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Path integral formulation of quantum mechanics: free particle propagator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8.

Although all the non-relativistic quantum mechanics we’ve done so far has started with the Schrödinger equation, a different approach was devised by Richard Feynman in the 1940s. The Schrödinger method requires us to find the eigenvalues (allowed energies) and eigenstates of the hamiltonian {H} and then use these to construct the unitary operator known as the propagator. For discrete energies, this propagator is

\displaystyle U\left(t\right)=\sum e^{-iEt/\hbar}\left|E\right\rangle \left\langle E\right| \ \ \ \ \ (1)

and for continuous energies, we have

\displaystyle U\left(t\right)=\int e^{-iEt/\hbar}\left|E\right\rangle \left\langle E\right|dE \ \ \ \ \ (2)

Given the state of the system at an initial time {t=0}, the general solution as a function of time is then

\displaystyle \left|\psi\left(t\right)\right\rangle =U\left(t\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (3)

Feynman’s method allows us to compute the propagator directly, without first solving the Schrödinger equation. It is known as the path integral forumulation.

The idea is based on the observation that the exponential {e^{-iEt/\hbar}} that appears in the propagator contains the ratio of two quantities with the dimensions of action, that is, energy times time. In classical mechanics, the actual trajectory of a particle is found by minimizing the action {S} over all possible paths available to the particle. The path integral formulation of quantum mechanics works in a similar way, although at first sight, it looks like a completely impractical method.

The formulation works like this, for a single particle:

  1. Find all paths available for the particle to travel between its initial point {\left(x^{\prime},t^{\prime}\right)} and its final point {\left(x,t\right)}. This is actually similar to what we do in classical mechanics, where {S} is defined as {S=\int L\;dt} where {L} is the Lagrangian. We then use the functional derivative to minimize {S} over all these paths and find the path that gives the minimum action.
  2. For each path, calculate the action {S}. (This is where things sound terribly impractical, since there are an infinite number of paths of all possible shape, so how can we find the action for all these paths? It turns out that, in most cases, we don’t need to.)
  3. Calculate the propagator as

\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=A\sum_{all\;paths}e^{iS\left[x\left(t\right)\right]/\hbar} \ \ \ \ \ (4)

 

The notation {S\left[x\left(t\right)\right]} indicates that {S} is a functional of the path {x\left(t\right)}.

The key to the success of this method is that since the action is real, the exponential {e^{iS\left[x\left(t\right)\right]/\hbar}} is an oscillatory function, so we can expect contributions from the actions for different paths to cancel each other to some extent. Although the quantum path of a particle can’t be defined precisely due to the uncertainty principle, we expect that the particle is much more likely to be found following a path that is close to the classical path, and the classical path occurs when {S\left[x\left(t\right)\right]} is a minimum. Paths sufficiently far from this minimum will tend to cancel each other, so for practical purposes, we need calculate 4 only for paths near to the classical path.

The example given by Shankar is of a particle of mass 1 gram moving from {\left(x,t\right)=\left(0,0\right)} to {\left(1,1\right)} by two different paths. In the first path, the particle moves with constant speed so {x=t}. The action is

\displaystyle S \displaystyle = \displaystyle \int_{0}^{1}L\;dt\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle \int_{0}^{1}\left(T-V\right)dt\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle \int_{0}^{1}\frac{1}{2}mv^{2}dt\ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle \frac{m}{2}\int_{0}^{1}\left(\frac{dx}{dt}\right)^{2}dt\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle \frac{m}{2}\int_{0}^{1}dt\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle \frac{m}{2} \ \ \ \ \ (10)

In the second path, we have {x=t^{2}}, so the velocity is

\displaystyle v=\frac{dx}{dt}=2t \ \ \ \ \ (11)

with associated action

\displaystyle S \displaystyle = \displaystyle \int_{0}^{1}\frac{1}{2}mv^{2}dt\ \ \ \ \ (12)
\displaystyle \displaystyle = \displaystyle 2m\int_{0}^{1}t^{2}dt\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle \frac{2m}{3} \ \ \ \ \ (14)

The guideline for when the phases of the paths start to cancel each other is when {S/\hbar} is about {\pi} out of phase with {S_{cl}/\hbar}. In this example, the second path is {\pi}out of phase with the first when

\displaystyle \left(\frac{2m}{3}-\frac{m}{2}\right)=\pi\hbar\approx3\times10^{-34}\mbox{ m}^{2}\mbox{kg s}^{-1} \ \ \ \ \ (15)

Thus for any mass larger than about {6\pi\hbar\approx1.8\times10^{-33}\mbox{ kg}} the second path will contribute essentially nothing to 4 and can be ignored. This mass is smaller than the mass of the electron.

For the free particle, we worked out the propagator earlier and found that (where we’ve generalized the earlier result for an arbitrary initial time {t^{\prime}}):

\displaystyle U\left(t,t^{\prime}\right)=\int_{-\infty}^{\infty}e^{-ip^{2}\left(t-t^{\prime}\right)/2m\hbar}\left|p\right\rangle \left\langle p\right|dp \ \ \ \ \ (16)

The matrix elements of {U} in the {x} basis are worked out by evaluating a Gaussian integral

\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right) \displaystyle = \displaystyle \int_{-\infty}^{\infty}e^{-ip^{2}\left(t-t^{\prime}\right)/2m\hbar}\left\langle x\left|p\right.\right\rangle \left\langle p\left|x^{\prime}\right.\right\rangle dp\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}e^{ip\left(x-x^{\prime}\right)/\hbar}e^{-ip^{2}\left(t-t^{\prime}\right)/2m\hbar}dp\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{m}{2\pi\hbar i\left(t-t^{\prime}\right)}}e^{im\left(x-x^{\prime}\right)^{2}/2\hbar\left(t-t^{\prime}\right)} \ \ \ \ \ (19)

We can try to estimate {U} using the path integral approach by assuming that only the classical path contributes to the propagator. For a free particle travelling between {\left(x^{\prime},t^{\prime}\right)} to {\left(x,t\right)}, the constant velocity is

\displaystyle v=\frac{x-x^{\prime}}{t-t^{\prime}} \ \ \ \ \ (20)

The Lagrangian is a constant

\displaystyle L=\frac{mv^{2}}{2}=\frac{m}{2}\left(\frac{x-x^{\prime}}{t-t^{\prime}}\right)^{2} \ \ \ \ \ (21)

The classical action is thus

\displaystyle S_{cl} \displaystyle = \displaystyle \int_{t^{\prime}}^{t}L\;dt^{\prime\prime}\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \frac{m}{2}\left(\frac{x-x^{\prime}}{t-t^{\prime}}\right)^{2}\int_{t^{\prime}}^{t}dt^{\prime\prime}\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle \frac{m}{2}\left(\frac{x-x^{\prime}}{t-t^{\prime}}\right)^{2}\left(t-t^{\prime}\right)\ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle \frac{m}{2}\frac{\left(x-x^{\prime}\right)^{2}}{t-t^{\prime}} \ \ \ \ \ (25)

The propagator in this approximation is

\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=A\exp\left[\frac{im}{2\hbar}\frac{\left(x-x^{\prime}\right)^{2}}{\left(t-t^{\prime}\right)}\right] \ \ \ \ \ (26)

 

Comparing with 19 we see that the exponential factors match; all that is left is to determine the constant {A}. To do this, we require {\lim_{t\rightarrow t^{\prime}}U\left(x,t;x^{\prime},t^{\prime}\right)=\delta\left(x-x^{\prime}\right)}, since if the time interval {t^{\prime}-t} goes to zero, the particle cannot move so must be in the same place. By comparing 26 with the form of a delta function as the limit of a gaussian integral, which is

\displaystyle \lim_{\Delta^{2}\rightarrow0}\frac{1}{\left(\pi\Delta^{2}\right)^{1/2}}\int_{-\infty}^{\infty}e^{-\left(x-x^{\prime}\right)/\Delta^{2}}dx=\delta\left(x-x^{\prime}\right) \ \ \ \ \ (27)

we see that

\displaystyle \Delta^{2}=\frac{2\hbar i\left(t-t^{\prime}\right)}{m} \ \ \ \ \ (28)

so the final propagator is the same as 19.

