Tag Archives: free particle

Propagator for a Gaussian wave packet for the free particle

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.1, Exercise 5.1.3.

The propagator for the free particle is

\displaystyle  U\left(t\right)=\int_{-\infty}^{\infty}e^{-ip^{2}t/2m\hbar}\left|p\right\rangle \left\langle p\right|dp \ \ \ \ \ (1)

We can find its matrix elements in position space by using the position space form of the momentum

\displaystyle  \left\langle x\left|p\right.\right\rangle =\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar} \ \ \ \ \ (2)

Taking the matrix element of 1 we have

\displaystyle   U\left(x,t;x^{\prime}\right) \displaystyle  = \displaystyle  \left\langle x\left|U\left(t\right)\right|x^{\prime}\right\rangle \ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \int\left\langle x\left|p\right.\right\rangle \left\langle p\left|x^{\prime}\right.\right\rangle e^{-ip^{2}t/2m\hbar}dp\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2\pi\hbar}\int e^{ip\left(x-x^{\prime}\right)/\hbar}e^{-ip^{2}t/2m\hbar}dp\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{m}{2\pi\hbar it}}e^{im\left(x-x^{\prime}\right)^{2}/2\hbar t} \ \ \ \ \ (6)

The final integral can be done by combining the exponents in the third line, completing the square and using the standard formula for Gaussian integrals. We won’t go through that here, as our main goal is to explore the evolution of an initial wave packet using the propagator. Given 6, we can in principle find the wave function for all future times given an initial wave function, by using the propagator:

\displaystyle  \psi\left(x,t\right)=\int U\left(x,t;x^{\prime}\right)\psi\left(x^{\prime},0\right)dx^{\prime} \ \ \ \ \ (7)

Here, we’re assuming that the initial time is {t=0}. Shankar uses the standard example where the initial wave packet is a Gaussian:

\displaystyle  \psi\left(x^{\prime},0\right)=e^{ip_{0}x^{\prime}/\hbar}\frac{e^{-x^{\prime2}/2\Delta^{2}}}{\left(\pi\Delta^{2}\right)^{1/4}} \ \ \ \ \ (8)

This is a wave packet distributed symmetrically about the origin, so that {\left\langle X\right\rangle =0}, and with mean momentum given by {\left\langle P\right\rangle =p_{0}}. By plugging this and 6 into 7, we can work out the time-dependent version of the wave packet, which Shankar gives as

\displaystyle  \psi\left(x,t\right)=\left[\sqrt{\pi}\left(\Delta+\frac{i\hbar t}{m\Delta}\right)\right]^{-1/2}\exp\left[\frac{-\left(x-p_{0}t/m\right)^{2}}{2\Delta^{2}\left(1+i\hbar t/m\Delta^{2}\right)}\right]\exp\left[\frac{ip_{0}}{\hbar}\left(x-\frac{p_{0}t}{2m}\right)\right] \ \ \ \ \ (9)

Again, we won’t go through the derivation of this result as it involves a messy calculation with Gaussian integrals again. The main problem we want to solve here is to use our alternative form of the propagator in terms of the Hamiltonian:

\displaystyle  U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (10)

For the free particle

\displaystyle  H=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} \ \ \ \ \ (11)

so if we expand {U\left(t\right)} as a power series, we have

\displaystyle  U\left(t\right)=\sum_{s=0}^{\infty}\frac{1}{s!}\left(\frac{i\hbar t}{2m}\right)^{s}\frac{d^{2s}}{dx^{2s}} \ \ \ \ \ (12)

To see how we can use this form to generate the time-dependent wave function, we’ll consider a special case of 8 with {p_{0}=0} and {\Delta=1}, so that

\displaystyle   \psi_{0}\left(x\right) \displaystyle  = \displaystyle  \frac{e^{-x^{2}/2}}{\pi^{1/4}}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\pi^{1/4}}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{2^{n}n!} \ \ \ \ \ (14)

We therefore need to apply one power series 12 to the other 14. This is best done by examining a few specific terms and then generalizing to the main result. To save writing, we’ll work with the following

\displaystyle   \alpha \displaystyle  \equiv \displaystyle  \frac{i\hbar t}{m}\ \ \ \ \ (15)
\displaystyle  \psi_{\pi}\left(x\right) \displaystyle  \equiv \displaystyle  \pi^{1/4}\psi_{0}\left(x\right) \ \ \ \ \ (16)

The {s=0} term in 12 is just 1, so we’ll look at the {s=1} term and apply it to 14:

\displaystyle   \frac{\alpha}{2}\frac{d^{2}}{dx^{2}}\left[\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{2^{n}n!}\right] \displaystyle  = \displaystyle  \frac{\alpha}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)\left(2n-1\right)x^{2n-2}}{2^{n}n!}\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  \frac{\alpha}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)!x^{2n-2}}{2^{n}n!\left(2n-2\right)!} \ \ \ \ \ (18)

We can simplify this by using an identity involving factorials:

\displaystyle   \frac{\left(2n\right)!}{n!} \displaystyle  = \displaystyle  \frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)\ldots\left(2\right)\left(1\right)}{n\left(n-1\right)\left(n-2\right)\ldots\left(2\right)\left(1\right)}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \frac{2^{n}\left[n\left(n-1\right)\left(n-2\right)\ldots\left(2\right)\left(1\right)\right]\left[\left(2n-1\right)\left(2n-3\right)\ldots\left(3\right)\left(1\right)\right]}{n!}\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  \frac{2^{n}n!\left(2n-1\right)!!}{n!}\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  2^{n}\left(2n-1\right)!! \ \ \ \ \ (22)

The ‘double factorial’ notation is defined as

\displaystyle  \left(2n-1\right)!!\equiv\left(2n-1\right)\left(2n-3\right)\ldots\left(3\right)\left(1\right) \ \ \ \ \ (23)

That is, it’s the product of every other term from {n} down to 1. Using this result, we can write 18 as

\displaystyle  \frac{\alpha}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)!x^{2n-2}}{2^{n}n!\left(2n-2\right)!}=\alpha\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(2n-1\right)!!x^{2n-2}}{2\left(2n-2\right)!} \ \ \ \ \ (24)

Now look at the {s=2} term from 12.

