3.2.2 Schrödinger Equation#

Worked solutions for the homework problems in the 3.2.2 Schrödinger Equation lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Zeroth Gaussian moment. Use the polar-coordinate trick to derive \(\int\mathrm{e}^{-r^{2}}\mathrm{d}^{2}r = \pi\) in the plane.

(a) From this, derive \(\int\mathrm{e}^{-bx^{2}}\,\mathrm{d}x = \sqrt{\pi/b}\) for \(\operatorname{Re}b>0\) by separating the 2D integral and rescaling.

(b) Analytically continue \(b\to-\mathrm{i}\alpha\) with \(\alpha>0\) (rotate the contour by \(\pi/4\) and check convergence) to obtain \(M_{0} = \sqrt{\mathrm{i}\pi/\alpha}\).

Solution.

The polar trick. Consider the 2D integral over the whole plane,

\[ J = \int_{\mathbb{R}^{2}}\mathrm{e}^{-r^{2}}\,\mathrm{d}^{2}r = \int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty}\mathrm{e}^{-(x^{2}+y^{2})}\,\mathrm{d}x\,\mathrm{d}y, \]

where \(r^{2} = x^{2}+y^{2}\). Switch to polar coordinates \((r,\theta)\) with area element \(\mathrm{d}^{2}r = r\,\mathrm{d}r\,\mathrm{d}\theta\):

\[ J = \int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{\infty}\mathrm{e}^{-r^{2}}\,r\,\mathrm{d}r = 2\pi\left[-\tfrac{1}{2}\mathrm{e}^{-r^{2}}\right]_{0}^{\infty} = 2\pi\cdot\tfrac{1}{2} = \pi. \]

The substitution \(u = r^{2}\), \(\mathrm{d}u = 2r\,\mathrm{d}r\) makes the radial integral elementary. Hence \(\int\mathrm{e}^{-r^{2}}\mathrm{d}^{2}r = \pi\).

(a) Reduction to 1D and rescaling. Because the integrand factorizes, \(\mathrm{e}^{-(x^{2}+y^{2})} = \mathrm{e}^{-x^{2}}\mathrm{e}^{-y^{2}}\), the 2D integral is the square of a single 1D integral:

\[ \pi = J = \left(\int_{-\infty}^{\infty}\mathrm{e}^{-x^{2}}\,\mathrm{d}x\right)^{2}, \]

so \(\int_{-\infty}^{\infty}\mathrm{e}^{-x^{2}}\,\mathrm{d}x = \sqrt{\pi}\).

The positive root is correct since the integrand is positive. Now rescale: for \(\operatorname{Re}b>0\) substitute \(x = u/\sqrt{b}\) (principal branch of the root), so \(\mathrm{d}x = \mathrm{d}u/\sqrt{b}\) and \(bx^{2} = u^{2}\):

\[ \int_{-\infty}^{\infty}\mathrm{e}^{-bx^{2}}\,\mathrm{d}x = \frac{1}{\sqrt{b}}\int_{-\infty}^{\infty}\mathrm{e}^{-u^{2}}\,\mathrm{d}u = \sqrt{\frac{\pi}{b}}. \]

The condition \(\operatorname{Re}b>0\) guarantees the original integrand decays at infinity so the integral converges and the substitution is legitimate.

(b) Analytic continuation to the oscillatory Gaussian. We want \(M_{0} = \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\mathrm{d}(\delta x)\) with \(\alpha>0\), i.e. the formula above evaluated at \(b = -\mathrm{i}\alpha\) (since \(-bx^{2} = \mathrm{i}\alpha x^{2}\)). This value of \(b\) sits on the boundary \(\operatorname{Re}b = 0\), so we must justify the continuation by an explicit contour rotation.