Dirac delta function as limit of a gaussian integral

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 1.10.

Yet another form of the Dirac delta function is as the limit of a Gaussian integral. We start with

\displaystyle  g_{\Delta}\left(x-x^{\prime}\right)=\frac{1}{\left(\pi\Delta^{2}\right)^{1/2}}e^{-\left(x-x^{\prime}\right)^{2}/\Delta^{2}} \ \ \ \ \ (1)

If {\Delta^{2}} is real and positive, we have

\displaystyle  \frac{1}{\left(\pi\Delta^{2}\right)^{1/2}}\int_{-\infty}^{\infty}e^{-\left(x-x^{\prime}\right)^{2}/\Delta^{2}}dx=1 \ \ \ \ \ (2)

Thus the area under the curve is always 1, for any real value of {\Delta^{2}}. Now as {\Delta^{2}\rightarrow0} the exponential becomes zero except when {x=x^{\prime}}. The factor {1/\left(\pi\Delta^{2}\right)^{1/2}} tends to infinity as {\Delta^{2}\rightarrow0}, but the exponential always tends to zero faster than any power of {\Delta}, so {g_{\Delta}\left(x-x^{\prime}\right)} tends to zero everywhere except at {x=x^{\prime}}. Thus it satisfies the requirements of a delta function: it is zero everywhere except when {x-x^{\prime}=0} and has an integral of 1. Thus

\displaystyle  \lim_{\Delta\rightarrow0}g_{\Delta}\left(x-x^{\prime}\right)=\delta\left(x-x^{\prime}\right) \ \ \ \ \ (3)

However, if we plug the integral into Maple without any restrictions on {\Delta^{2}}, it informs us that the integral is still 1 even if {\Delta^{2}} is pure imaginary, provided that the imaginary number is positive, that is, we can write {\Delta^{2}=i\beta^{2}} for real {\beta}. Thus it would appear that {g_{\Delta}} still gives a delta function in the limit {\Delta^{2}\rightarrow0} even if {\Delta^{2}} is a positive imaginary number.

Shankar provides a rationale for this in his footnote to equation 1.10.19. In terms of {\beta} we can integrate some smooth function {f\left(x^{\prime}\right)} multiplied by {g_{\Delta}} over a region that includes {x^{\prime}=x}.

\displaystyle  \frac{1}{\left(\pi i\beta^{2}\right)^{1/2}}\int_{-\infty}^{\infty}e^{i\left(x-x^{\prime}\right)^{2}/\beta^{2}}f\left(x^{\prime}\right)dx \ \ \ \ \ (4)

As {\beta^{2}\rightarrow0}, the exponent becomes a very large positive imaginary number everywhere except at {x=x^{\prime}}, so the exponential oscillates very rapidly. Provided that {f\left(x^{\prime}\right)} doesn’t vary as rapidly, the integral will contain equal positive and negative contributions everywhere except at {x=x^{\prime}} so in the limit of {\beta^{2}=0}, only the point {x=x^{\prime}} contributes, which means we can pull {f\left(x\right)} out of the integral and get

\displaystyle  \lim_{\beta^{2}\rightarrow0}\frac{1}{\left(\pi i\beta^{2}\right)^{1/2}}\int_{-\infty}^{\infty}e^{i\left(x-x^{\prime}\right)^{2}/\beta^{2}}f\left(x^{\prime}\right)dx=f\left(x\right) \ \ \ \ \ (5)

Thus 3 is valid for all real {\Delta} and for {\Delta^{2}} positive imaginary.

Thermodynamics of harmonic oscillators – classical and quantum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.5, Exercise 7.5.4.

One application of harmonic oscillator theory is in the behaviour of crystals as a function of temperature. A reasonable model of a crystal is of a number of atoms that vibrate as harmonic oscillators. From statistical mechanics, the probability {P\left(i\right)} of finding a system in a state {i} is given by the Boltzmann formula

\displaystyle  P\left(i\right)=\frac{e^{-\beta E\left(i\right)}}{Z} \ \ \ \ \ (1)

where {\beta=1/kT}, with {k} being Boltzmann’s constant and {T} the absolute temperature, and {Z} is the partition function

\displaystyle  Z=\sum_{i}e^{-\beta E\left(i\right)} \ \ \ \ \ (2)

The thermal average energy of the system is then

\displaystyle   \bar{E} \displaystyle  = \displaystyle  \sum_{i}E\left(i\right)P\left(i\right)\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \frac{\sum_{i}E\left(i\right)e^{-\beta E\left(i\right)}}{Z}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\partial\left(\ln Z\right)}{\partial\beta} \ \ \ \ \ (5)

For a classical harmonic oscillator, the energy is a continuous function of the position {x} and momentum {p}:

\displaystyle  E_{cl}=\frac{p^{2}}{2m}+\frac{1}{2}m\omega^{2}x^{2} \ \ \ \ \ (6)

The classical partition function is then

\displaystyle   Z_{cl} \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\beta p^{2}/2m}e^{-\beta m\omega^{2}x^{2}/2}dp\;dx\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}e^{-\beta p^{2}/2m}dp\int_{-\infty}^{\infty}e^{-\beta m\omega^{2}x^{2}/2}dx\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta m\omega^{2}}}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{2\pi}{\omega\beta} \ \ \ \ \ (10)

Where we used the standard formula for Gaussian integrals to get the third line. The average classical energy is, from 5

\displaystyle  \bar{E}_{cl}=-\frac{\partial\left(\ln Z_{cl}\right)}{\partial\beta}=\frac{1}{\beta}=kT \ \ \ \ \ (11)

The average energy of a classical oscillator thus depends only on the temperature, and not on the frequency {\omega}.

