# Free particle moving in the z direction

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.10.

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The radial function for a free particle can be either a spherical Bessel function ${j_{l}}$ or a spherical Neumann function ${n_{l}}$. If the solution space includes the origin, then only ${j_{l}}$ is acceptable since the ${n_{l}}$ functions diverge as ${r\rightarrow0}$.

In rectangular coordinates, a free particle wave function has the form

$\displaystyle \psi_{E}\left(x,y,z\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{i\mathbf{p}\cdot\mathbf{r}/\hbar} \ \ \ \ \ (1)$

where the energy ${E}$ is

$\displaystyle E=\frac{p^{2}}{2\mu}=\frac{\hbar^{2}k^{2}}{2\mu} \ \ \ \ \ (2)$

For a free particle travelling in the ${z}$ direction, this becomes

$\displaystyle \psi_{E}\left(r,\theta,\phi\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{ikr\cos\theta} \ \ \ \ \ (3)$

since ${z=r\cos\theta}$.

Since the solutions of the free-particle Schrödinger equation in spherical coordinations form a complete set, we must be able to express this wave function as a linear combination of these solutions, so that

$\displaystyle e^{ikr\cos\theta}=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}C_{l}^{m}j_{l}\left(kr\right)Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (4)$

where the ${C_{l}^{m}}$ are constants. Because we’re looking at motion in the ${z}$ direction, there is no angular momentum about the ${z}$ axis, which is reflected in the fact that ${\psi_{E}}$ does not depend on ${\phi}$. Thus ${L_{z}=m\hbar=0}$ and ${m=0}$. We therefore have

 $\displaystyle e^{ikr\cos\theta}$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}^{0}j_{l}\left(kr\right)Y_{l}^{0}\left(\theta,\phi\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}\sqrt{\frac{2l+1}{4\pi}}C_{l}^{0}j_{l}\left(kr\right)P_{l}\left(\cos\theta\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}j_{l}\left(kr\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (7)$

where

$\displaystyle C_{l}\equiv\sqrt{\frac{2l+1}{4\pi}}C_{l}^{0} \ \ \ \ \ (8)$

The problem, of course, is to find these constants. We can do this using the identities given by Shankar in his problem 12.6.10, which are

 $\displaystyle \int_{-1}^{1}P_{l}\left(x\right)P_{l^{\prime}}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{2\delta_{ll^{\prime}}}{2l+1}\ \ \ \ \ (9)$ $\displaystyle P_{l}\left(x\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2^{l}l!}\frac{d^{l}\left(x^{2}-1\right)^{l}}{dx^{l}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{l}}{2^{l}l!}\frac{d^{l}\left(1-x^{2}\right)^{l}}{dx^{l}}\ \ \ \ \ (11)$ $\displaystyle \int_{0}^{1}\left(1-x^{2}\right)^{m}dx$ $\displaystyle =$ $\displaystyle \frac{\left(2m\right)!!}{\left(2m+1\right)!!}\ \ \ \ \ (12)$ $\displaystyle \int_{-1}^{1}\left(1-x^{2}\right)^{m}dx$ $\displaystyle =$ $\displaystyle \frac{2\left(2m\right)!!}{\left(2m+1\right)!!} \ \ \ \ \ (13)$

The last line follows because ${\left(1-x^{2}\right)^{m}}$ is an even function and is therefore symmetric about ${x=0}$.

We can use the standard procedure for isolating ${C_{l}}$ by multiplying both sides by ${C_{a}}$ and using 9.

 $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}j_{l}\left(kr\right)\int_{-1}^{1}P_{a}\left(x\right)P_{l}\left(x\right)dx\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{2a+1}C_{a}j_{a}\left(kr\right) \ \ \ \ \ (15)$

This relation must be true for all values of ${r}$, so we can look at the limit of small (but not zero, since both sides are then zero) ${r}$. We have the asymptotic relation for the spherical Bessel functions

 $\displaystyle j_{l}$ $\displaystyle \underset{\rho\rightarrow0}{\longrightarrow}$ $\displaystyle \frac{\rho^{l}}{\left(2l+1\right)!!} \ \ \ \ \ (16)$

We thus have

$\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx=\underset{r\rightarrow0}{\longrightarrow}\frac{2}{2a+1}\frac{k^{a}r^{a}}{\left(2a+1\right)!!}C_{a} \ \ \ \ \ (17)$

We can then look at the integral on the LHS and hope that, when we expand the exponential, that the terms in ${\left(kr\right)^{n}}$ for ${n vanish. We can then match the coefficients of ${\left(kr\right)^{a}}$ on both sides to find ${C_{a}}$.

