I am working through Shankar's Principles of Quantum Mechanics and I am confused about his derivation for the propagator of a free particle from the Schrodinger Equation (Section 5.1).

He states that the propagator is given by the following equation (in terms of an integral over $p$):

$$U(t) = \int_{-\infty}^\infty \lvert p\rangle\langle p\rvert \exp(-ip^2t/2m\hbar)dp\tag{5.1.9}$$

and in Exercise 5.1.1 he states that the propagator, in terms of an integral over $E$, is:

$$U(t) = \sum_{\alpha = \pm} \int_0^\infty \biggl[\frac{m}{\sqrt{2mE}}\biggr]\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}dE \tag{1}$$

However, from 4.3, the propagator appears to be defined as a sum over $E$ (the eigenvalues of the Hamiltonian):

$$U(t) = \sum_{E} \lvert E\rangle\langle E\rvert e^{-iEt/{\hbar}}\tag{4.3.13}$$

which, when $E$ is continuous, can presumably be written in integral form:

$$ U(t) = \int_{-\infty}^{\infty} \lvert E\rangle\langle E\rvert e^{-iEt/\hbar}dE \tag{2}$$

I am having difficulty reconciling these two forms of the propagator (1 and 2 above), as they appear to contradict. Surely, based on the definition given in 4.3, the propagator for a free particle should be given by:

$$U(t) = \sum_{\alpha = \pm} \int_0^\infty \lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}dE$$

and the extra $\bigl[\frac{m}{\sqrt{2mE}}\bigr]$ term should be present in the $p$ version of the equation?

Note 1:

I am aware of an existing question Free particle Schrodinger equation: propagator that is similar. However, I find the accepted answer not very satisfactory. It's quite brief and doesn't seem to explain the whole story. It suggests there should be an additional term in the propagator integrand to account for a 'Density of States'.

So, what is happening here? Has Shankar glossed over this, in his earlier description of the propagator? If so, why would he do that? Does he address this point anywhere else in his book (as it seems fairly important)?

Can anyone explain more clearly why this Density of States is required and where it comes from? I thought the degeneracy of $H$ was being accounted for by summing over the +/- values of $p$. Why would an extra term be required?

Note 2:

Also, to clarify, I am aware that the extra term comes from changing the integral from $dp$ to $dE$:

$$p = \pm (2mE)^{1/2}$$ $$dp = \frac{m}{(2mE)^{1/2}}dE$$

However, I am confused as to why this factor appears in the integral over $E$, rather then the integral over $p$.

  • 1
    $\begingroup$ The formula with the "density of states" is the most general, correct formula. In the case when you have a discrete spectrum with eigenvalues $E_n$ you have $\rho(E) = \sum_n \delta(E-E_n)$ which gives you your equation between $(1)$ and $(2)$. For the free particle, the density of states is the ``extra" term you wrote. $\endgroup$ – childofsaturn Jan 27 at 0:56
  • $\begingroup$ Hint: how do you change the measure from p to E? Do it in your question to reassure the reader that's not where you are stuck. $\endgroup$ – Cosmas Zachos Jan 27 at 1:10
  • $\begingroup$ @CosmasZachos I am aware that term comes from changing the integral from dp to dE. I have edited the question to clarify, but I don't think that is where I am stuck. $\endgroup$ – Time4Tea Jan 27 at 2:03
  • $\begingroup$ @childofsaturn ok, thanks for your comment. Can you point to a good reference that derives the form of the propagator that includes the $\rho (E)$? $\endgroup$ – Time4Tea Jan 27 at 2:05
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    $\begingroup$ Your Eqn 2 is simply incorrect. Is that an equation from Shankar, or did you come up with it yourself? If you came up with it yourself, try working through why you think it’s true. $\endgroup$ – Jahan Claes Jan 27 at 2:45

Overall, I think that this is a matter of normalization and dimensions. The propagator, of course, is dimensionless.

When we have a system with discreet energy levels, we use the convention $\langle E | E' \rangle = \delta_{E,E'}$ which is dimensionless, and also $\sum_E |E\rangle \langle E| = 1$. Then, writing the propagator using the sum expression you have (numberless) is correct, as can be seen by inserting the identity operator when evaluating the matrix element.

When we have a continuous system, more care is needed. In your Eq.(1) (taken from Shankar, I assume), the notation $|E\rangle$ denotes an eigenstate of the energy but is also an eigenstate of the momentum, and the normalization is $\langle E | E' \rangle = \delta(p-p')$. That is, the dimensions of the state is ${p}^{-1/2}$ and when taking the integral (which itself has dimensions due to the $dp$ factor) something has to come to take care for normalization. In your Eq.(2), it is implicitly assumed that $\langle E | E' \rangle = \delta(E-E')$, which is different normalization, which will work well as $|E\rangle\langle E|dE$ is dimensionless.

The transformation between these two conventions of normalization is exactly the density of state, as we always require $\int\! dx \rho(x)|x\rangle \langle x|= I$. This is where the factor is coming from - we can change the integration measure "by hand" or by incorporating it into the normalization of the states.

Moving from sums to integrals can be tricky, and one should always note the dimensions and normalization.

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  • $\begingroup$ Thanks, this has been very helpful. I think I understand it now. So, basically, it is assumed that $|p\rangle$ is normalized, and then $|E\rangle$ is normalized based on that, because they are the same eigenstates? $\endgroup$ – Time4Tea Jan 27 at 22:03
  • $\begingroup$ Yes, for a free particle the energy eigenstate is also a momentum eigenstate, and the normalization with respect to the momentum basis was kept. $\endgroup$ – user245141 Jan 28 at 8:04

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