For a quantum oscillator, the energies are quantized with values of

\displaystyle  E\left(n\right)=\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (12)

The quantum partition function is therefore

\displaystyle  Z_{qu}=e^{-\beta\hbar\omega/2}\sum_{n=0}^{\infty}e^{-\beta\hbar\omega n} \ \ \ \ \ (13)

The sum is a geometric series, so we can use the standard result for {\left|x\right|<1}:

\displaystyle  \sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x} \ \ \ \ \ (14)

This gives

\displaystyle  Z_{qu}=\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}} \ \ \ \ \ (15)

The mean quantum energy is again found from 5, although this time the derivative is a bit messier, so is most easily done using Maple. However, by hand, you’d get

\displaystyle   \bar{E}_{qu} \displaystyle  = \displaystyle  -\frac{\partial\left(\ln Z_{qu}\right)}{\partial\beta}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \frac{1-e^{-\beta\hbar\omega}}{e^{-\beta\hbar\omega/2}}\left[-\frac{1}{2}\frac{\hbar\omega e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}-\frac{\hbar\omega e^{-\beta\hbar\omega/2}e^{-\beta\hbar\omega}}{\left(1-e^{-\beta\hbar\omega}\right)^{2}}\right]\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar\omega}{2}\left(\frac{1+e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\right)\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar\omega}{2}\left(\frac{1-e^{-\beta\hbar\omega}+2e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\right)\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \hbar\omega\left(\frac{1}{2}+\frac{1}{e^{\beta\hbar\omega}-1}\right) \ \ \ \ \ (20)

The average energy is the ground state energy {\hbar\omega/2} plus a quantity that increases with increasing temperature (decreasing {\beta}). For small {\beta} we have

\displaystyle   \bar{E}_{qu} \displaystyle  \rightarrow \displaystyle  \hbar\omega\left(\frac{1}{2}+\frac{1}{1+\beta\hbar\omega-1}\right)\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar\omega}{2}+\frac{1}{\beta}\ \ \ \ \ (22)
\displaystyle  \displaystyle  \rightarrow \displaystyle  kT \ \ \ \ \ (23)

since as {\beta\rightarrow0}, {\frac{1}{\beta}\gg\frac{\hbar\omega}{2}}. Thus the quantum energy reduces to the classical energy 11 for high temperatures. The ‘high temperature’ condition is that

\displaystyle   \frac{1}{\beta} \displaystyle  \gg \displaystyle  \frac{\hbar\omega}{2}\ \ \ \ \ (24)
\displaystyle  T \displaystyle  \gg \displaystyle  \frac{\hbar\omega}{2k} \ \ \ \ \ (25)

So far, we’ve considered the average behaviour of only one oscillator. Suppose we now have a 3-d crystal with {N_{0}} atoms. Assuming small oscillations we can approximate its behaviour by a system of {3N_{0}} decoupled oscillators. In the classical case, the average energy is found from 11:

\displaystyle  \bar{\mathcal{E}}_{cl}=3N_{0}\bar{E}_{cl}=3N_{0}kT \ \ \ \ \ (26)

The heat capacity per atom is the amount of heat (energy) {\Delta E} required to raise the temperature by {\Delta T}, so

\displaystyle  C_{cl}=\frac{1}{N_{0}}\frac{\partial\bar{\mathcal{E}}_{cl}}{\partial T}=3k \ \ \ \ \ (27)

For the quantum system, we have from 20

\displaystyle   \bar{\mathcal{E}}_{qu} \displaystyle  = \displaystyle  3N_{0}\bar{E}_{qu}\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  3N_{0}\hbar\omega\left(\frac{1}{2}+\frac{1}{e^{\beta\hbar\omega}-1}\right) \ \ \ \ \ (29)

The quantum heat capacity is therefore

\displaystyle   C_{qu} \displaystyle  = \displaystyle  \frac{1}{N_{0}}\frac{\partial\bar{\mathcal{E}}_{qu}}{\partial T}\ \ \ \ \ (30)
\displaystyle  \displaystyle  = \displaystyle  3\hbar\omega\frac{\partial}{\partial\beta}\left(\frac{1}{e^{\beta\hbar\omega}-1}\right)\frac{d\beta}{dT}\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  3\frac{\hbar^{2}\omega^{2}}{kT^{2}}\frac{e^{\hbar\omega/kT}}{\left(e^{\beta\hbar\omega}-1\right)^{2}} \ \ \ \ \ (32)

We can define the Einstein temperature as

\displaystyle  \theta_{E}\equiv\frac{\hbar\omega}{k} \ \ \ \ \ (33)

which gives the heat capacity as

\displaystyle  C_{qu}=3k\frac{\theta_{E}^{2}}{T^{2}}\frac{e^{\theta_{E}/T}}{\left(e^{\theta_{E}/T}-1\right)^{2}} \ \ \ \ \ (34)

For large temperatures, the exponent {\theta_{E}/T} becomes small, so we have

\displaystyle   C_{qu} \displaystyle  \underset{T\gg\theta_{E}}{\longrightarrow} \displaystyle  3k\frac{\theta_{E}^{2}}{T^{2}}\frac{1+\theta_{E}/T}{\left(1+\theta_{E}/T-1\right)^{2}}\ \ \ \ \ (35)
\displaystyle  \displaystyle  \rightarrow \displaystyle  3k \ \ \ \ \ (36)

For low temperatures {e^{\theta_{E}/T}\gg1} so we have

\displaystyle   C_{qu} \displaystyle  \underset{T\ll\theta_{E}}{\longrightarrow} \displaystyle  3k\frac{\theta_{E}^{2}}{T^{2}}\frac{e^{\theta_{E}/T}}{e^{2\theta_{E}/T}}\ \ \ \ \ (37)
\displaystyle  \displaystyle  = \displaystyle  3k\frac{\theta_{E}^{2}}{T^{2}}e^{-\theta_{E}/T} \ \ \ \ \ (38)

The heat capacity again reduces to the classical value for high temperatures. The observed behaviour at low temperatures is that {C_{qu}\rightarrow T^{3}}, so this simple model fails for very low temperatures. However, as is shown by Shankar’s figure 7.3 Einstein’s quantum model is actually quite good for all but the lowest temperatures.

Harmonic oscillator: momentum space functions and Hermite polynomial recursion relations from raising and lowering operators

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.5, Exercises 7.5.1 – 7.5.3.

Earlier, we found the position space energy eigenfunctions of the harmonic oscillator to be

\displaystyle \psi_{n}(y) \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2}\ \ \ \ \ (1)
\displaystyle \psi_{n}(x) \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (2)

where {y} in the first equation is shorthand for

\displaystyle y=\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (3)

It turns out that an alternative method for deriving these functions uses the lowering operator {a}. Shankar gives the derivation of {\psi_{n}\left(x\right)} in his section 7.5, but we can use the same technique to derive the momentum space functions. We start with the ground state and use

\displaystyle a\left|0\right\rangle =0 \ \ \ \ \ (4)

In terms of {X} and {P}, we have

\displaystyle a=\sqrt{\frac{m\omega}{2\hbar}}X+i\frac{1}{\sqrt{2m\omega\hbar}}P \ \ \ \ \ (5)

 

To find the momentum space functions, we need to express {X} and {P} in terms of {p}:

\displaystyle X \displaystyle = \displaystyle i\hbar\frac{d}{dp}\ \ \ \ \ (6)
\displaystyle P \displaystyle = \displaystyle p \ \ \ \ \ (7)

We thus have

\displaystyle \left[i\hbar\sqrt{\frac{m\omega}{2\hbar}}\frac{d}{dp}+i\frac{1}{\sqrt{2m\omega\hbar}}p\right]\psi_{0}\left(p\right)=0 \ \ \ \ \ (8)

If we define the auxiliary variable

\displaystyle z\equiv\frac{p}{\sqrt{\hbar m\omega}} \ \ \ \ \ (9)

we get

\displaystyle \left(\frac{d}{dz}+z\right)\psi_{0}\left(z\right)=0 \ \ \ \ \ (10)

This has the solution

\displaystyle \psi_{0}\left(z\right)=Ae^{-z^{2}/2} \ \ \ \ \ (11)

for some normalization constant {A}. Thus in terms of {p} we have

\displaystyle \psi_{0}\left(p\right)=Ae^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (12)

Normalizing in the usual way, making use of the Gaussian integral, we have

\displaystyle \int_{-\infty}^{\infty}\psi_{0}^{2}\left(p\right)dp \displaystyle = \displaystyle A^{2}\int_{-\infty}^{\infty}e^{-p^{2}/\hbar m\omega}dp=1\ \ \ \ \ (13)
\displaystyle A \displaystyle = \displaystyle \frac{1}{\left(\pi\hbar m\omega\right)^{1/4}} \ \ \ \ \ (14)

This agrees with the earlier result which was obtained by solving a second-order differential equation.