We can see that this will work because the Legendre polynomials ${P_{l}}$ are a complete set of functions, and the polynomial ${P_{l}}$ has degree ${l}$. This means that any polynomial of degree ${a-1}$ can be written as a linear combination of the ${P_{l}}$, where ${l=0,\ldots,a-1}$. Because of 9, this means that

$\displaystyle \int_{-1}^{1}x^{l}P_{a}\left(x\right)dx=0\;\;\mbox{if \ensuremath{l

Therefore, when we expand ${e^{ikrx}}$ in a power series, we have

 $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx$ $\displaystyle =$ $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)\left(1+ikrx+\frac{\left(ikrx\right)^{2}}{2!}+\ldots\right)dx\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)\left(\frac{\left(ikrx\right)^{a}}{a!}+\ldots\right)dx \ \ \ \ \ (20)$

In the limit of small ${r}$, higher order terms in the sum on the RHS can be ignored, so we get

 $\displaystyle \frac{\left(ikr\right)^{a}}{a!}\int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{2}{2a+1}\frac{k^{a}r^{a}}{\left(2a+1\right)!!}C_{a}\ \ \ \ \ (21)$ $\displaystyle C_{a}$ $\displaystyle =$ $\displaystyle \frac{i^{a}\left(2a+1\right)\left(2a+1\right)!!}{2a!}\int_{-1}^{1}x^{a}P_{a}\left(x\right)dx \ \ \ \ \ (22)$

Now consider the integral in the last line. Using 11 we have

$\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx=\frac{\left(-1\right)^{a}}{2^{a}a!}\int_{-1}^{1}x^{a}\frac{d^{a}\left(1-x^{2}\right)^{a}}{dx^{a}}dx \ \ \ \ \ (23)$

We can integrate by parts repeatedly until the derivative in the integrand disappears. Note that the ${n}$th derivative of ${\left(1-x^{2}\right)^{a}}$ will always contain a factor of ${\left(1-x^{2}\right)}$ to some power for any ${n, and thus is zero at both limits of integration. Since the integrated term in the integration by parts always contains such a derivative, all integrated terms are zero at both limits. We therefore integrate ${\frac{d^{a}\left(1-x^{2}\right)^{a}}{dx^{a}}}$ (${a}$ times) and differentiate ${x^{a}}$ (${a}$ times) and keep only the residual integral after each iteration. The differentiation of ${x^{a}}$ (${a}$ times) introduces a factor of ${a!}$. Since the sign of the residual integral alternates as we perform each integration by parts, the final result is

 $\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{2a}}{2^{a}a!}a!\int_{-1}^{1}\left(1-x^{2}\right)^{a}dx\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2^{a}}\frac{2\left(2a\right)!!}{\left(2a+1\right)!!} \ \ \ \ \ (25)$

where we used 13 in the last line. The double factorial in the numerator can be written as

 $\displaystyle \left(2a\right)!!$ $\displaystyle =$ $\displaystyle \left(2a\right)\left(2a-2\right)\ldots\left(4\right)\left(2\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2^{a}a\left(a-1\right)\ldots\left(2\right)\left(1\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2^{a}a! \ \ \ \ \ (28)$

We therefore have

 $\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{1}{2^{a}}\frac{2\times2^{a}a!}{\left(2a+1\right)!!}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2a!}{\left(2a+1\right)!!} \ \ \ \ \ (30)$

Plugging this back into 22 we have

$\displaystyle C_{a}=i^{a}\left(2a+1\right) \ \ \ \ \ (31)$

The wave function for a free particle moving in the ${z}$ direction is therefore

$\displaystyle \psi_{E}\left(r,\theta,\phi\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}\sum_{l=0}^{\infty}i^{a}\left(2a+1\right)j_{l}\left(kr\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (32)$

# Free particle in spherical coordinates – finding the solutions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.6.

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In solving the Schrödinger equation for spherically symmetric potentials, we found that we could reduce the problem to the equation

