I'm facing some conceptual doubts related to the resolution of identity in Quantum mechanics while converting from Bra-ket notation to an integration.

For example, say we have an operator $\hat{A}$, such that we know $\langle\hat{A}\rangle = \frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle}$.

We know the Identity operator such that : $\space\hat{1} = \int dx|x\rangle\langle x|$.

We can input this, in the expression for expectation value, to get an integral of the form $\langle\hat{A}\rangle=\int\psi^*(x)f(x)\psi(x)dx$.

However, say we have a new basis $u=g(x).$, and our wavefunction has become $\psi(u)=\psi(g(x))$. I now want to find the new expectation value for the operator $\hat{A}$. One simple way would be to write the integral in $x$ basis and then use the chain rule to convert everything in terms of $u$ and evaluate the integral. However, I want to do the integral directly in terms of $u$.

Can we directly say $\hat{1} = \int du|u\rangle\langle u|$ ? If I want to write this integral in this $u$ basis, how should I proceed?

For example, $$\langle\hat{A}\rangle = \frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle} = \frac{\langle\psi|\int du |u\rangle\langle u|\hat{A}\int du'|u'\rangle\langle u'|\psi\rangle}{\langle\psi\int|u\rangle\langle u|\psi\rangle}=\frac{\int du'\int du \space\psi^*(u) h(u)\psi(u')\delta(u-u')}{\int du\space \psi^*(u)\psi(u)}$$

Here $h(u)$ is the representation of the operator in the $u$ basis. However, I run into a small problem here.

If I evaluate the above integral, I get the form :

$$\langle\hat{A}\rangle = \frac{\int \psi^*(u)h(u)\psi(u)du}{\int \psi^*(u)\psi(u) du}$$

This form doesn't seem to be correct at all. For example, I can derive the same expression in a different way i.e. using the chain rule. In that case, we would obtain :

$$\langle \hat{A}\rangle = \frac{\int\psi^*(g(x))f(x)\psi(g(x))dx}{\int\psi^*(g(x))\psi(g(x))dx}=\frac{\int\psi^*(u)f\space o\space g^{-1}(u)\psi(u)\frac{du}{g' \space o \space g^{-1}(u)}}{\int\psi^*(u)\psi(u)\frac{du}{g' \space o \space g^{-1}(u)}}$$

As we can clearly compare $h(u) = f\space o\space g^{-1}(u)$. However, where is the $g' \space o\space g^{-1}(u)$ term, present under the differential $du$ come from ?

Using the chain rule,I'm obtaining a different expression to the one I obtained by directly trying to represent the expectation value in the $u=g(x)$ basis, in Dirac notation.

This leads me to believe that my assertion that $\space\hat{1} = \int dx|x\rangle\langle x|$ automatically implies that $\hat{1} = \int du|u\rangle\langle u|$ is incorrect. Can anyone show me where that extra factor under the integral is coming from without using the chain rule. In a way, I'm asking how does Dirac notation encapsulate the Chain rule in the integration change of variables. I can easily notice how the Dirac notation encapsulates the change in the basis for the operator. The operator $f(x)$ in $x$ basis becomes $h(u)=f\space o\space g^{-1}(u)$ in the $u=g(x)$ basis.However, why does $dx$ doesn't directly become $du$. Instead, there is a factor dividing it. This is quite obvious using the chain rule. However, I don't see how this happens in Dirac notation, when I try to represent the integral in terms of $u$ directly.


1 Answer 1


the question is a question of regularization and normalization of the wave-functions and basis vectors, which for continuous sets is not immediatly clear.

Let's try to see what is needed for $1 = \int du |u\rangle \langle u|$ to happen $$ \int du |u\rangle \langle u| = \int dx dx' du |x\rangle \langle x | u \rangle \langle u | x' \rangle \langle x' | = \int dx dx' du u(x)u^*(x') |x\rangle \langle x'| \stackrel{?}{=} \int dx |x\rangle \langle x|$$ so we want $$\int du u(x) u^*(x') = \delta(x-x')$$ which already sets a normalization and orthogonality constraint on $u(x)$, not all $u$ will automatically maintain that!

Some such changes of basis will indeed only result in a constant factor, for example his is what happens if we will move to a momentum-basis like set of functions $u(x) = \exp(i u x)$ and then $\int du u(x) u^*(x') = 2\pi \delta(x-x')$. So we will just have to divide by $2\pi$ for every time we insert the "identity". However some such changes might be more problematic, so it really depends on your choice of $u(x)$.

  • $\begingroup$ Thanks ! In case of the momentum basis transformation, I'm dividing by this constant every time I insert the identity. In case of some complicated basis transformation, I'd have to insert some function ( which may be a constant ), every time I introduce the identity operator. Is that correct ? $\endgroup$ Oct 25, 2021 at 11:15
  • $\begingroup$ That is where that extra term under the differential $du$ comes from, right ? $\endgroup$ Oct 25, 2021 at 11:16
  • $\begingroup$ yes, the term comes from the Jacobian when we change variables of integration, or from the proper normalization of the basis of $|u\rangle$ states $\endgroup$
    – user275556
    Oct 25, 2021 at 12:17
  • $\begingroup$ Yeah, in the case of the jacobian, it is quite obvious when one uses the change of variables method. However, using the Dirac notation directly in deriving the integral, seems a lot more difficult. For example, I suppose we can say something like $\int dx |x\rangle\langle x|= \hat{1}$ implies that $\int \frac{du\space |u\rangle\langle u|}{h(u)} =\hat{1}$, where $h(u)$ is some function or (some constant as we saw in case of momentum basis). This is true when $u=g(x)$. Hence, any time we insert $\hat{1}$, we need to divide the integral by $h(u)$. Is this an acceptable summary? $\endgroup$ Oct 25, 2021 at 13:20
  • $\begingroup$ $\int du |u\rangle \langle u| \propto I$ only if $|u\rangle$ satisfies the condition that it is a complete (or over-complete) set. The coefficient as you write it $h(u)$ implies that it depends on $u$ but that cannot be as it was integrated over (unless it means that it depends on the properties of the entire set of $|u\rangle$). But yes, if you have $|u\rangle$ such that $\int du |u\rangle \langle u| = A I $ for some fixed $A$, you just need to divide by it every time you insert the identity $\endgroup$
    – user275556
    Oct 25, 2021 at 13:30

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