We can also use {a} and {a^{\dagger}} to verify a couple of recursion relations for Hermite polynomials. Reverting back to position space we have

\displaystyle X \displaystyle = \displaystyle x\ \ \ \ \ (15)
\displaystyle P \displaystyle = \displaystyle -i\hbar\frac{d}{dx} \ \ \ \ \ (16)

so 5 becomes

\displaystyle a=\sqrt{\frac{m\omega}{2\hbar}}x+\frac{\hbar}{\sqrt{2m\omega\hbar}}\frac{d}{dx} \ \ \ \ \ (17)

Also from 5 we have, since {X} and {P} are both hermitian operators

\displaystyle a^{\dagger} \displaystyle = \displaystyle \sqrt{\frac{m\omega}{2\hbar}}X-i\frac{1}{\sqrt{2m\omega\hbar}}P\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{m\omega}{2\hbar}}x-\frac{\hbar}{\sqrt{2m\omega\hbar}}\frac{d}{dx} \ \ \ \ \ (19)

Defining

\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (20)

we have

\displaystyle a \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left(y+\frac{d}{dy}\right)\ \ \ \ \ (21)
\displaystyle a^{\dagger} \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left(y-\frac{d}{dy}\right) \ \ \ \ \ (22)

We also recall the normalization conditions on the raising and lowering operators:

\displaystyle a\left|n\right\rangle \displaystyle = \displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (23)
\displaystyle a^{\dagger}\left|n\right\rangle \displaystyle = \displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (24)

Applying 23 to 1 we have, after cancelling common factors from each side:

\displaystyle \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2^{n}n!}}\left(y+\frac{d}{dy}\right)\left[H_{n}(y)e^{-y^{2}/2}\right] \displaystyle = \displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}\left(y\right)e^{-y^{2}/2}\ \ \ \ \ (25)
\displaystyle \frac{1}{2\sqrt{n}}\frac{1}{\sqrt{2^{n-1}\left(n-1\right)!}}e^{-y^{2}/2}\left[yH_{n}\left(y\right)-yH_{n}\left(y\right)+\frac{dH_{n}}{dy}\right] \displaystyle = \displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}\left(y\right)e^{-y^{2}/2}\ \ \ \ \ (26)
\displaystyle yH_{n}\left(y\right)-yH_{n}\left(y\right)+\frac{dH_{n}}{dy} \displaystyle = \displaystyle 2nH_{n-1}\left(y\right)\ \ \ \ \ (27)
\displaystyle H_{n}^{\prime}\left(y\right) \displaystyle = \displaystyle 2nH_{n-1}\left(y\right) \ \ \ \ \ (28)

Another recursion relation for Hermite polynomials can be found as follows. We start with 22 to get

\displaystyle a+a^{\dagger}=\sqrt{2}y \ \ \ \ \ (29)

We now apply 23 and 24 to 1. We can cancel common factors, including {e^{-y^{2}/2}}, from both sides to get

\displaystyle \left(a+a^{\dagger}\right)\psi_{n} \displaystyle = \displaystyle \sqrt{2}y\psi_{n}\ \ \ \ \ (30)
\displaystyle \frac{\sqrt{2}y}{\sqrt{2^{n}n!}}H_{n}(y) \displaystyle = \displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}(y)+\frac{\sqrt{n+1}}{\sqrt{2^{n+1}\left(n+1\right)!}}H_{n+1}(y)\ \ \ \ \ (31)
\displaystyle \frac{y}{\sqrt{2^{n-1}n\left(n-1\right)!}}H_{n}(y) \displaystyle = \displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}(y)+\frac{1}{2\sqrt{2^{n-1}n\left(n-1\right)!}}H_{n+1}(y)\ \ \ \ \ (32)
\displaystyle yH_{n}\left(y\right) \displaystyle = \displaystyle nH_{n-1}\left(y\right)+\frac{1}{2}H_{n+1}\left(y\right)\ \ \ \ \ (33)
\displaystyle H_{n+1}\left(y\right) \displaystyle = \displaystyle 2yH_{n}\left(y\right)-2nH_{n-1}\left(y\right) \ \ \ \ \ (34)

10 million hits

For anyone who is following such things, physicspages.com has just (in the past hour) had its 10 millionth hit. I’m still amazed and grateful that a site with so much mathematics on it has proved so popular. Many thanks to everyone who has visited.

Hamiltonian in non-rectangular coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.10.

The standard procedure for quantizing a classical hamiltonian is to write the classical hamiltonian in terms of position and momentum variables in rectangular coordinates and then convert the position and momentum variables to operators satisfying the usual commutation relations. However, in some cases, another coordinate system makes solving the differential equation resulting from the Schrödinger equation easier (as, for example, with the hydrogen atom, where the system has spherical symmetry).

As a 2-d example, suppose we have the classical hamiltonian

\displaystyle  H=\frac{p_{x}^{2}+p_{y}^{2}}{2m}+a\sqrt{x^{2}+y^{2}} \ \ \ \ \ (1)

for some constant {a}. Since the system has radial symmetry, polar coordinates should make things easier. That is, we’d like to transform to

\displaystyle   \rho \displaystyle  = \displaystyle  \sqrt{x^{2}+y^{2}}\ \ \ \ \ (2)
\displaystyle  \phi \displaystyle  = \displaystyle  \arctan\frac{y}{x} \ \ \ \ \ (3)

In the rectangular position basis, the quantized operators are

\displaystyle   P_{x} \displaystyle  = \displaystyle  -i\hbar\frac{\partial}{\partial x}\ \ \ \ \ (4)
\displaystyle  P_{y} \displaystyle  = \displaystyle  -i\hbar\frac{\partial}{\partial y}\ \ \ \ \ (5)
\displaystyle  X \displaystyle  = \displaystyle  x\ \ \ \ \ (6)
\displaystyle  Y \displaystyle  = \displaystyle  y \ \ \ \ \ (7)

so the quantum hamiltonian is

\displaystyle  H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)+a\sqrt{x^{2}+y^{2}} \ \ \ \ \ (8)

The first term contains the Laplacian derivative operator, which can be written in polar coordinates as

\displaystyle  \frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}=\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}} \ \ \ \ \ (9)

Thus the quantum hamiltonian in polar coordinates is

\displaystyle  H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+a\rho \ \ \ \ \ (10)

The question is: can we instead convert the hamiltonian 1 to polar coordinates and then quantize the result, rather than converting the rectangular coordinates after the hamiltonian is written? The answer turns out to be surprisingly complicated, and I’m not sure I follow everything Shankar says, but here’s the argument anyway. Comments, as usual, are welcome.