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (1)$

where ${U_{El}\left(r\right)}$ is related to the radial function by

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (2)$

For a free particle, ${V=0}$ and ${E>0}$, so we have

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (3)$

Defining

 $\displaystyle k^{2}$ $\displaystyle \equiv$ $\displaystyle \frac{2\mu E}{\hbar^{2}}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle kr \ \ \ \ \ (5)$

we convert the equation to

$\displaystyle \left(-\frac{d^{2}}{d\rho^{2}}+\frac{l\left(l+1\right)}{\rho^{2}}\right)U_{l}=U_{l} \ \ \ \ \ (6)$

This equation can be solved by a method similar to that for the harmonic oscillator and its raising and lowering operators. The entire solution is fairly involved, so we’ll start out here by showing how the new raising and lowering operators are defined.

We define

$\displaystyle d_{l}\equiv\frac{d}{d\rho}+\frac{l+1}{\rho} \ \ \ \ \ (7)$

$\displaystyle d_{l}^{\dagger}=-\frac{d}{d\rho}+\frac{l+1}{\rho} \ \ \ \ \ (8)$

To see where the minus sign comes from on the RHS, we need to recall that the momentum operator is defined in one dimension as

$\displaystyle P=-i\hbar\frac{\partial}{\partial x} \ \ \ \ \ (9)$

Since ${P}$ is an observable, it is hermitian, so that ${P^{\dagger}=P}$. Under the hermitian operation ${i\rightarrow-i}$, so we must also have ${\frac{\partial}{\partial x}\rightarrow-\frac{\partial}{\partial x}}$. Thus the first derivative with respect to a position variable is anti-hermitian. If this doesn’t convince you, you can also work out the integral:

$\displaystyle \int_{0}^{\infty}\psi_{2}^*\frac{d}{d\rho}\psi_{1}d\rho=\left.\psi_{2}^*\psi_{1}\right|_{0}^{\infty}-\int_{0}^{\infty}\psi_{1}\frac{d}{d\rho}\psi_{2}^*d\rho \ \ \ \ \ (10)$

Under the usual assumption that ${\psi\rightarrow0}$ at the limits, the integrated term is zero and we have

 $\displaystyle \int_{0}^{\infty}\psi_{2}^*\frac{d}{d\rho}\psi_{1}d\rho$ $\displaystyle =$ $\displaystyle -\int_{0}^{\infty}\psi_{1}\frac{d}{d\rho}\psi_{2}^*d\rho\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left[\int_{0}^{\infty}\psi_{1}^*\frac{d}{d\rho}\psi_{2}d\rho\right]^* \ \ \ \ \ (12)$

In bracket notation, this is

$\displaystyle \left\langle \psi_{2}\left|\frac{d}{d\rho}\psi_{1}\right.\right\rangle =-\left\langle \frac{d}{d\rho}\psi_{2}\left|\psi_{1}\right.\right\rangle \ \ \ \ \ (13)$

which shows that ${\frac{d}{d\rho}}$ is an anti-hermitian operator.

Returning to 7 and 8, we have

 $\displaystyle d_{l}d_{l}^{\dagger}U_{l}$ $\displaystyle =$ $\displaystyle \left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)U_{l}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(-U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}-\frac{l+1}{\rho^{2}}U_{l}+\frac{l+1}{\rho}U_{l}^{\prime}-\frac{l+1}{\rho}U_{l}^{\prime}+\frac{\left(l+1\right)^{2}}{\rho^{2}}U_{l}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{l\left(l+1\right)}{\rho^{2}}U_{l} \ \ \ \ \ (17)$

Comparing with 6 we see that

$\displaystyle d_{l}d_{l}^{\dagger}U_{l}=U_{l} \ \ \ \ \ (18)$

We can also show that

 $\displaystyle d_{l}^{\dagger}d_{l}U_{l}$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)U_{l}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{l+1}{\rho^{2}}U_{l}-\frac{l+1}{\rho}U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}^{\prime}+\frac{\left(l+1\right)^{2}}{\rho^{2}}U_{l}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{\left(l+1\right)^{2}+l+1}{\rho^{2}}U_{l}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{\left(l+1\right)\left(l+2\right)}{\rho^{2}}U_{l}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle d_{l+1}d_{l+1}^{\dagger}U_{l} \ \ \ \ \ (24)$