\displaystyle   \frac{1}{2!}\frac{\alpha^{2}}{2^{2}}\frac{d^{4}}{dx^{4}}\left[\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{2^{n}n!}\right] \displaystyle  = \displaystyle  \frac{1}{2!}\frac{\alpha^{2}}{2^{2}}\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)x^{2n-4}}{2^{n}n!}\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2!}\frac{\alpha^{2}}{2^{2}}\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)!x^{2n-4}}{2^{n}n!\left(2n-4\right)!}\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  \frac{\alpha^{2}}{2^{2}2!}\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}\left(2n-1\right)!!x^{2n-4}}{\left(2n-4\right)!} \ \ \ \ \ (27)

We can see the pattern for the general term for arbitrary {s} from 12 (we could prove it by induction, but hopefully the pattern is fairly obvious):

\displaystyle   \frac{1}{s!}\frac{\alpha^{s}}{2^{s}}\frac{d^{2s}}{dx^{2s}}\left[\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{2^{n}n!}\right] \displaystyle  = \displaystyle  \frac{1}{s!}\frac{\alpha^{s}}{2^{s}}\sum_{n=s}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)!x^{2n-2s}}{2^{n}n!\left(2n-2s\right)!}\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  \frac{\alpha^{s}}{2^{s}s!}\sum_{n=s}^{\infty}\frac{\left(-1\right)^{n}\left(2n-1\right)!!x^{2n-2s}}{\left(2n-2s\right)!} \ \ \ \ \ (29)

Now we can collect terms for each power of {x}. The constant term (for {x^{0}}) is the first term from each series for each value of {s}, so we have, using the general term 29 and taking the first term where {n=s}:

\displaystyle  \sum_{s=0}^{\infty}\frac{\left(-1\right)^{s}\alpha^{s}\left(2s-1\right)!!}{2^{s}s!}=1-\frac{\alpha}{2}+\frac{\alpha^{2}}{2!}\frac{3}{2}\frac{1}{2}-\frac{\alpha^{3}}{3!}\frac{5}{2}\frac{3}{2}\frac{1}{2}+\ldots \ \ \ \ \ (30)

[The {\left(2s-1\right)!!} factor is 1 when {s=0} as we can see from the result 22.] The series on the RHS is the Taylor expansion of {\left(1+\alpha\right)^{-1/2}}, as can be verified using tables.

In general, to get the coefficient of {x^{2r}} (only even powers of {x} occur in the series), we take the term where {n=s+r} from 29 and sum over {s}. This gives

\displaystyle   \sum_{s=0}^{\infty}\frac{\alpha^{s}}{2^{s}s!}\frac{\left(-1\right)^{s+r}\left(2s+2r-1\right)!!}{\left(2r\right)!} \displaystyle  = \displaystyle  \frac{\left(-1\right)^{r}}{2^{r}r!}\sum_{s=0}^{\infty}\frac{\alpha^{s}}{2^{s}s!}\frac{\left(-1\right)^{s}\left(2s+2r-1\right)!!}{\left(2r-1\right)!!} \ \ \ \ \ (31)

where we used 22 to get the RHS. Expanding the sum gives

\displaystyle   \sum_{s=0}^{\infty}\frac{\alpha^{s}}{2^{s}s!}\frac{\left(-1\right)^{s}\left(2s+2r-1\right)!!}{\left(2r-1\right)!!} \displaystyle  = \displaystyle  1-\alpha\frac{2r+1}{2}+\frac{\alpha^{2}}{2!}\left(\frac{2r+3}{2}\right)\left(\frac{2r+1}{2}\right)-\ldots\ \ \ \ \ (32)
\displaystyle  \displaystyle  = \displaystyle  1-\alpha\left(r+\frac{1}{2}\right)+\frac{\alpha^{2}}{2!}\left(r+\frac{3}{2}\right)\left(r+\frac{1}{2}\right)-\ldots\ \ \ \ \ (33)
\displaystyle  \displaystyle  = \displaystyle  \left(1+\alpha\right)^{-r-\frac{1}{2}} \ \ \ \ \ (34)

where again we’ve used a standard series from tables (given by Shankar in the problem) to get the last line. Combining this with 31, we see that the coefficient of {x^{2r}} is

\displaystyle  \frac{\left(-1\right)^{r}}{2^{r}r!}\left(1+\alpha\right)^{-r-\frac{1}{2}} \ \ \ \ \ (35)

Thus the time-dependent wave function can be written as a single series as:

\displaystyle   \psi\left(x,t\right) \displaystyle  = \displaystyle  U\left(t\right)\psi\left(x,0\right)\ \ \ \ \ (36)
\displaystyle  \displaystyle  = \displaystyle  e^{-iHt/\hbar}\psi\left(x,0\right)\ \ \ \ \ (37)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\pi^{1/4}}\sum_{r=0}^{\infty}\frac{\left(-1\right)^{r}}{2^{r}r!}\left(1+\alpha\right)^{-r-\frac{1}{2}}x^{2r}\ \ \ \ \ (38)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\pi^{1/4}\sqrt{1+\alpha}}\sum_{r=0}^{\infty}\frac{\left(-1\right)^{r}}{2^{r}\left(1+\alpha\right)^{r}r!}x^{2r}\ \ \ \ \ (39)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\pi^{1/4}\sqrt{1+\alpha}}\exp\left[\frac{-x^{2}}{2\left(1+\alpha\right)}\right]\ \ \ \ \ (40)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\pi^{1/4}\sqrt{1+i\hbar t/m}}\exp\left[\frac{-x^{2}}{2\left(1+i\hbar t/m\right)}\right] \ \ \ \ \ (41)

This agrees with 9 when {p_{0}=0} and {\Delta=1}, though it does take a fair bit of work!

Free particle in the position basis

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.1, Exercise 5.1.2.

In quantum mechanics, the free particle has degenerate energy eigenstates for each energy

\displaystyle  E=\frac{p^{2}}{2m} \ \ \ \ \ (1)

where {p} is the momentum. The degeneracy arises because the momentum can be either positive (for a particle moving to the right) or negative (to the left):

\displaystyle  p=\pm\sqrt{2mE} \ \ \ \ \ (2)

Thus the most general energy eigenstate is a linear combination of the two momentum states:

\displaystyle  \left|E\right\rangle =\beta\left|p=\sqrt{2mE}\right\rangle +\gamma\left|p=-\sqrt{2mE}\right\rangle \ \ \ \ \ (3)

This bizarre feature of quantum mechanics means that a particle in such a state could be moving either left or right, and if we make a measurement of the momentum we force the particle into one or other of the two momentum states.