We first convert the rectangular momentum coordinates to polar momentum coordinates by means of the substitutions

\displaystyle   p_{\rho} \displaystyle  = \displaystyle  \hat{\mathbf{r}}\cdot\mathbf{p}=\frac{xp_{x}+yp_{y}}{\sqrt{x^{2}+y^{2}}}\ \ \ \ \ (11)
\displaystyle  p_{\phi}=\ell_{z} \displaystyle  = \displaystyle  xp_{y}-yp_{x} \ \ \ \ \ (12)

Note that the two components of polar momentum have different units: {p_{\rho}} has the dimensions of linear momentum while {p_{\phi}} is actually the angular momentum about the {z} axis {\ell_{z}}. In terms of these new momenta, the classical hamiltonian 1 becomes

\displaystyle  H=\frac{p_{\rho}^{2}}{2m}+\frac{p_{\phi}^{2}}{2m\rho^{2}}+a\rho \ \ \ \ \ (13)

This can be verified either by inverting equations 11 and 12 to get {p_{x}} and {p_{y}} in terms of {p_{\rho}} and {p_{\phi}} and then plugging these into 1 (very messy), or else just starting with 13 and showing it reduces to 1. We’ll do the latter.

\displaystyle   p_{\rho}^{2}+\frac{p_{\phi}^{2}}{\rho^{2}} \displaystyle  = \displaystyle  \left[\frac{xp_{x}+yp_{y}}{\sqrt{x^{2}+y^{2}}}\right]^{2}+\frac{\left(xp_{y}-yp_{x}\right)^{2}}{x^{2}+y^{2}}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\rho^{2}}\left(x^{2}p_{x}^{2}+y^{2}p_{y}^{2}+2xyp_{x}p_{y}+x^{2}p_{y}^{2}+y^{2}p_{x}^{2}-2xyp_{x}p_{y}\right)\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\rho^{2}}\left(x^{2}+y^{2}\right)\left(p_{x}^{2}+p_{y}^{2}\right)\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  p_{x}^{2}+p_{y}^{2} \ \ \ \ \ (17)

We can now try quantizing 13 by creating a couple of quantum momentum operators according to the standard rule:

\displaystyle   P_{\rho} \displaystyle  = \displaystyle  -i\hbar\frac{\partial}{\partial\rho}\ \ \ \ \ (18)
\displaystyle  P_{\phi} \displaystyle  = \displaystyle  -i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (19)

These operators satisfy the usual commutation rule, in the sense that

\displaystyle  \left[\rho,P_{\rho}\right]=\left[\phi,P_{\phi}\right]=i\hbar \ \ \ \ \ (20)

However, substituting them into 13 gives

\displaystyle  H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+a\rho \ \ \ \ \ (21)

Comparing with 10 we see that the middle term with the first order derivative is missing. The problem is due to the fact that 18 is actually not a hermitian operator, which we can see by calculating the bracket as follows:

\displaystyle  \left\langle \psi_{1}\left|P_{\rho}\right|\psi_{2}\right\rangle =-i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\psi_{1}^*\frac{\partial\psi_{2}}{\partial\rho}\rho\;d\rho\;d\phi \ \ \ \ \ (22)

We can do the {\rho} integral by parts and, assuming that {\rho\psi_{1}^*\psi_{2}\rightarrow0} at both {\rho\rightarrow0} and {\rho\rightarrow\infty}, we have

\displaystyle   \int_{0}^{\infty}\psi_{1}^*\frac{\partial\psi_{2}}{\partial\rho}\rho\;d\rho \displaystyle  = \displaystyle  \left.\rho\psi_{1}^*\psi_{2}\right|_{0}^{\infty}-\int_{0}^{\infty}\psi_{2}\frac{\partial\left(\rho\psi_{1}^*\right)}{\partial\rho}\;d\rho\ \ \ \ \ (23)
\displaystyle  \displaystyle  = \displaystyle  -\int_{0}^{\infty}\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho\;d\rho-\int_{0}^{\infty}\psi_{1}^*\psi_{2}d\rho \ \ \ \ \ (24)

Substituting back into 22 we get

\displaystyle  \left\langle \psi_{1}\left|P_{\rho}\right|\psi_{2}\right\rangle =i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\left[\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho+\psi_{1}^*\psi_{2}\right]d\rho\;d\phi \ \ \ \ \ (25)

If {P_{\rho}} is to be hermitian, we need to satisfy

\displaystyle   \left\langle \psi_{1}\left|P_{\rho}\right|\psi_{2}\right\rangle \displaystyle  = \displaystyle  \left\langle P_{\rho}\psi_{1}\left|\psi_{2}\right.\right\rangle \ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho\;d\rho\;d\phi \ \ \ \ \ (27)

We can see that the presence of the second term in the integrand of 25 messes things up. This term arises from the presence of the extra factor of {\rho} that is present in a polar area integral.

We can, in fact, attempt to fix this by defining the radial momentum operator to be, instead of 18:

\displaystyle  P_{\rho}=-i\hbar\left(\frac{\partial}{\partial\rho}+\frac{1}{2\rho}\right) \ \ \ \ \ (28)

We first verify that this is hermitian:

\displaystyle   \left\langle \psi_{1}\left|P_{\rho}\right|\psi_{2}\right\rangle \displaystyle  = \displaystyle  -i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\psi_{1}^*\left[\frac{\partial\psi_{2}}{\partial\rho}\rho+\frac{\psi_{2}}{2}\right]\;d\rho\;d\phi\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\left[\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho+\psi_{1}^*\psi_{2}-\frac{1}{2}\psi_{1}^*\psi_{2}\right]d\rho\;d\phi\ \ \ \ \ (30)
\displaystyle  \displaystyle  = \displaystyle  i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\left[\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho+\frac{1}{2}\psi_{1}^*\psi_{2}\right]d\rho\;d\phi\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \left\langle P_{\rho}\psi_{1}\left|\psi_{2}\right.\right\rangle \ \ \ \ \ (32)

In the second line we did the same integration by parts on the first term and used the result in 24. Thus this new {P_{\rho}} is indeed hermitian. If we now insert this along with the old {P_{\phi}} from 19 into 13 we get

\displaystyle  H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial}{\partial\rho}+\frac{1}{2\rho}\right)^{2}+a\rho \ \ \ \ \ (33)

To work out the differential part of the hamiltonian we can apply it to a test function.