Starting from 18 we multiply on the left by ${d_{l}^{\dagger}}$ to get

$\displaystyle d_{l}^{\dagger}d_{l}\left(d_{l}^{\dagger}U_{l}\right)=d_{l}^{\dagger}U_{l} \ \ \ \ \ (25)$

Comparing this with 24 we see that

$\displaystyle d_{l}^{\dagger}U_{l}=c_{l}U_{l+1} \ \ \ \ \ (26)$

where ${c_{l}}$ is a constant.

Thus ${d_{l}^{\dagger}}$ is a raising operator, in that it raises the angular momentum number ${l}$ by 1 when it acts on ${U_{l}}$. By convention, ${c_{l}=1}$ (any adjustments to the constant can be made when normalizing).

We can start the process by looking at 6 with ${l=0}$ which is

$\displaystyle \frac{d^{2}}{d\rho^{2}}U_{l}=-U_{l} \ \ \ \ \ (27)$

This has the two solutions

 $\displaystyle U_{0}^{A}\left(\rho\right)$ $\displaystyle =$ $\displaystyle \sin\rho\ \ \ \ \ (28)$ $\displaystyle U_{0}^{B}\left(\rho\right)$ $\displaystyle =$ $\displaystyle -\cos\rho \ \ \ \ \ (29)$

The minus sign in front of ${\cos\rho}$ is just conventional. Since we require ${U_{0}\left(0\right)=0}$, ${U_{0}^{B}}$ is unacceptable if the region we’re considering include ${\rho=0}$, so we have

$\displaystyle U_{0}\left(\rho\right)=\sin\rho \ \ \ \ \ (30)$

For the general case that excludes ${\rho=0}$, we must include the cosine term as well.

From here, we can generate solutions for higher values of ${l}$ by applying 26. Actually, the radial function that appears in the wave function is given by 2, so it is ${R_{l}}$ that we really want. That is, we want

$\displaystyle R_{l}=\frac{U_{l}}{r}=k\frac{U_{l}}{\rho} \ \ \ \ \ (31)$

As with the constant ${c_{l}}$ in 26, we can absorb ${k}$ into normalization to be done later, so we can generate functions

$\displaystyle R_{l}=\frac{U_{l}}{\rho} \ \ \ \ \ (32)$

Applying 26 we have

 $\displaystyle \rho R_{l+1}$ $\displaystyle =$ $\displaystyle d_{l}^{\dagger}\left(\rho R_{l}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(\rho R_{l}\right)\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{l}-\rho R_{l}^{\prime}+\left(l+1\right)R_{l}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho R_{l}^{\prime}+lR_{l}\ \ \ \ \ (36)$ $\displaystyle R_{l+1}$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l}{\rho}\right)R_{l}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho^{l}\frac{d}{d\rho}\left(\frac{R_{l}}{\rho^{l}}\right) \ \ \ \ \ (38)$

We can convert this into a general formula by writing

$\displaystyle \frac{R_{l+1}}{\rho^{l+1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{l}}{\rho^{l}} \ \ \ \ \ (39)$

Starting at ${l=0}$, we have

$\displaystyle \frac{R_{1}}{\rho^{1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (40)$

For the next step, we have

 $\displaystyle \frac{R_{2}}{\rho^{2}}$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{1}}{\rho^{1}}\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{0}}{\rho^{0}}\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{2}\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (43)$

Thus in general

$\displaystyle \frac{R_{l+1}}{\rho^{l+1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{l+1}\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (44)$

Note that

$\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{l+1}\ne\left(-\frac{1}{\rho}\right)^{l+1}\frac{d^{l+1}}{d\rho^{l+1}} \ \ \ \ \ (45)$

since the factor of ${\frac{1}{\rho}}$ has to be included when taking the derivative.

We’ll explore the nature of these solutions in the next post.

# Free particle propagator from a complete path integral

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.4.

We’ve seen that the free-particle propagator can be obtained in the path integral approach by using only the classical path in the sum over paths. It turns out that it’s not too hard to calculate the propagator for a free particle properly, by summing over all possible paths. The notation used by Shankar is as follows.