We obtained this solution by working in the momentum basis, but we can also find the solution in the position basis. In that basis, the momentum operator has the form

\displaystyle  P=-i\hbar\frac{d}{dx} \ \ \ \ \ (4)

The matrix elements of this operator in the position basis are

\displaystyle  \left\langle x\left|P\right|x^{\prime}\right\rangle =-i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (5)

where {\delta^{\prime}\left(x-x^{\prime}\right)} is the derivative of the delta function with respect to the {x}, not the {x^{\prime}}. We can use the properties of this derivative to get a solution in the {X} basis. To be completely formal about it, the derivation of the matrix elements of {P^{2}} in the {X} basis is:

\displaystyle   \left\langle x\left|P^{2}\right|\psi\right\rangle \displaystyle  = \displaystyle  \int\int\left\langle x\left|P\right|x^{\prime}\right\rangle \left\langle x^{\prime}\left|P\right|x^{\prime\prime}\right\rangle \left\langle x^{\prime\prime}\left|\psi\right.\right\rangle dx^{\prime}dx^{\prime\prime}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \int\int\left\langle x\left|P\right|x^{\prime}\right\rangle \left(-i\hbar\delta^{\prime}\left(x^{\prime}-x^{\prime\prime}\right)\right)\psi\left(x^{\prime\prime}\right)dx^{\prime}dx^{\prime\prime}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -i\hbar\int\left\langle x\left|P\right|x^{\prime}\right\rangle \frac{d\psi\left(x^{\prime}\right)}{dx^{\prime}}dx^{\prime}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  -i\hbar\int\int\left(-i\hbar\delta^{\prime}\left(x-x^{\prime}\right)\right)\frac{d\psi\left(x^{\prime}\right)}{dx^{\prime}}dx^{\prime}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -\hbar^{2}\frac{d^{2}}{dx^{2}}\psi\left(x\right) \ \ \ \ \ (10)

In this basis, the Schrödinger equation is therefore the familiar one:

\displaystyle   \frac{P^{2}}{2m}\left|\psi\right\rangle \displaystyle  = \displaystyle  E\left|\psi\right\rangle \ \ \ \ \ (11)
\displaystyle  \left\langle x\left|\frac{P^{2}}{2m}\right|\psi\right\rangle \displaystyle  = \displaystyle  E\psi\left(x\right)\ \ \ \ \ (12)
\displaystyle  -\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi\left(x\right) \displaystyle  = \displaystyle  E\psi\left(x\right)\ \ \ \ \ (13)
\displaystyle  \frac{d^{2}}{dx^{2}}\psi\left(x\right) \displaystyle  = \displaystyle  -\frac{2mE}{\hbar^{2}}\psi\left(x\right) \ \ \ \ \ (14)

This has the general solution

\displaystyle  \psi\left(x\right)=\beta e^{ix\sqrt{2mE}/\hbar}+\gamma e^{-ix\sqrt{2mE}/\hbar} \ \ \ \ \ (15)

[Shankar extracts a factor of {1/\sqrt{2\pi\hbar}} but as he notes, this is arbitrary and can be absorbed into the constants {\beta} and {\gamma} as we’ve done here.]

In this derivation we’ve implicitly assumed that {E>0}, since there is no potential so a free particle can’t really have a negative energy. However, if you follow through the derivation, you’ll see that it works even if {E<0}. In that case, we’d get

\displaystyle  \psi\left(x\right)=\beta e^{-x\sqrt{2m\left|E\right|}/\hbar}+\gamma e^{x\sqrt{2m\left|E\right|}/\hbar} \ \ \ \ \ (16)

That is, the exponents in both terms are now real instead of imaginary. The problem with this is that the first term blows up for {x\rightarrow-\infty} while the second blows up for {x\rightarrow+\infty}. Thus this function is not normalizable, even to a delta function (as was the case when {E>0}), so functions such as these when {E<0} are not in the Hilbert space.

Free particle revisited: solution in terms of a propagator

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.1, Exercise 5.1.1.

Having reviewed the background mathematics and postulates of quantum mechanics as set out by Shankar, we can now revisit some of the classic problems in non-relativistic quantum mechanics using Shankar’s approach, as opposed to that of Griffiths that we’ve already studied.

The first problem we’ll look it is that of the free particle. Following the fourth postulate, we write down the classical Hamiltonian for a free particle, which is

\displaystyle  H=\frac{p^{2}}{2m} \ \ \ \ \ (1)

where {p} is the momentum (we’re working in one dimension) and {m} is the mass. To get the quantum version, we replace {p} by the momentum operator {P} and insert the result into the Schrödinger equation:

\displaystyle   i\hbar\left|\dot{\psi}\right\rangle \displaystyle  = \displaystyle  H\left|\psi\right\rangle \ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{P^{2}}{2m}\left|\psi\right\rangle \ \ \ \ \ (3)

Since {H} is time-independent, the solution can be written using a propagator:

\displaystyle  \left|\psi\left(t\right)\right\rangle =U\left(t\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (4)

To find {U}, we need to solve the eigenvalue equation for the stationary states

\displaystyle  \frac{P^{2}}{2m}\left|E\right\rangle =E\left|E\right\rangle \ \ \ \ \ (5)

where {E} is an eigenvalue representing the allowable energies. Since the Hamiltonian is {P^{2}/2m}, and an eigenstate of {P} with eigenvalue {p} is also an eigenstate of {P^{2}} with eigenvalue {p^{2}}, we can write this equation in terms of the momentum eigenstates {\left|p\right\rangle }:

\displaystyle  \frac{P^{2}}{2m}\left|p\right\rangle =E\left|p\right\rangle \ \ \ \ \ (6)

Using {P^{2}\left|p\right\rangle =p^{2}\left|p\right\rangle } this gives

\displaystyle  \left(\frac{p^{2}}{2m}-E\right)\left|p\right\rangle =0 \ \ \ \ \ (7)

Assuming that {\left|p\right\rangle } is not a null vector gives the relation between momentum and energy:

\displaystyle  p=\pm\sqrt{2mE} \ \ \ \ \ (8)

Thus each allowable energy {E} has two possible momenta. Once we specify the momentum, we also specify the energy and since each energy state is two-fold degenerate, we can eliminate the ambiguity by specifying only the momentum. Therefore the propagator can be written as

\displaystyle  U\left(t\right)=\int_{-\infty}^{\infty}e^{-ip^{2}t/2m\hbar}\left|p\right\rangle \left\langle p\right|dp \ \ \ \ \ (9)

We can convert this to an integral over the energy by using 8 to change variables, and by splittling the integral into two parts. For {p>0} we have

\displaystyle  dp=\sqrt{\frac{m}{2E}}dE \ \ \ \ \ (10)

and for {p<0} we have

\displaystyle  dp=-\sqrt{\frac{m}{2E}}dE \ \ \ \ \ (11)