\displaystyle   \left(\frac{\partial}{\partial\rho}+\frac{1}{2\rho}\right)^{2}\psi \displaystyle  = \displaystyle  \left(\frac{\partial}{\partial\rho}+\frac{1}{2\rho}\right)\left(\frac{\partial\psi}{\partial\rho}+\frac{\psi}{2\rho}\right)\ \ \ \ \ (34)
\displaystyle  \displaystyle  = \displaystyle  \frac{\partial^{2}\psi}{\partial\rho^{2}}+\frac{1}{2\rho}\frac{\partial\psi}{\partial\rho}-\frac{\psi}{2\rho^{2}}+\frac{1}{2\rho}\frac{\partial\psi}{\partial\rho}+\frac{\psi}{4\rho^{2}}\ \ \ \ \ (35)
\displaystyle  \displaystyle  = \displaystyle  \frac{\partial^{2}\psi}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial\psi}{\partial\rho}-\frac{\psi}{4\rho^{2}} \ \ \ \ \ (36)

The hamiltonian then becomes

\displaystyle  H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}-\frac{1}{4\rho^{2}}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+a\rho \ \ \ \ \ (37)

Comparing this with 10 we see that now we have an extra term {-\frac{1}{4\rho^{2}}}. Shankar doesn’t really explain in detail what the problem is, except to state that when converting from a classical to a quantum hamiltonian, terms of order {\hbar} or higher may be present in the quantum version that are absent in the classical version. Presumably he means terms of order {\hbar} that don’t involve derivatives, since the entire momentum-dependent part of the hamiltonian is multiplied by a factor of {\hbar^{2}}. In any case, we’ll have to leave it at that.

Angular momentum – Poisson bracket to commutator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.8.

The classical angular momentum components are

\displaystyle   \ell_{x} \displaystyle  = \displaystyle  yp_{z}-zp_{y}\ \ \ \ \ (1)
\displaystyle  \ell_{y} \displaystyle  = \displaystyle  zp_{x}-xp_{z}\ \ \ \ \ (2)
\displaystyle  \ell_{z} \displaystyle  = \displaystyle  xp_{y}-yp_{x} \ \ \ \ \ (3)

In the position basis, we can replace each coordinate by its quantum operator {x\rightarrow X}, {y\rightarrow Y} and {z\rightarrow Z}, and each momentum component by the derivative {p_{i}\rightarrow-i\hbar\partial/\partial q_{i}}, where {q_{i}} is the {i}th coordinate. This gives

\displaystyle   L_{x} \displaystyle  = \displaystyle  -i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\ \ \ \ \ (4)
\displaystyle  L_{y} \displaystyle  = \displaystyle  -i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\ \ \ \ \ (5)
\displaystyle  L_{z} \displaystyle  = \displaystyle  -i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) \ \ \ \ \ (6)

Because coordinates always commute with momentum components of other coordinates ({x} commutes with {p_{y}} and {p_{z}}, etc), there is no ordering ambiguity in making the transition from classical to quantum mechanics. That is, we could place the coordinate on either side of the momentum in each term for all components {L_{i}}.

Classically, we can calculate the Poisson brackets for the angular momentum components. For example

\displaystyle   \left\{ \ell_{x},\ell_{y}\right\} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\ell_{x}}{\partial q_{i}}\frac{\partial\ell_{y}}{\partial p_{i}}-\frac{\partial\ell_{x}}{\partial p_{i}}\frac{\partial\ell_{y}}{\partial q_{i}}\right)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -p_{y}\left(-x\right)-yp_{z}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  xp_{y}-yp_{z}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \ell_{z} \ \ \ \ \ (10)

According to the rule for converting classical Poisson brackets to quantum commutators, we should get (since there is no ordering ambiguity)

\displaystyle  \left[L_{x},L_{y}\right]=i\hbar L_{z} \ \ \ \ \ (11)

As we’ve seen earlier, this is verified by direct calculation using the position-momentum commutator

\displaystyle  \left[q_{i},p_{j}\right]=i\hbar\delta_{ij} \ \ \ \ \ (12)

Poisson brackets to commutators: classical to quantum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.7.

The postulates of quantum mechanics that we described earlier included specifications for the matrix elements of position {X} and momentum {P} in position space:

\displaystyle \left\langle x\left|X\right|x^{\prime}\right\rangle \displaystyle = \displaystyle x\delta\left(x-x^{\prime}\right)\ \ \ \ \ (1)
\displaystyle \left\langle x\left|P\right|x^{\prime}\right\rangle \displaystyle = \displaystyle -i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (2)

A more fundamental form of this postulate is to specify the commutation relation between {X} and {P}, which is independent of the basis and is

\displaystyle \left[X,P\right]=i\hbar \ \ \ \ \ (3)

 

This allows the construction of explicit forms of the operators in other bases, such as the momentum basis, where

\displaystyle X \displaystyle = \displaystyle i\hbar\frac{d}{dp}\ \ \ \ \ (4)
\displaystyle P \displaystyle = \displaystyle p \ \ \ \ \ (5)

We can verify this by calculating the commutator by applying it to a function {f\left(p\right)}:

\displaystyle \left[X,P\right]f \displaystyle = \displaystyle i\hbar\frac{d}{dp}\left(pf\left(p\right)\right)-i\hbar p\frac{d}{dp}f\left(p\right)\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle i\hbar f\left(p\right)+i\hbar p\frac{d}{dp}f\left(p\right)-i\hbar p\frac{d}{dp}f\left(p\right)\ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle i\hbar f\left(p\right) \ \ \ \ \ (8)

Thus 3 is satisfied in the momentum basis as well.

The standard recipe for converting a classical system to a quantum one is to first calculate the Poisson bracket for two physical quantities in the classical system, which gives

\displaystyle \left\{ \omega,\lambda\right\} =\sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right) \ \ \ \ \ (9)

where {q_{i}} and {p_{i}} are the canonical coordinates and momenta. To convert to a quantum commutator, we replace the classical quantities by their quantum operator equivalents and the Poisson bracket by {i\hbar} times the corresponding commutator. That is

\displaystyle \left[\Omega,\Lambda\right]=i\hbar\left\{ \omega,\lambda\right\} \ \ \ \ \ (10)

For the case of {X} and {P}, we have, in classical mechanics in one dimension

\displaystyle \left\{ x,p\right\} =\frac{\partial x}{\partial x}\frac{\partial p}{\partial p}-\frac{\partial x}{\partial p}\frac{\partial p}{\partial x}=1 \ \ \ \ \ (11)

 

so the quantum commutator is given by 3.

For other quantities, we can use the theorems on the Poisson brackets to reduce them:

\displaystyle \left\{ \omega,\lambda\right\} \displaystyle = \displaystyle -\left\{ \lambda,\omega\right\} \ \ \ \ \ (12)
\displaystyle \left\{ \omega,\lambda+\sigma\right\} \displaystyle = \displaystyle \left\{ \omega,\lambda\right\} +\left\{ \omega,\sigma\right\} \ \ \ \ \ (13)
\displaystyle \left\{ \omega,\lambda\sigma\right\} \displaystyle = \displaystyle \left\{ \omega,\lambda\right\} \sigma+\left\{ \omega,\sigma\right\} \lambda \ \ \ \ \ (14)

Quantum commutators obey similar rules

\displaystyle \left[\Omega,\Lambda\right] \displaystyle = \displaystyle -\left[\Lambda,\Omega\right]\ \ \ \ \ (15)
\displaystyle \left[\Omega,\Lambda+\Gamma\right] \displaystyle = \displaystyle \left[\Omega,\Lambda\right]+\left[\Omega,\Gamma\right]\ \ \ \ \ (16)
\displaystyle \left[\Omega\Lambda,\Gamma\right] \displaystyle = \displaystyle \Omega\left[\Lambda,\Gamma\right]+\left[\Omega,\Gamma\right]\Lambda \ \ \ \ \ (17)

