# Differentiation operator with respect to observable acting as a function of the observable?

In his Principles of Quantum Mechanics Dirac writes: $$\int \langle \phi \frac{d}{dq}|q'\rangle dq' \psi(q')=\int \phi(q') dq' \frac{d\psi(q')}{dq'}.$$

To me it is rather strange, and it seems as if he was treating the operator $\frac{d}{dq}$ as a function of the observable canonical coordinate $q$, beacuse for functions of observables he gave the definition: $$f(\xi)|\xi'\rangle=f(\xi')|\xi'\rangle,$$ where $|\xi'\rangle$ is an eigenket of the observable $\xi$. Using this analogy, taking $f(q)=\frac{d}{dq}$ one could write $$\frac{d}{dq}|q'\rangle=\frac{d}{dq'}|q'\rangle$$ but then we would also need to differentiate $|q'\rangle$, since it is a function in $q'$.

So what happens here? Please explain!

Also, as a side question: why does $$\int \langle \phi \frac{d}{dq}|q'\rangle dq' \psi(q')=-\int \frac{d\phi(q')}{dq'} dq' \psi(q')$$ imply $$\langle \phi \frac{d}{dq}|q'\rangle=-\frac{d\phi(q')}{dq'}~?$$ Simply because the results of the integrations equal, that doesn't mean that their arguments also equal.

The book can be accessed here. The formulas are on page 90, using the books original numbering.

You have to interpret $|\frac{d}{dq} \psi\rangle$. Knowing that decomposition of the basis $|q'\rangle$ gives :

$$|\psi\rangle = \int dq' \psi(q') |q'\rangle \tag{1}$$ You have :

$$|\frac{d}{dq}\psi\rangle = \int dq' \frac {d\psi(q')}{dq'} |q'\rangle\tag{2}$$

So, applying it to $|\psi\rangle = |q"\rangle = \int dq' \delta(q"-q') |q'\rangle$, you get :

$$|\frac{d}{dq}q"\rangle = - \int dq' \delta'(q"-q') |q'\rangle\tag{3}$$

So, for instance, you have :

$$\int dq" \langle \phi |\frac{d}{dq} q"\rangle \psi(q") = -\int dq"dq'\delta'(q"- q') \langle \phi|q'\rangle \psi(q")\\= -\int dq"dq'\delta'(q"- q') \phi(q') \psi(q")=-\int dq"dq'\delta(q"- q') \frac{d\phi(q')}{dq'} \psi(q")\\=-\int dq' \frac{d\phi(q')}{dq'} \psi(q') \tag{4}$$

This corresponds to the formula $(15)$ in the Dirac paper, and we have made use of integration by parts relatively to the variable $q'$ on the second line of the equations.

For your last side question, as soon as you have the equality $\int dq' f(q')\psi(q')=\int dq' g(q')\psi(q')$, for all possible functions $\psi(q')$, the only possibility is $f(q')=g(q')$

The operator $d/dx$ isn't a "function" and Dirac surely never claims so. It's an operator, something that changes one function to another. By a function, we mean something that maps one number to another.

Functions of $x$, like $f(x)$, may also be connected with operators on the space of (wave) functions. The wave function $\psi(x)$ is mapped to $f(x)\psi(x)$, a product of two functions of $x$, by this operator.

All operators may be written as something on the left side from $\psi(x)$. But that doesn't mean that all operators are functions. In other words, the sentence "[it is a function] because for functions of observables he gave the definition..." is based on faulty logic. The fact that two objects share a property isn't sufficient to say that they're the same.

The operator $d/dx$ is an operator differentiating with respect to a particular variable $x$ but of course that if we use it as a momentum (after we multiply it by $-i\hbar$ etc.), we want to differentiate with respect to relevant variable one which the relevant wave function depends. Sometimes it's called $q$, sometimes $q'$.

The "equality of the integrals" implies the "equality between the integrands" because the "equality between the integrals" has been proven for every function $\psi(q)$. That's why the coefficient in front of every $\psi(q')$ has to vanish separately – because you may always assume $\psi(q)$ to be a delta-function localized at the point $q'$. The second equation simply extracts the coefficients in front of $\psi(q')$.

• I've written "function of observable". Functions can definitely be used not only to map numbers to one another but observables, or basically anything. Functions are generally meant as mappings between the elements of two sets with the property that to one element of the domain only one element of the codomain may belong and not more (the converse is not required, ie. it may not be invertible). Eg. operators are special kind of functions. I wanted to ask whether $\frac{d}{dq}$ may be treated as a function of the observable $q$. The answer to myself is not really, but I haven't had a better idea – user3237992 Aug 1 '14 at 10:54
• Dear @user3237992, I know that the word function in modern algebra has this general meaning, but if you want to follow Dirac's book, you should probably avoid clashing terminology. In Dirac's book, a function is a function of a number giving a number. ... There exists no formalism or axiomatic system in which $d/dq$ would be a "function of the observable $q$. It's simply not. All functions of $q$ commute with $q$ but $d/dq$ doesn't which proves that it is not a function of $q$. – Luboš Motl Aug 1 '14 at 11:56
• Dirac uses functions which act on operators, altough it's true he introduces them via functions of numbers. If $\xi$ is an observable with eigenkets $|\xi'\rangle$, let $f(\xi)$ be the observable that has the following property: $f(\xi)|\xi'\rangle=f(\xi')|\xi'\rangle$. This is enough to define $f(\xi)$, because for every ket $|P\rangle$: $f(\xi)|P\rangle=f(\xi)(\sum_r |\xi^r\rangle+\int|\xi'\rangle d\xi')=\sum_r f(\xi^r)|\xi^r\rangle+\int f(\xi')|\xi'\rangle d\xi'=|Q\rangle$. $|Q\rangle$ is linearly dependent on $|P\rangle$, thus it can be viewed as a linear operator acting on $|P\rangle$ – user3237992 Aug 4 '14 at 8:02

First you realize:

$$\int\limits |q' \rangle dq' \langle q'|=1$$

Then using the definition of a linear operator on a bra:

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

Applying the first equation to both sides of the second one (at the right of the expression in parenthesis for the left side of the equation and at the left of the expression in parenthesis for the right hand side) yields:

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

You consider the derivative as a function of the q with diracs definition. The second part and the last part of the last equation form the exact equation you were asking about.