Therefore, we get

\displaystyle   U\left(t\right) \displaystyle  = \displaystyle  \int_{0}^{\infty}e^{-iEt/\hbar}\left|E,+\right\rangle \left\langle E,+\right|\sqrt{\frac{m}{2E}}dE+\int_{\infty}^{0}e^{-iEt/\hbar}\left|E,-\right\rangle \left\langle E,-\right|\left(-\sqrt{\frac{m}{2E}}\right)dE\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \int_{0}^{\infty}e^{-iEt/\hbar}\left|E,+\right\rangle \left\langle E,+\right|\sqrt{\frac{m}{2E}}dE+\int_{0}^{\infty}e^{-iEt/\hbar}\left|E,-\right\rangle \left\langle E,-\right|\sqrt{\frac{m}{2E}}dE\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \sum_{\alpha=\pm}\int_{0}^{\infty}\frac{m}{\sqrt{2mE}}e^{-iEt/\hbar}\left|E,\alpha\right\rangle \left\langle E,\alpha\right|dE \ \ \ \ \ (14)

Here, {\left|E,+\right\rangle } is the state with energy {E} and momentum {p=+\sqrt{2mE}} and similarly for {\left|E,-\right\rangle }. In the first line, the first integral is for {p>0} and corresponds to the {\int_{0}^{\infty}} part of 9. The second integral is for {p<0} and corresponds to the {\int_{-\infty}^{0}} part of 9, which is why the limits on the second integral have {\infty} at the bottom and 0 at the top. Reversing the order of integration cancels out the minus sign in {-\sqrt{\frac{m}{2E}}}, which allows us to add the two integrals together to get the final answer.

Free particle in momentum space

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.40.

Since the Hamiltonian for a free particle is {H=p^{2}/2m}, the Schrodinger equation in momentum space is

\displaystyle  i\hbar\frac{\partial\Phi}{\partial t}=\frac{p^{2}}{2m}\Phi \ \ \ \ \ (1)

so the solution can be found by simply integrating with respect to {t}:

\displaystyle  \Phi(p,t)=e^{-ip^{2}t/2m\hbar}\Phi(p,0) \ \ \ \ \ (2)

We looked at the travelling Gaussian wave packet in free space earlier. Its initial state in position space is

\displaystyle  \Psi(x,0)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^{2}}e^{ilx} \ \ \ \ \ (3)

To find {\Phi(p,0)} we use the conversion to momentum space we found earlier:

\displaystyle  \Phi(p,0)=\frac{1}{\sqrt{2\pi\hbar}}\left(\frac{2a}{\pi}\right)^{1/4}\int_{-\infty}^{\infty}e^{-ax^{2}+ilx-ipx/\hbar}dx \ \ \ \ \ (4)

From the analysis of the travelling Gaussian packet we see that the integral is the same as that done when calculating {\phi(k)} if we replace {k} with {p/\hbar}. Therefore

\displaystyle  \Phi(p,0)=\left(\frac{2}{\pi a}\right)^{1/4}\frac{1}{\sqrt{2\hbar}}e^{-(p/\hbar-l)^{2}/4a} \ \ \ \ \ (5)

Using 2, we have the full solution for {\Phi(p,t)}:

\displaystyle  \Phi(p,t)=\left(\frac{2}{\pi a}\right)^{1/4}\frac{1}{\sqrt{2\hbar}}e^{-ip^{2}t/2m\hbar}e^{-(p/\hbar-l)^{2}/4a} \ \ \ \ \ (6)

Also

\displaystyle  |\Phi(p,t)|^{2}=\frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}e^{-(p/\hbar-l)^{2}/2a} \ \ \ \ \ (7)

which is independent of time. (As a check, we can integrate this over all {p} and verify that this integral is 1.)

We can calculate the means for momentum in the usual way:

\displaystyle   \langle p\rangle \displaystyle  = \displaystyle  \frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}\int_{-\infty}^{\infty}pe^{-(p/\hbar-l)^{2}/2a}dp\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \hbar l\ \ \ \ \ (9)
\displaystyle  \langle p^{2}\rangle \displaystyle  = \displaystyle  \frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}\int_{-\infty}^{\infty}p^{2}e^{-(p/\hbar-l)^{2}/2a}dp\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \hbar^{2}(l^{2}+a) \ \ \ \ \ (11)

Both results agree with those in the analysis of the travelling Gaussian packet.

For the mean energy, we have

\displaystyle   \langle H\rangle \displaystyle  = \displaystyle  \left\langle \frac{p^{2}}{2m}\right\rangle \ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}}{2m}(l^{2}+a)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{\langle p\rangle^{2}}{2m}+\frac{a\hbar^{2}}{2m} \ \ \ \ \ (14)

Referring back to the stationary Gaussian wave packet in free space, we see that {\langle p^{2}\rangle=a\hbar^{2}}, so the energy is the sum of that for a stationary Gaussian wave packet and the term {\langle p\rangle^{2}/2m}. For the travelling packet, there is a net non-zero average momentum, so {\langle p\rangle} is non-zero. Thus the energy arises from the inherent energy of the wave packet, plus the kinetic energy of motion of the packet.

Energy-time uncertainty principle: Gaussian free particle

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.19.

As another example of the energy-time uncertainty relation, we can look again at the example of a travelling free particle with a Gaussian wave packet. We have already worked out most of what we need to test the uncertainty relation:

\displaystyle   \langle x\rangle \displaystyle  = \displaystyle  \frac{l\hbar t}{m}\ \ \ \ \ (1)
\displaystyle  \langle x^{2}\rangle \displaystyle  = \displaystyle  \frac{1+(2a\hbar t/m)^{2}+a(2\hbar lt/m)^{2}}{4a}\ \ \ \ \ (2)
\displaystyle  \langle p^{2}\rangle \displaystyle  = \displaystyle  \hbar^{2}(a+l^{2}) \ \ \ \ \ (3)

From this we get

\displaystyle   \left\langle H\right\rangle \displaystyle  = \displaystyle  \frac{\left\langle p^{2}\right\rangle }{2m}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}(a+l^{2})}{2m} \ \ \ \ \ (5)

We still need {\left\langle H^{2}\right\rangle }. From our previous calculations, we have the wave function:

\displaystyle  \Psi(x,t)=\left(\frac{2a}{\pi}\right)^{1/4}\frac{e^{\left(-ax^{2}+i(lx-\hbar l^{2}t/2m)\right)/(1+2i\hbar at/m)}}{\sqrt{1+2i\hbar at/m}} \ \ \ \ \ (6)