We want to evaluate the path integral

$\displaystyle \int_{x_{0}}^{x_{N}}e^{iS\left[x\left(t\right)\right]/\hbar}\mathfrak{D}\left[x\left(t\right)\right] \ \ \ \ \ (1)$

The notation ${\mathfrak{D}\left[x\left(t\right)\right]}$ means an integration over all possible paths from ${x_{0}}$ to ${x_{N}}$ in the given time interval. This includes paths where the particle might move to the right for a while, then jog back to the left, then back to the right again and so on. This might seem like a hopeless task, but we can make sense of this method by splitting the time interval between ${t_{0}}$ and ${t_{N}}$ into ${N}$ small intervals, each of length ${\varepsilon}$. Thus an intermediate time ${t_{n}=t_{0}+n\varepsilon}$, and the final time is ${t_{N}=t_{0}+N\varepsilon}$.

For a free particle, there is no potential energy so the Lagrangian is just the kinetic energy:

$\displaystyle L=\frac{1}{2}m\dot{x}^{2} \ \ \ \ \ (2)$

We can estimate the velocity in each time slice by

$\displaystyle \dot{x}_{i}=\frac{x_{i+1}-x_{i}}{\varepsilon} \ \ \ \ \ (3)$

Note that this assumes that the velocity within each time slice is constant, but as we make ${\varepsilon}$ smaller and smaller, this is increasingly accurate. Also note that it is possible for ${\dot{x}_{i}}$ to be both positive (if the particle moves to the right in the interval) or negative (if it moves to the left).

The action for a given path is given by the integral of the Lagrangian:

$\displaystyle S=\int_{t_{0}}^{t_{N}}L\left(t\right)dt \ \ \ \ \ (4)$

In our discretized approximation, we evaluate ${L}$ within each time slice, and ${dt}$ becomes the interval length ${\varepsilon}$, so the action becomes a sum:

 $\displaystyle S$ $\displaystyle =$ $\displaystyle \sum_{i=0}^{N-1}L\left(t_{i}\right)\varepsilon\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\sum_{i=0}^{N-1}\left(\frac{x_{i+1}-x_{i}}{\varepsilon}\right)^{2}\varepsilon\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\sum_{i=0}^{N-1}\frac{\left(x_{i+1}-x_{i}\right)^{2}}{\varepsilon} \ \ \ \ \ (7)$

The key point here is to notice that we can label any given path by choosing values for all the ${x_{i}}$s between the two times, and that each ${x_{i}}$ can vary independently of the others, over a range from ${-\infty}$ to ${+\infty}$. We can therefore implement the multiple integration required by ${\mathfrak{D}\left[x\left(t\right)\right]}$ by integrating over all the ${x_{i}}$ variables separately. That is,

$\displaystyle \int_{x_{0}}^{x_{N}}e^{iS\left[x\left(t\right)\right]/\hbar}\mathfrak{D}\left[x\left(t\right)\right]=A\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty}\exp\left[\frac{im}{2\hbar}\sum_{i=0}^{N-1}\frac{\left(x_{i+1}-x_{i}\right)^{2}}{\varepsilon}\right]dx_{1}dx_{2}\ldots dx_{N-1} \ \ \ \ \ (8)$

where ${A}$ is some constant to make the scale come out right.

We don’t integrate over ${x_{0}}$ or ${x_{N}}$ since these are fixed as the end points of the path. To get the final version, we need to take the limit of this expression as ${N\rightarrow\infty}$ and ${\varepsilon\rightarrow0}$. This still looks pretty scary, but in fact it is doable. We define the variable

 $\displaystyle y_{i}$ $\displaystyle \equiv$ $\displaystyle \sqrt{\frac{m}{2\hbar\varepsilon}}x_{i}\ \ \ \ \ (9)$ $\displaystyle dx_{i}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\hbar\varepsilon}{m}}dy_{i} \ \ \ \ \ (10)$

This gives us

$\displaystyle A\left(\frac{2\hbar\varepsilon}{m}\right)^{\left(N-1\right)/2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty}\exp\left[i\sum_{i=0}^{N-1}\left(y_{i+1}-y_{i}\right)^{2}\right]dy_{1}dy_{2}\ldots dy_{N-1} \ \ \ \ \ (11)$