The main difference between Poisson brackets and commutators is that, for the latter, the order of the operators in the last equation can make a difference. That is, in 14 we could also have written

\displaystyle \left\{ \omega,\lambda\sigma\right\} =\sigma\left\{ \omega,\lambda\right\} +\lambda\left\{ \omega,\sigma\right\} \ \ \ \ \ (18)

since all three quantities are numerical (not operators), so multiplication commutes. In 17 it is not true in general that, for example

\displaystyle \Omega\left[\Lambda,\Gamma\right]+\left[\Omega,\Gamma\right]\Lambda=\left[\Lambda,\Gamma\right]\Omega+\left[\Omega,\Gamma\right]\Lambda \ \ \ \ \ (19)

The conversion from classical to quantum mechanics can then be achieved in general by replacing

\displaystyle \left\{ \omega\left(x,p\right),\lambda\left(x,p\right)\right\} =\gamma\left(x,p\right) \ \ \ \ \ (20)

by

\displaystyle \left[\Omega\left(X,P\right),\Lambda\left(X,P\right)\right]=i\hbar\Gamma\left(X,P\right) \ \ \ \ \ (21)

where each of the operators in the last equation is obtained by replacing {x} in the first equation by {X} and {p} by {P}. We do need to be careful with the ordering of the operators in the quantum version, however.

As an example, suppose we have

\displaystyle \Omega \displaystyle = \displaystyle X\ \ \ \ \ (22)
\displaystyle \Lambda \displaystyle = \displaystyle X^{2}+P^{2} \ \ \ \ \ (23)

In the classical version, we calculate the Poisson bracket

\displaystyle \left\{ \omega,\lambda\right\} \displaystyle = \displaystyle \left\{ x,x^{2}+p^{2}\right\} \ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle \left\{ x,x^{2}\right\} +\left\{ x,p^{2}\right\} \ \ \ \ \ (25)
\displaystyle \displaystyle = \displaystyle 0+2\left\{ x,p\right\} p\ \ \ \ \ (26)
\displaystyle \displaystyle = \displaystyle 2p \ \ \ \ \ (27)

Thus, by our rule above, the quantum version should be

\displaystyle \left[\Omega,\Lambda\right]=2i\hbar P \ \ \ \ \ (28)

We can verify this using 17

\displaystyle \left[X,X^{2}+P^{2}\right] \displaystyle = \displaystyle \left[X,X^{2}\right]+\left[X,P^{2}\right]\ \ \ \ \ (29)
\displaystyle \displaystyle = \displaystyle 0-\left[P^{2},X\right]\ \ \ \ \ (30)
\displaystyle \displaystyle = \displaystyle -P\left[P,X\right]-\left[P,X\right]P\ \ \ \ \ (31)
\displaystyle \displaystyle = \displaystyle -P\left(-i\hbar\right)-\left(-i\hbar\right)P\ \ \ \ \ (32)
\displaystyle \displaystyle = \displaystyle 2i\hbar P \ \ \ \ \ (33)

In this case, there is no ordering ambiguity in the quantum version, since {\left[X,P\right]=i\hbar} is just a number.

For a second example, suppose we have

\displaystyle \Omega \displaystyle = \displaystyle X^{2}\ \ \ \ \ (34)
\displaystyle \Lambda \displaystyle = \displaystyle P^{2} \ \ \ \ \ (35)

The classical version gives us, using the relations 14, 11 and 27

\displaystyle \left\{ x^{2},p^{2}\right\} \displaystyle = \displaystyle -\left\{ p^{2},x^{2}\right\} \ \ \ \ \ (36)
\displaystyle \displaystyle = \displaystyle -2\left\{ p^{2},x\right\} x\ \ \ \ \ (37)
\displaystyle \displaystyle = \displaystyle 2\left\{ x,p^{2}\right\} x\ \ \ \ \ (38)
\displaystyle \displaystyle = \displaystyle 4px \ \ \ \ \ (39)

In the classical case, this result is the same as {4xp}, but because {X} and {P} don’t commute in the quantum form, we need to be careful about the ordering.

We can do the calculation:

\displaystyle \left[X^{2},P^{2}\right]=X\left[X,P^{2}\right]+\left[X,P^{2}\right]X \ \ \ \ \ (40)

From 33 we have

\displaystyle \left[X,P^{2}\right]=2i\hbar P \ \ \ \ \ (41)

so we get

\displaystyle \left[X^{2},P^{2}\right]=2i\hbar\left(XP+PX\right) \ \ \ \ \ (42)

Thus if the Poisson bracket involves a product of {p} and {x}, this should be replaced by

\displaystyle xp\mbox{ or }px\rightarrow\frac{1}{2}\left(XP+PX\right) \ \ \ \ \ (43)

in the quantum version.

Changing the position basis with a unitary transformation

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.9.

The standard representation of the position and momentum operators in the position basis is

\displaystyle   X \displaystyle  \rightarrow \displaystyle  x\ \ \ \ \ (1)
\displaystyle  P \displaystyle  \rightarrow \displaystyle  -i\hbar\frac{d}{dx} \ \ \ \ \ (2)

It turns out it’s possible to modify this definition by adding some arbitrary function of position {f\left(x\right)} to {P} so we have

\displaystyle   X^{\prime} \displaystyle  \rightarrow \displaystyle  x\ \ \ \ \ (3)
\displaystyle  P^{\prime} \displaystyle  \rightarrow \displaystyle  -i\hbar\frac{d}{dx}+f\left(x\right) \ \ \ \ \ (4)

Since any function of {x} commutes with {X}, the commutation relations remain unchanged, so we have

\displaystyle  \left[X^{\prime},P^{\prime}\right]=i\hbar \ \ \ \ \ (5)

Another way of interpreting this change in operators is by using the unitary transformation of the {X} basis, in the form

\displaystyle  \left|x\right\rangle \rightarrow\left|\tilde{x}\right\rangle =e^{ig\left(X\right)/\hbar}\left|x\right\rangle =e^{ig\left(x\right)/\hbar}\left|x\right\rangle  \ \ \ \ \ (6)

where

\displaystyle  g\left(x\right)\equiv\int^{x}f\left(x^{\prime}\right)dx^{\prime} \ \ \ \ \ (7)

The last equality in 6 comes from the fact that operating on {\left|x\right\rangle } with any function of the {X} operator (provided the function can be expanded in a power series) results in multiplying {\left|x\right\rangle } by the same function, but with the operator {X} replaced by the numeric position value.

To verify this works, we can calcuate the matrix elements of the old {X} and {P} operators in the new basis. We have

\displaystyle  \left\langle \tilde{x}\left|X\right|\tilde{x}^{\prime}\right\rangle =\left\langle x\left|e^{-ig\left(x\right)/\hbar}Xe^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle  \ \ \ \ \ (8)

At this stage, since the two exponentials are numerical functions and not operators, we can take them outside the bracket to

\displaystyle   \left\langle \tilde{x}\left|X\right|\tilde{x}^{\prime}\right\rangle \displaystyle  = \displaystyle  e^{-ig\left(x\right)/\hbar}e^{ig\left(x^{\prime}\right)/\hbar}\left\langle x\left|X\right|x^{\prime}\right\rangle \ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  e^{-ig\left(x\right)/\hbar}e^{ig\left(x^{\prime}\right)/\hbar}x^{\prime}\delta\left(x-x^{\prime}\right)\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  x\delta\left(x-x^{\prime}\right) \ \ \ \ \ (11)

The exponentials cancel in the last line since the delta function is non-zero only when {x=x^{\prime}}.