We can calculate {\left\langle H^{2}\right\rangle } by direct integration, using Maple:

\displaystyle   \left\langle H^{2}\right\rangle \displaystyle  = \displaystyle  \frac{\hbar^{4}}{4m^{2}}\int_{-\infty}^{\infty}\left|\frac{d^{2}\Psi(x,t)}{dx^{2}}\right|^{2}dx\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{4}}{4m^{2}}\left(3a^{2}+6al^{2}+l^{4}\right) \ \ \ \ \ (8)

As a check on this result, we can work out the units (always a good test to make sure you haven’t dropped a factor somewhere). From the original wave function, since exponents must be dimensionless, we know that {a} has dimensions {distance^{-2}} and {l} has {distance^{-1}}. Planck’s constant has dimensions of {energy\times time}, so the expression above has overall units of {energy^{4}\times time^{4}\times mass^{-2}\times distance^{-4}=energy^{2}}. (Recall kinetic energy is {mv^{2}/2}.) It’s also worth noting that {\left\langle H^{2}\right\rangle } is independent of time.

From here, we can get {\sigma_{H}^{2}}:

\displaystyle   \sigma_{H}^{2} \displaystyle  = \displaystyle  \left\langle H^{2}\right\rangle -\left\langle H\right\rangle ^{2}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{a\hbar^{4}}{2m^{2}}\left(a+2l^{2}\right) \ \ \ \ \ (10)

We also have, from above

\displaystyle   \sigma_{x}^{2} \displaystyle  = \displaystyle  \left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \frac{m^{2}+4\left(at\hbar\right)^{2}}{4am^{2}} \ \ \ \ \ (12)

We’re trying to show that {\sigma_{H}\sigma_{x}\ge\frac{\hbar}{2}\frac{d\left\langle x\right\rangle }{dt}=l\hbar^{2}/2m} from above. So we want

\displaystyle   \sigma_{H}^{2}\sigma_{x}^{2} \displaystyle  \ge \displaystyle  \frac{l^{2}\hbar^{4}}{4m^{2}}\ \ \ \ \ (13)
\displaystyle  \frac{a\hbar^{4}}{2m^{2}}\left(a+2l^{2}\right)\frac{m^{2}+4\left(at\hbar\right)^{2}}{4am^{2}} \displaystyle  \ge \displaystyle  \frac{l^{2}\hbar^{4}}{4m^{2}} \ \ \ \ \ (14)

The minimum of the LHS occurs at {t=0} so if the inequality is true there, it is true always. In this case, it reduces to

\displaystyle   a+2l^{2} \displaystyle  \ge \displaystyle  2l^{2}\ \ \ \ \ (15)
\displaystyle  a \displaystyle  \ge \displaystyle  0 \ \ \ \ \ (16)

This final condition is certainly true (it is required for the Gaussian wave form to converge at large {x}), so the uncertainty condition is verified.

Free particle – travelling wave packet

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.43.

We’ve looked at the stationary Gaussian wave packet for the free particle. The initial wave function in that case was

\displaystyle  \Psi(x,0)=Ae^{-ax^{2}} \ \ \ \ \ (1)

We can turn this into a travelling Gaussian wave by adding a factor to the wave function:

\displaystyle  \Psi(x,0)=Ae^{-ax^{2}}e^{ilx} \ \ \ \ \ (2)

where {l} is a real constant.

Since we have added only a complex exponential, the normalization condition is the same as for the stationary case:

\displaystyle  A=\left(\frac{2a}{\pi}\right)^{1/4} \ \ \ \ \ (3)

To find {\Psi(x,t)} we follow the same procedure as in the stationary case. So we get

\displaystyle  \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk \ \ \ \ \ (4)

Given the initial wave function, we can find {\phi(k)} via Plancherel’s theorem:

\displaystyle   \phi(k) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}dx\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\left(\frac{2a}{\pi}\right)^{1/4}\int_{-\infty}^{\infty}e^{-ax^{2}-ikx+ilx}dx\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{1}{2\pi a}\right)^{1/4}e^{-(k-l)^{2}/4a} \ \ \ \ \ (7)

So we can now find the general solution:

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{i(kx-\hbar k^{2}t/2m)}dk\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{2a}{\pi}\right)^{1/4}\frac{e^{\left(-ax^{2}+i(lx-\hbar l^{2}t/2m)\right)/(1+2i\hbar at/m)}}{\sqrt{1+2i\hbar at/m}} \ \ \ \ \ (9)

where Maple was used for the integral.

Calculating {|\Psi(x,t)|^{2}} can be done using Maple, with the result:

\displaystyle  |\Psi(x,t)|^{2}=\sqrt{\frac{2}{\pi}}we^{-2w^{2}(\hbar lt/m-x)^{2}} \ \ \ \ \ (10)

with

\displaystyle  w\equiv\left(\frac{a}{1+(2\hbar at/m)^{2}}\right)^{1/2} \ \ \ \ \ (11)

The results above reduce to the stationary wave packet when {l=0}.

At {t=0}, {w=\sqrt{a}}, so {|\Psi(x,0)|^{2}=\sqrt{2a/\pi}e^{-2ax^{2}}} which is correct. The wave packet at {t=0} is therefore a Gaussian centred at {x=0}. As {t} increases, {w} gets smaller but in this case, the peak of the Gaussian moves according to {x_{peak}=\hbar lt/m}. The speed of the peak is {x/t=\hbar l/m}.

By direct integration we find, {\langle x\rangle=\hbar lt/m}. Calculating the other means requires a bit of effort but we can use Maple to do most of it. The results are:

\displaystyle   \langle p\rangle \displaystyle  = \displaystyle  l\hbar\ \ \ \ \ (12)
\displaystyle  \langle x^{2}\rangle \displaystyle  = \displaystyle  \frac{1+(2a\hbar t/m)^{2}+a(2\hbar lt/m)^{2}}{4a}\ \ \ \ \ (13)
\displaystyle  \langle p^{2}\rangle \displaystyle  = \displaystyle  \hbar^{2}(a+l^{2}) \ \ \ \ \ (14)

All these results reduce to those for the stationary wave packet from problem 2.22 when {l=0}.

