mathrm{D}=\mathrm{d}/\mathrm{d}q$. Note the first equation you wrote. It is just saying that $\mathrm{D}$ acting "backwards" on the bra is equivalent to $\mathrm{D}$ acting "forwards" on the ket. He is trying to derive an expression for $\mathrm{D}$ acting backwards, cf. Eq. (16). So insert in the first equation the completeness relation $$I=\int |q\rangle\langle q|\,\mathrm{d}q$$ In the LHS of the first equation, insert this right after the $\}$ and on the RHS insert this right before the $\{$. This gives the desired equation.

Edit in response to comment:

($\mathrm{D}$ is defined above.) I think my answer is in the spirit of the Dirac, it's just that his notation is strange to modern eyes. I skimmed the section and will try to emulate his notation.

What does the first equation say? We have a ket (vector) $\psi\rangle$, a linear operator $\mathrm{D}$ and a bra (dual vector) $\langle\phi$. Now we act on the vector with the linear operator, this gives $\mathrm{D}\psi\rangle$. Now we take the dot product with the bra, which gives $\langle\phi\{\mathrm{D}\psi\rangle\}$. Here the braces don't mean anything special, they just make clear that $\mathrm{D}$ is acting on $\psi\rangle$. Ok, it is obvious how to define $\mathrm{D}\psi\rangle$, cf. Eq. (11) in Dirac's book. What is less obvious is how to define $\langle\phi \mathrm{D}$. We define it in such a way that $\{\langle\phi\mathrm{D}\}$ is the bra that when acted on $\psi\rangle$ gives $\langle\phi\{\mathrm{D}\psi\rangle\}$. This is then written as $$\tag{1}\{\langle\phi\mathrm{D}\}\psi\rangle=\langle\phi\{\mathrm{D}\psi\rangle\}$$ which is your first equation. In modern notation, this is actually trivial, namely $$\langle\phi|\mathrm{D}|\psi\rangle(=\{\langle\phi\mathrm{D}\}\psi\rangle)=\langle\phi|\mathrm{D}|\psi\rangle(=\langle\phi\{\mathrm{D}\psi\rangle\})$$ To continue, we use Eq. (24) on page 63, which reads $$\int|\xi'\rangle\,\mathrm{d}\xi'\,\langle \xi'|=1$$ with the $1$ to be understood in the matrix sense. Here, the continuous set of eigenvectors we are interested in are the position eigenvectors. Thus, we have $$\int|q'\rangle\,\mathrm{d}q'\,\langle q'|=1$$ We now insert this into $(1)$. On the LHS, insert it right before the $\}$ and on the RHS insert it right after the $\{$. We obtain $$\int\{\langle\phi\mathrm{D}|q'\rangle\,\mathrm{d}q'\,\langle q'|\}\psi\rangle =\langle\phi\left\{\int|q'\rangle\,\mathrm{d}q'\,\langle q'|\mathrm{D}\psi\rangle\right\}$$ Removing the braces and using $\langle x|f\rangle=f(x)$, we get $$\int\langle\phi\frac{\mathrm{d}}{\mathrm{d}q}|q'\rangle \,\mathrm{d}q'\,\psi(q')=\int\phi(q')\,\mathrm{d}q'\,\frac{\mathrm{d}\psi(q')}{\mathrm{d}q}$$ where we used (11) on page 89. Now we integrate by parts on the RHS: $$\int\langle\phi\frac{\mathrm{d}}{\mathrm{d}q}|q'\rangle \,\mathrm{d}q'\,\psi(q')=-\int\frac{\mathrm{d}\phi(q')}{\mathrm{d}q}\,\mathrm{d}q'\,\psi(q')$$ By inspection, we have $$\langle\phi\frac{\mathrm{d}}{\mathrm{d}q}|q'\rangle=-\frac{\mathrm{d}\phi(q')}{\mathrm{d}q}$$ If this is to hold for all $|q'\rangle$, we must have $$\langle\phi\frac{\mathrm{d}}{\mathrm{d}q}=-\langle\frac{\mathrm{d}\phi}{\mathrm{d}q}$$ I hope this is what you meant by "in the spirit of Dirac's method".

  • $\begingroup$ Archaic notation or not, I'm looking for an answer in the spirit of Dirac's method; but thanks for your effort. $\endgroup$ – Physiks lover Apr 7 '15 at 16:35
  • $\begingroup$ @Physikslover Please see my updated answer. $\endgroup$ – Ryan Unger Apr 7 '15 at 19:33

Given a densely defined pre-closed operator $T$ on a Hilbert space $H$, you can define its transpose (more properly called the adjoint) $T^*$ by requiring it to be the operator with the property that $$(\eta,T\psi)=(T^*\eta,\psi)$$ for any $\eta$ in the domain of $T^*$ (which is dense) and $\psi$ in the domain of $T$. Using Dirac's notation we can rewrite this in the equivalent forms $$\langle\eta|T|\psi\rangle = \langle\eta|T\psi\rangle=\langle T^*\eta|\psi\rangle$$ so that we can formally define the action of $T$ on bras by $$\langle \eta|T:=\langle T^*\eta|.$$ This is to be intended as an equality between linear functionals on the (rigged) Hilbert space and therefore it should hold in the weak sense $$\langle\eta|T=\langle T^*\eta|\qquad\text{iff}\qquad\langle\eta|T|\psi\rangle =\langle T^*\eta|\psi\rangle,\qquad\forall |\psi\rangle\in H.$$ Recall that the inner product is defined as $$(\eta,\psi) = \langle\eta|\psi\rangle = \int\overline{\eta(x)}\psi(x)\text dx,$$ so that, with $T=\frac{\text d}{\text dx}$, we have $$\begin{align} \langle\eta|T|\psi\rangle &=\langle\eta|T\psi\rangle\\ &=\int\overline{\eta(x)}\psi'(x)\text dx\\ &=\langle T^*\eta|\psi\rangle. \end{align}$$ This can be paraphrased by saying that the action of $T$ on the linear functional identified by $\langle\eta|$, i.e. the map $$|\psi\rangle\mapsto\int\overline{\eta(x)}\psi(x)\text dx$$ is sent to the linear function identified by $\langle\eta|T\equiv\langle T^*\eta|$ given by $$|\psi\rangle\mapsto\int\overline{\eta(x)}\psi'(x)\text dx.$$

  • $\begingroup$ Thanks for your effort, but I was hoping for an answer based more around Dirac's procedure. Maybe you could have a quick glance at an online version of his book available on scribd.com? ;) $\endgroup$ – Physiks lover Apr 7 '15 at 16:34
  • $\begingroup$ Well this is what Dirac does, except that I have tried to use a more mathematically correct notation. On your second equation you should think of the LHS as a formal expression that defines the kernel of the functional $\langle\eta|T$, with the RHS giving you the way this formal functional acts on $\psi$. $\endgroup$ – Phoenix87 Apr 7 '15 at 16:38

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