We can do the integral in stages in order to spot a pattern. Consider first the integral over ${y_{1}}$, which involves only two of the factors in the integrand:

$\displaystyle \int_{-\infty}^{\infty}e^{i\left[\left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}dy_{1} \ \ \ \ \ (12)$

We first simplify the exponent

 $\displaystyle \left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$ $\displaystyle =$ $\displaystyle y_{2}^{2}+y_{0}^{2}+2\left(y_{1}^{2}-y_{0}y_{1}-y_{1}y_{2}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle y_{2}^{2}+y_{0}^{2}+2y_{1}^{2}-2\left(y_{0}+y_{2}\right)y_{1} \ \ \ \ \ (14)$

We get

$\displaystyle \int_{-\infty}^{\infty}e^{i\left[\left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}dy_{1}=e^{i\left(y_{2}^{2}+y_{0}^{2}\right)}\int_{-\infty}^{\infty}e^{2i\left[y_{1}^{2}-\left(y_{0}+y_{2}\right)y_{1}\right]}dy_{1} \ \ \ \ \ (15)$

We can evaluate this using a standard Gaussian integral

$\displaystyle \int_{-\infty}^{\infty}e^{-ax^{2}+bx}dx=e^{b^{2}/4a}\sqrt{\frac{\pi}{a}} \ \ \ \ \ (16)$

This gives

 $\displaystyle \int_{-\infty}^{\infty}e^{i\left[\left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}dy_{1}$ $\displaystyle =$ $\displaystyle e^{i\left(y_{2}^{2}+y_{0}^{2}\right)}e^{4\left(y_{0}+y_{2}\right)^{2}/8i}\sqrt{-\frac{\pi}{2i}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\left(y_{2}^{2}+y_{0}^{2}\right)}e^{\left(y_{0}+y_{2}\right)^{2}/2i}\sqrt{\frac{\pi i}{2}} \ \ \ \ \ (18)$

To simplify the exponents on the RHS:

 $\displaystyle i\left(y_{2}^{2}+y_{0}^{2}\right)+\frac{\left(y_{0}+y_{2}\right)^{2}}{2i}$ $\displaystyle =$ $\displaystyle \frac{1}{2i}\left[\left(y_{0}+y_{2}\right)^{2}-2y_{2}^{2}-2y_{0}^{2}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2i}\left(y_{0}-y_{2}\right)^{2} \ \ \ \ \ (20)$

Thus we have

$\displaystyle \int_{-\infty}^{\infty}e^{i\left[\left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}dy_{1}=\sqrt{\frac{\pi i}{2}}e^{-\left(y_{0}-y_{2}\right)^{2}/2i} \ \ \ \ \ (21)$

Having eliminated ${y_{1}}$ we can now do the integral over ${y_{2}}$:

$\displaystyle \sqrt{\frac{\pi i}{2}}\int_{-\infty}^{\infty}e^{-\left(y_{3}-y_{2}\right)^{2}/i-\left(y_{2}-y_{0}\right)^{2}/2i}dy_{2} \ \ \ \ \ (22)$

Again, we can simplify the exponent:

$\displaystyle -\frac{\left(y_{3}-y_{2}\right)^{2}}{i}-\frac{\left(y_{2}-y_{0}\right)^{2}}{2i}=\frac{1}{2i}\left[-\left(2y_{3}^{2}+y_{0}^{2}\right)-3y_{2}^{2}+y_{2}\left(4y_{3}+2y_{0}\right)\right] \ \ \ \ \ (23)$

The integral now becomes

 $\displaystyle \sqrt{\frac{\pi i}{2}}\int_{-\infty}^{\infty}e^{-\left(y_{3}-y_{2}\right)^{2}/i-\left(y_{2}-y_{0}\right)^{2}/2i}dy_{2}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\pi i}{2}}e^{-\left(2y_{3}^{2}+y_{0}^{2}\right)/2i}\int_{-\infty}^{\infty}e^{\left(-3y_{2}^{2}+y_{2}\left(4y_{3}+2y_{0}\right)\right)/2i}dy_{2} \ \ \ \ \ (24)$

Doing the Gaussian integral on the RHS using 16:

 $\displaystyle \int_{-\infty}^{\infty}e^{\left(-3y_{2}^{2}+y_{2}\left(4y_{3}+2y_{0}\right)\right)/2i}dy_{2}$ $\displaystyle =$ $\displaystyle e^{-\left(4y_{3}+2y_{0}\right)^{2}i/24}\sqrt{\frac{2\pi i}{3}}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{\left(2y_{3}+y_{0}\right)^{2}/6i}\sqrt{\frac{2\pi i}{3}} \ \ \ \ \ (26)$

Thus the combined integral over ${y_{1}}$ and ${y_{2}}$ is

 $\displaystyle \sqrt{\frac{\pi i}{2}}e^{-\left(2y_{3}^{2}+y_{0}^{2}\right)/2i}e^{\left(2y_{3}+y_{0}\right)^{2}/6i}\sqrt{\frac{2\pi i}{3}}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\left(\pi i\right)^{2}}{3}}e^{\left(-6y_{3}^{2}-3y_{0}^{2}+\left(2y_{3}+y_{0}\right)^{2}\right)/6i}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\left(\pi i\right)^{2}}{3}}e^{-\left(y_{3}-y_{0}\right)^{2}/3i} \ \ \ \ \ (28)$

The general pattern after ${N-1}$ integrations is (presumably this could be proved by induction, but we’ll accept the result):

$\displaystyle \frac{\left(\pi i\right)^{\left(N-1\right)/2}}{\sqrt{N}}e^{-\left(y_{N}-y_{0}\right)^{2}/Ni}=\frac{\left(\pi i\right)^{\left(N-1\right)/2}}{\sqrt{N}}e^{-m\left(x_{N}-x_{0}\right)^{2}/2\hbar\varepsilon Ni} \ \ \ \ \ (29)$

where we reverted back to ${x_{i}}$ using 9.

Going back to 11, we must multiply the result by ${A\left(\frac{2\hbar\varepsilon}{m}\right)^{\left(N-1\right)/2}}$ to get the final expression for the propagator:

 $\displaystyle U$ $\displaystyle =$ $\displaystyle A\left(\frac{2\hbar\varepsilon}{m}\right)^{\left(N-1\right)/2}\frac{\left(\pi i\right)^{\left(N-1\right)/2}}{\sqrt{N}}e^{-m\left(x_{N}-x_{0}\right)^{2}/2\hbar\varepsilon Ni}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\left(\frac{2\pi\hbar\varepsilon i}{m}\right)^{N/2}\sqrt{\frac{m}{2\pi\hbar iN\varepsilon}}e^{im\left(x_{N}-x_{0}\right)^{2}/2\hbar\varepsilon N} \ \ \ \ \ (31)$

In the limit as ${N\rightarrow\infty}$ and ${\varepsilon\rightarrow0}$, ${N\varepsilon=t_{N}-t_{0}}$ so we have

$\displaystyle U=A\left(\frac{2\pi\hbar\varepsilon i}{m}\right)^{N/2}\sqrt{\frac{m}{2\pi\hbar i\left(t_{N}-t_{0}\right)}}e^{im\left(x_{N}-x_{0}\right)^{2}/2\hbar\left(t_{N}-t_{0}\right)} \ \ \ \ \ (32)$

The expression we got earlier using the Schrödinger method is

$\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=\sqrt{\frac{m}{2\pi\hbar i\left(t-t^{\prime}\right)}}e^{im\left(x-x^{\prime}\right)^{2}/2\hbar\left(t-t^{\prime}\right)} \ \ \ \ \ (33)$

Thus the full path integral gives the same result, with ${t^{\prime}=t_{0}}$ and ${t=t_{N}}$ (similarly for ${x}$), provided that we can set

$\displaystyle A=\left(\frac{m}{2\pi\hbar\varepsilon i}\right)^{N/2}\equiv B^{-N} \ \ \ \ \ (34)$

Shankar then says that it is conventional to associate one factor of ${B^{-1}}$ with each integration over an ${x_{i}}$, and the remaining factor with the overall process. This seems to overlook a basic problem, in that as ${N\rightarrow\infty}$ and ${\varepsilon\rightarrow0}$, ${A\rightarrow\infty}$, so we seem to be cancelling two infinities when we multiply the path integral by ${A}$.

# 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.

# 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.

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

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.