The uncertainty principle thus becomes

\displaystyle   \sigma_{x}\sigma_{p} \displaystyle  = \displaystyle  \sqrt{(\langle x^{2}\rangle-\langle x\rangle^{2})(\langle p^{2}\rangle-\langle p\rangle^{2})}\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar}{2}\sqrt{1+(2\hbar at/m)^{2}} \ \ \ \ \ (16)

which is the same result as in the stationary wave packet. Thus although the packet here travels with a constant speed, it spreads out at the same rate as the stationary packet.

Free particle: Gaussian wave packet

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.22.

While analyzing the free particle, we saw that we could construct a normalizable combination of stationary states by writing

\displaystyle  \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk \ \ \ \ \ (1)

Given the initial wave function, we can find {\phi(k)} via Plancherel’s theorem:

\displaystyle  \phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}dx \ \ \ \ \ (2)

The integral 1 cannot usually be done in closed form, but one case where it can is if the initial wave function is a Gaussian, of the form

\displaystyle  \Psi(x,0)=Ae^{-ax^{2}} \ \ \ \ \ (3)

To find {A}, we normalize:

\displaystyle  A^{2}\int_{-\infty}^{\infty}e^{-2ax^{2}}dx=1 \ \ \ \ \ (4)

The integral comes out to {\sqrt{\pi/2a}} from which we get

\displaystyle  A=\left(\frac{2a}{\pi}\right)^{1/4} \ \ \ \ \ (5)

Finding {\Psi(x,t)} requires finding {\phi(k)} via equation 2. So we get (using Maple to do the integral):

\displaystyle   \phi(k) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\left(\frac{2a}{\pi}\right)^{1/4}\int_{-\infty}^{\infty}e^{-ax^{2}-ikx}dx\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{1}{2\pi a}\right)^{1/4}e^{-k^{2}/4a} \ \ \ \ \ (7)

We can now use 1 again using Maple to do the integral:

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\left(\frac{1}{2\pi a}\right)^{1/4}\int_{-\infty}^{\infty}e^{-k^{2}/4a}e^{ikx}e^{-i\hbar k^{2}t/2m}dk\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{2a}{\pi}\right)^{1/4}\frac{e^{-ax^{2}/(1+2i\hbar at/m)}}{\sqrt{1+2i\hbar at/m}} \ \ \ \ \ (10)

where Maple was used for the integral.

Calculating {|\Psi(x,t)|^{2}} can be done using Maple, but it seems to require a bit of help. First we write out the complex conjugate:

\displaystyle  \Psi^*(x,t)=\left(\frac{2a}{\pi}\right)^{1/4}\frac{e^{-ax^{2}/(1-2i\hbar at/m)}}{\sqrt{1-2i\hbar at/m}} \ \ \ \ \ (11)

Then we calculate {\Psi^*\Psi} using the Maple command simplify(evalc({\Psi^*\Psi})) assuming positive and we get

\displaystyle   \left|\Psi(x,t)\right|^{2} \displaystyle  = \displaystyle  \sqrt{\frac{2a}{\pi}}\frac{e^{-2ax^{2}/\left[1+\left(2\hbar at/m\right)^{2}\right]}}{\sqrt{1+\left(2\hbar at/m\right)^{2}}}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{2}{\pi}}we^{-2w^{2}x^{2}} \ \ \ \ \ (13)

with

\displaystyle  w\equiv\left(\frac{a}{1+(2\hbar at/m)^{2}}\right)^{1/2} \ \ \ \ \ (14)

At {t=0}, {w=\sqrt{a}}, so {|\Psi(x,t)|^{2}=\sqrt{2a/\pi}e^{-2ax^{2}}} which is correct. The wave packet at {t=0} is therefore a Gaussian centred at {x=0}. As {t} increases, {w} gets smaller but the centre of the Gaussian does not move from {x=0} so the packet spreads out. A couple of plots show this behaviour. We’ve set {a=1} in both plots. In the red plot {t=0} so {w=1} and in the green plot {t} is larger, at a value such that {w=0.1}.

We can get the mean values of position and momentum by integration, although it takes a bit of work. By symmetry, {\langle x\rangle=\langle p\rangle=0}. To get the other two average values, we use integration with Maple.

\displaystyle   \langle x^{2}\rangle \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}x^{2}|\Psi(x,t)|^{2}dx\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{4w^{2}}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \frac{1+(2\hbar at/m)^{2}}{4a} \ \ \ \ \ (17)

This shows that the wave function spreads out with time. At {t=0} {\left\langle x^{2}\right\rangle =1/4a}, but it then increases quadratically with {t}.

Calculating {\langle p^{2}\rangle} starts with:

\displaystyle   \langle p^{2}\rangle \displaystyle  = \displaystyle  -\hbar^{2}\int_{-\infty}^{\infty}\Psi^*\frac{\partial^{2}}{\partial x^{2}}\Psi dx \ \ \ \ \ (18)

This can be evaluated with the Maple command simplify(evalc(int(-h^2*simplify(evalc(psixtconj*(diff(psixt, x$2)))), x = -infinity .. infinity))) assuming positive where psixtconj and psixt are the Maple expressions for {\Psi^*} and {\Psi} respectively. The result is:

\displaystyle  \langle p^{2}\rangle=a\hbar^{2} \ \ \ \ \ (19)

The uncertainty principle thus becomes

\displaystyle   \sigma_{x}\sigma_{p} \displaystyle  = \displaystyle  \sqrt{\langle x^{2}\rangle\langle p^{2}\rangle}\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar}{2}\sqrt{1+(2\hbar at/m)^{2}} \ \ \ \ \ (21)

The system has the least uncertainty at {t=0}. Uncertainty increases with time as the wave packet spreads out.

Plancherel's theorem

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David
J. (2005), Introduction to Quantum
Mechanics, 2nd Edition; Pearson Education – Chapter 2, Post 20.