The above result can also be obtained by inserting a couple of identity operators into 8:

\displaystyle   \left\langle x\left|e^{-ig\left(x\right)/\hbar}Xe^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle \displaystyle  = \displaystyle  \int\int\left\langle x\left|e^{-ig\left(x\right)/\hbar}\right|y\right\rangle \left\langle y\left|X\right|z\right\rangle \left\langle z\left|e^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle dy\;dz\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \int\int\left\langle x\left|e^{-ig\left(x\right)/\hbar}\right|y\right\rangle z\delta\left(y-z\right)\left\langle z\left|e^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle dy\;dz\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \int\left\langle x\left|e^{-ig\left(x\right)/\hbar}\right|z\right\rangle z\left\langle z\left|e^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle dz\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \int e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\left\langle x\left|z\right.\right\rangle z\left\langle z\left|x^{\prime}\right.\right\rangle dz\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \int e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\delta\left(x-z\right)z\delta\left(z-x^{\prime}\right)dz\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}x^{\prime}\delta\left(x-x^{\prime}\right)\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  x\delta\left(x-x^{\prime}\right) \ \ \ \ \ (18)

The momentum operator works as follows. Using the original definition 2 on the modified basis we have

\displaystyle   \left\langle \tilde{x}\left|P\right|\tilde{x}^{\prime}\right\rangle \displaystyle  = \displaystyle  -i\hbar\left\langle x\left|e^{-ig\left(x\right)/\hbar}\frac{d}{dx^{\prime}}e^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle \ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  -i\hbar\left\langle x\left|e^{-ig\left(x\right)/\hbar}\frac{i}{\hbar}e^{ig\left(x^{\prime}\right)/\hbar}\frac{dg\left(x^{\prime}\right)}{dx^{\prime}}\right|x^{\prime}\right\rangle -\ \ \ \ \ (20)
\displaystyle  \displaystyle  \displaystyle  i\hbar\left\langle x\left|e^{-ig\left(x\right)/\hbar}e^{ig\left(x^{\prime}\right)/\hbar}\frac{d}{dx^{\prime}}\right|x^{\prime}\right\rangle \ \ \ \ \ (21)

From 7 we have

\displaystyle  \frac{dg\left(x\right)}{dx}=\frac{d}{dx}\int^{x}f\left(x^{\prime}\right)dx^{\prime}=f\left(x\right) \ \ \ \ \ (22)

This gives

\displaystyle   \left\langle \tilde{x}\left|P\right|\tilde{x}^{\prime}\right\rangle \displaystyle  = \displaystyle  \left\langle x\left|e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\left[f\left(x^{\prime}\right)-i\hbar\frac{d}{dx^{\prime}}\right]\right|x^{\prime}\right\rangle \ \ \ \ \ (23)
\displaystyle  \displaystyle  = \displaystyle  e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\left[f\left(x^{\prime}\right)-i\hbar\frac{d}{dx^{\prime}}\right]\left\langle x\left|x^{\prime}\right.\right\rangle \ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\left[f\left(x^{\prime}\right)-i\hbar\frac{d}{dx^{\prime}}\right]\delta\left(x-x^{\prime}\right)\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  \left[f\left(x\right)-i\hbar\frac{d}{dx}\right]\delta\left(x-x^{\prime}\right) \ \ \ \ \ (26)

This shows that by a unitary change of {X} basis 6, we transform the position and momentum operators (well, just the momentum operator, really) according to 3. We’ve multiplied the original {\left|x\right\rangle } states by a phase factor which depends on some function {f\left(x\right)}. This doesn’t change the matrix elements of {X}, but it does add {f\left(x\right)} to the matrix elements of {P}. The commonly used definition of {P} is thus with {f\left(x\right)=0}.

Harmonic oscillator – raising and lowering operators as functions of time

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.6.

We’ll consider here the problem of finding the averages of the raising and lowering operators (from the harmonic oscillator) as functions of time, that is, we want to find {\left\langle a\left(t\right)\right\rangle } and {\left\langle a^{\dagger}\left(t\right)\right\rangle }. At first glance we might think they are both zero, since they are defined in terms of position and momentum as

\displaystyle   a^{\dagger} \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\hbar m\omega}}\left[-iP+m\omega X\right]\ \ \ \ \ (1)
\displaystyle  a \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\hbar m\omega}}\left[iP+m\omega X\right] \ \ \ \ \ (2)

and the averages of {P} and {X} in any of the energy eigenstates of the harmonic oscillator are all zero. However, suppose we have a mixed state {\left|\psi\right\rangle } which can be written as a sum over the eigenstates as

\displaystyle   \psi\left(t\right) \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}c_{n}e^{-iE_{n}t/\hbar}\left|n\right\rangle \ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left|n\right\rangle \ \ \ \ \ (4)

where in the second line we used the energies of the oscillator as

\displaystyle  E_{n}=\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (5)

We now have

\displaystyle   \left\langle a\left(t\right)\right\rangle \displaystyle  = \displaystyle  \left\langle \psi\left|a\right|\psi\right\rangle \ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*e^{i\left(2m+1\right)\omega t/2}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left\langle m\left|a\right|n\right\rangle \ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\left\langle m\left|a\right|n\right\rangle \ \ \ \ \ (8)

We can now use the formula

\displaystyle  a\left|n\right\rangle =\sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (9)

This gives

\displaystyle   \left\langle a\left(t\right)\right\rangle \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n}\left\langle m\left|n-1\right.\right\rangle \ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n}\delta_{m,n-1}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  e^{-i\omega t}\sum_{n=0}^{\infty}c_{n-1}^*c_{n}\sqrt{n}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  e^{-i\omega t}\left\langle a\left(0\right)\right\rangle \ \ \ \ \ (13)

Note that if {\left|\psi\right\rangle } is an eigenstate, then only one of the coefficients {c_{n}} is non-zero, so {\left\langle a\left(0\right)\right\rangle =0} as we’d expect.

The derivation for {\left\langle a^{\dagger}\left(t\right)\right\rangle } is similar:

\displaystyle   \left\langle a^{\dagger}\left(t\right)\right\rangle \displaystyle  = \displaystyle  \left\langle \psi\left|a^{\dagger}\right|\psi\right\rangle \ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*e^{i\left(2m+1\right)\omega t/2}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left\langle m\left|a^{\dagger}\right|n\right\rangle \ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\left\langle m\left|a^{\dagger}\right|n\right\rangle \ \ \ \ \ (16)

We can now use the formula

\displaystyle  a^{\dagger}\left|n\right\rangle =\sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (17)

This gives

\displaystyle   \left\langle a^{\dagger}\left(t\right)\right\rangle \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n+1}\left\langle m\left|n+1\right.\right\rangle \ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n+1}\delta_{m,n+1}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  e^{i\omega t}\sum_{n=0}^{\infty}c_{n+1}^*c_{n}\sqrt{n+1}\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  e^{i\omega t}\left\langle a^{\dagger}\left(0\right)\right\rangle \ \ \ \ \ (21)