While analyzing the free particle, we saw that we could construct a normalizable combination of stationary states by writing

\displaystyle \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk \ \ \ \ \ (1)

We can find the function {\phi(k)} by specifying the initial wave function:

\displaystyle \Psi(x,0) \displaystyle = \displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}dk \ \ \ \ \ (2)

This relation can be inverted by using Plancherel’s theorem, which states

\displaystyle \phi(k) \displaystyle = \displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}dx \ \ \ \ \ (3)

Here we run through a plausibility argument which is a sort of physicist’s proof of Plancherel’s theorem. We start with Dirichlet’s theorem which says that any (physically realistic, anyway) function can be written as a Fourier series. We can show that this is equivalent to a series in complex exponentials. That is

\displaystyle f(x) \displaystyle = \displaystyle \sum_{n=-\infty}^{\infty}c_{n}e^{in\pi x/a}\ \ \ \ \ (4)
\displaystyle \displaystyle = \displaystyle \sum_{n=-\infty}^{\infty}c_{n}\left[\cos\frac{n\pi x}{a}+i\sin\frac{n\pi x}{a}\right]\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle c_{0}+\sum_{n=1}^{\infty}\left(c_{n}+c_{-n}\right)\cos\frac{n\pi x}{a}+i\sum_{n=1}^{\infty}\left(c_{n}-c_{-n}\right)\sin\frac{n\pi x}{a} \ \ \ \ \ (6)

We’ve used the facts that cosine is even and sine is odd. This is equivalent to a Fourier series:

\displaystyle f(x)=\sum_{n=0}^{\infty}\left[a_{n}\sin\frac{n\pi x}{a}+b_{n}\cos\frac{n\pi x}{a}\right] \ \ \ \ \ (7)

where the coefficients are related by

\displaystyle b_{0} \displaystyle = \displaystyle c_{0}\ \ \ \ \ (8)
\displaystyle b_{n} \displaystyle = \displaystyle c_{n}+c_{-n}\ \ \ \ \ (9)
\displaystyle a_{n} \displaystyle = \displaystyle i\left(c_{n}-c_{-n}\right) \ \ \ \ \ (10)

Inverting the relations we get, for {n>0}

\displaystyle c_{n} \displaystyle = \displaystyle \frac{1}{2}\left(b_{n}-ia_{n}\right)\ \ \ \ \ (11)
\displaystyle c_{-n} \displaystyle = \displaystyle \frac{1}{2}\left(b_{n}+ia_{n}\right) \ \ \ \ \ (12)

We can get the coefficients in terms of {f(x)} by integration:

\displaystyle \frac{1}{2a}\int_{-a}^{a}f(x)e^{-in\pi x/a}dx=\frac{1}{2a}\sum_{m=-\infty}^{\infty}c_{m}\int_{-a}^{a}e^{i\left(m-n\right)\pi x/a}dx \ \ \ \ \ (13)

The integral is zero if {m\ne n} and {2a} if {m=n}, so the right hand side comes out to just {c_{n}} and we get

\displaystyle c_{n}=\frac{1}{2a}\int_{-a}^{a}f(x)e^{-in\pi x/a}dx \ \ \ \ \ (14)

Now we can make the substitutions

\displaystyle k \displaystyle \equiv \displaystyle \frac{n\pi}{a}\ \ \ \ \ (15)
\displaystyle F(k) \displaystyle \equiv \displaystyle \sqrt{\frac{2}{\pi}}ac_{n} \ \ \ \ \ (16)

If {\Delta k} is the increment in {k} from one {n} to the next, then {\Delta k=\pi/a}. We can then write the original series as

\displaystyle f(x) \displaystyle = \displaystyle \sum_{n=-\infty}^{\infty}\sqrt{\frac{\pi}{2}}\frac{1}{a}F(k)e^{ikx}\left(\frac{a}{\pi}\Delta k\right)\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle \frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}F(k)e^{ikx}\Delta k \ \ \ \ \ (18)

The formula for {c_{n}} now becomes

\displaystyle \sqrt{\frac{\pi}{2}}\frac{1}{a}F(k) \displaystyle = \displaystyle \frac{1}{2a}\int_{-a}^{a}f(x)e^{-ikx}dx\ \ \ \ \ (19)
\displaystyle F(k) \displaystyle = \displaystyle \frac{1}{\sqrt{2\pi}}\int_{-a}^{a}f(x)e^{-ikx}dx \ \ \ \ \ (20)

Now we can take the limit as {a\rightarrow\infty}. In this case, {\Delta k\rightarrow dk} (that is, it becomes a differential) and the sum becomes an integral, so we get

\displaystyle f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(k)e^{ikx}dk \ \ \ \ \ (21)

In the second formula, the limits on the integral become infinite, and we get the other half of Plancherel’s theorem:

\displaystyle F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx \ \ \ \ \ (22)

Complex exponentials and trig functions

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.18.

Solutions to the Schrödinger equation are written either as complex exponentials or a sum of sine and cosine functions. For example, we used the former in the solution for the free particle and the latter in the infinite square well. In fact these two ways are equivalent for writing a function, as we can demonstrate.

We start with

\displaystyle  Ae^{ikx}+Be^{-ikx} \ \ \ \ \ (1)

and we want to write this as {C\cos kx+D\sin kx}.

We can expand the exponentials in terms of sine and cosine to get

\displaystyle   Ae^{ikx}+Be^{-ikx} \displaystyle  = \displaystyle  \left(A+B\right)\cos kx+i\left(A-B\right)\sin x\ \ \ \ \ (2)
\displaystyle  \displaystyle  \equiv \displaystyle  C\cos kx+D\sin kx \ \ \ \ \ (3)

So

\displaystyle   C \displaystyle  = \displaystyle  A+B\ \ \ \ \ (4)
\displaystyle  D \displaystyle  = \displaystyle  i\left(A-B\right) \ \ \ \ \ (5)

or, the other way round:

\displaystyle   A \displaystyle  = \displaystyle  \frac{1}{2}\left(C-iD\right)\ \ \ \ \ (6)
\displaystyle  B \displaystyle  = \displaystyle  \frac{1}{2}\left(C+iD\right) \ \ \ \ \ (7)

The free particle

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 2.4, Problem 2.21.

In a system in which there is no potential ({V(x)=0} everywhere), all particles are free particles. In principle, the solution of the Schrödinger equation is the easiest in this case:

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}} \displaystyle  = \displaystyle  E\psi\ \ \ \ \ (1)
\displaystyle  \frac{d^{2}\psi}{dx^{2}} \displaystyle  = \displaystyle  -\frac{2mE}{\hbar^{2}}\psi\ \ \ \ \ (2)
\displaystyle  \displaystyle  \equiv \displaystyle  -k^{2}\psi\ \ \ \ \ (3)
\displaystyle  k \displaystyle  \equiv \displaystyle  \frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (4)

The stationary states, in terms of complex exponentials (the solutions can also be written in terms of sines and cosines by expanding the exponentials) are:

\displaystyle   \psi(x) \displaystyle  = \displaystyle  Ae^{ikx}+Be^{-ikx}\ \ \ \ \ (5)
\displaystyle  \Psi(x,t) \displaystyle  = \displaystyle  (Ae^{ikx}+Be^{-ikx})e^{-iEt/\hbar} \ \ \ \ \ (6)

Normally, at this point, we would impose some boundary conditions in an attempt to determine the constants {A} and {B} and also the allowed energies {E}. However, there are no boundary conditions in this problem since the potential is uniform everywhere, so we really can’t go any further in specifying the stationary states. For a free particle, any positive energy {E} is allowable.

However, there is one big problem with the stationary states: they aren’t normalizable. We can see this by calculating {|\Psi|^{2}} directly:

\displaystyle   |\Psi|^{2} \displaystyle  = \displaystyle  \Psi^*\Psi\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  (A^*e^{-ikx}+B^*e^{ikx})(Ae^{ikx}+Be^{-ikx})\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  |A|^{2}+|B|^{2}+A^*Be^{-2ikx}+B^*Ae^{2ikx} \ \ \ \ \ (9)

Whatever the constants {A} and {B} are, this quantity must be real and positive (since it is the square modulus of a complex number) and, since the complex exponentials simply oscillate for all values of {x}, it is also clear that {|\Psi|^{2}} does not decrease to zero at infinite distances, thus its integral over all {x} is infinite.

Thus we get the first important conclusion about a free particle: a free particle cannot exist in a stationary state. So what does this mean? Does quantum mechanics predict that particles can’t exist without some sort of potential to constrain them? If it did, this would be a fatal flaw in the theory, but before we panic, we need to remember another property of solutions to the Schrödinger equation: it is linear, so all linear combinations of solutions are also solutions. That is, even though a stationary state corresponding to one particular energy cannot on its own be a solution, maybe if we add up a bunch of stationary states, we can get a solution which is normalizable, and therefore physically acceptable.

Since any energy is allowable here, the linear combination of stationary states will be, in general, an integral rather than a sum. That is, the solution we want to try will look something like this:

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-iEt/\hbar}dk\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk \ \ \ \ \ (11)

where in the second line we’ve substituted for {E=\hbar^{2}k^{2}/2m} using the result above.

There are a few things we’ve done in this equation. First, we’ve replaced the sum {Ae^{ikx}+Be^{-ikx}} by a single term {\phi(k)e^{ikx}/\sqrt{2\pi}}, but we’re taking {k} over the entire set of real numbers, positive and negative. This has the same effect as integrating {Ae^{ikx}+Be^{-ikx}} over {k} from 0 to {\infty}.

Second, we’ve introduced a function {\phi(k)/\sqrt{2\pi}} (the {\sqrt{2\pi}} is introduced to make this result more symmetric with another result that we’ll get to in a minute; since {\phi(k)} hasn’t yet been specified, this is perfectly allowable). The function {\phi(k)} plays the role of the {c_{n}} constants in the general solution of the particle in a box problem. It specifies how much of the stationary state with value {k} contributes to the overall solution {\Psi(x,t)}.

This solution is acceptable, provided {\Psi(x,t)} can be normalized. To see that it can, we can use the standard procedure for specifying and solving the Schrödinger equation: specify some initial conditions and use them to determine the initial distribution of stationary states. That is, we specify {\Psi(x,0)} (which, to be physically realistic, must be normalizable) and use it to find the function {\phi(k)}. That is, the problem is, given:

\displaystyle   \Psi(x,0) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}dk

find {\phi(k)}.

Remember that in the particle in a box problem, we solved the analogous problem using the orthonormal properties of the stationary states. In the continuous case, the solution isn’t quite as easy to prove, but the results are similar in spirit. The result is known as Plancherel’s theorem, and it states that if we know the function {\Psi(x,0)} and want to find {\phi(k)} then we can invert the relation between {\Psi(x,0)} and {\phi(k)} to get:

\displaystyle   \phi(k) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}dx \ \ \ \ \ (12)

(This was the reason for introducing the weird factor of {\sqrt{2\pi}} in the original equation – it makes Plancherel’s theorem work out symmetrically.) For readers familiar with Fourier transforms, the two functions {\Psi(x,0)} and {\phi(k)} are Fourier transforms of each other.

Once we have {\phi(k)}, we can get the general time-dependent solution by plugging it back into the equation above:

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk \ \ \ \ \ (13)

although this integral is often not easy to work out(!)

Example: Suppose we have {\Psi(x,0)=Ae^{-a|x|}}. How far can we get in working out this system’s behaviour as a free particle?

First, we need to normalize the initial condition, so we do the usual integral

\displaystyle   1 \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}|\Psi(x,0)|^{2}dx\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  A^{2}\int_{-\infty}^{\infty}e^{-2a|x|}dx\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  A^{2}/a\ \ \ \ \ (16)
\displaystyle  A \displaystyle  = \displaystyle  \sqrt{a} \ \ \ \ \ (17)

Next, we can use Plancherel’s theorem to work out {\phi(x)}:

\displaystyle   \phi(k) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}dx\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{a}{2\pi}}\int_{-\infty}^{+\infty}e^{-a|x|}e^{-ikx}dx\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{2}{\pi}}\frac{a^{3/2}}{a^{2}+k^{2}} \ \ \ \ \ (20)

The full solution, including time-dependence, is therefore

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\pi}a^{3/2}\int_{-\infty}^{\infty}\frac{e^{ikx-i\hbar k^{2}t/2m}}{a^{2}+k^{2}}dk \ \ \ \ \ (22)

This integral is clearly not easy and probably doesn’t have a closed form, but the idea of how a problem is solved should be clear from this example.

Despite the intractability of the integral, there are a few things we can say about this result. First, the initial condition {\Psi(x,0)} defines a wave function that is peaked at {x=0} and falls off symmetrically as {x} gets further from 0. The rate of fall-off is determined by the constant {a}; for small {a} the function falls off very slowly and for large {a} the function is a very sharp spike at {x=0}.

When translated into {\phi(k)}, we see that for small {a},

\displaystyle   \phi(k) \displaystyle  = \displaystyle  \sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{a}+k^{2}/a^{3/2}} \ \ \ \ \ (23)

which has a peak around {k=0} and falls off fairly rapidly once {k} gets larger. Thus when the wave function is spread out over {x} the values of {k} (and hence of energy {E}) that contribute to this function are quite restricted.

Conversely, if {a} is large, the {k^{2}} term in the denominator of {\phi(k)} becomes negligible and {\phi(k)} becomes more or less constant. Thus when the particle is restricted to a narrow spatial area ({x} is peaked), pretty well all energies contribute to the wave function.

This is an illustration of the uncertainty principle: when the location is well known, the energy (and hence the momentum, since {E=p^{2}/2m}) is uncertain; conversely when the location is uncertain, the energy and momentum are known. You can’t measure both position and momentum accurately at the same time.