Treat \(\delta x\) as complex and write \(\delta x = \mathrm{e}^{\mathrm{i}\pi/4}u\) with \(u\) real. Then

\[ \mathrm{i}\alpha(\delta x)^{2} = \mathrm{i}\alpha\,\mathrm{e}^{\mathrm{i}\pi/2}u^{2} = \mathrm{i}\alpha\cdot\mathrm{i}\,u^{2} = -\alpha u^{2}, \]

which is a convergent real Gaussian for \(\alpha>0\). The integrand \(\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\) is entire, so by Cauchy’s theorem the integral over the real axis equals the integral over the ray \(\arg(\delta x) = \pi/4\), provided the connecting circular arcs at infinity contribute nothing. On an arc \(\delta x = R\mathrm{e}^{\mathrm{i}\phi}\) the exponent has real part

\[ \operatorname{Re}\!\left[\mathrm{i}\alpha R^{2}\mathrm{e}^{2\mathrm{i}\phi}\right] = -\alpha R^{2}\sin(2\phi) \le 0 \quad\text{for } 0\le\phi\le\tfrac{\pi}{4}, \]

with strict decay in the interior, so the arc contribution vanishes as \(R\to\infty\). Therefore

\[ M_{0} = \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\mathrm{d}(\delta x) = \mathrm{e}^{\mathrm{i}\pi/4}\int_{-\infty}^{\infty}\mathrm{e}^{-\alpha u^{2}}\,\mathrm{d}u = \mathrm{e}^{\mathrm{i}\pi/4}\sqrt{\frac{\pi}{\alpha}}. \]

Since \(\mathrm{e}^{\mathrm{i}\pi/4} = \sqrt{\mathrm{i}}\), this is exactly

\[ M_{0} = \sqrt{\frac{\mathrm{i}\pi}{\alpha}}, \]

in agreement with the naive substitution \(\sqrt{\pi/(-\mathrm{i}\alpha)} = \sqrt{\mathrm{i}\pi/\alpha}\) (using \(1/(-\mathrm{i}) = \mathrm{i}\)). Substituting \(\alpha = m/(2\hbar\,\delta t)\) gives the form used in the lecture, \(M_{0} = \sqrt{2\pi\mathrm{i}\hbar\,\delta t/m}\).

2. Second Gaussian moment. Starting from \(M_{0}(\alpha) = \sqrt{\mathrm{i}\pi/\alpha}\), use the derivative trick \(\partial_{\alpha}M_{0} = \mathrm{i}M_{2}\) to compute \(M_{2}\) and verify the result \(M_{2} = (\mathrm{i}\hbar\,\delta t/m)\,M_{0}\) stated in \(M_0 = \sqrt{2\pi\mathrm{i}\hbar\delta t/m},\; M_1=0,\;M_2 = (\mathrm{i}\hbar\delta t/m)\,M_0\).

Solution.

The moment \(M_{0}(\alpha) = \int\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\mathrm{d}(\delta x)\) depends on \(\alpha\) only through the exponent. Differentiating under the integral sign brings down a factor \(\mathrm{i}(\delta x)^{2}\):

\[ \frac{\partial M_{0}}{\partial\alpha} = \int_{-\infty}^{\infty}\mathrm{i}(\delta x)^{2}\,\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\mathrm{d}(\delta x) = \mathrm{i}\int_{-\infty}^{\infty}(\delta x)^{2}\,\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\mathrm{d}(\delta x) = \mathrm{i}\,M_{2}. \]

On the other hand, differentiating the closed form \(M_{0} = \sqrt{\mathrm{i}\pi}\;\alpha^{-1/2}\) directly,

\[ \frac{\partial M_{0}}{\partial\alpha} = \sqrt{\mathrm{i}\pi}\left(-\tfrac{1}{2}\right)\alpha^{-3/2} = -\frac{1}{2\alpha}\,\sqrt{\mathrm{i}\pi}\;\alpha^{-1/2} = -\frac{1}{2\alpha}\,M_{0}. \]

Equate the two expressions for \(\partial_{\alpha}M_{0}\) and solve for \(M_{2}\):

\[ \mathrm{i}\,M_{2} = -\frac{M_{0}}{2\alpha}, \]

so \(M_{2} = -(1/\mathrm{i})\,M_{0}/(2\alpha) = \mathrm{i}\,M_{0}/(2\alpha)\),

using \(-1/\mathrm{i} = \mathrm{i}\). Finally substitute the kinetic-phase coefficient \(\alpha = m/(2\hbar\,\delta t)\), so that \(1/(2\alpha) = \hbar\,\delta t/m\):

\[ M_{2} = \frac{\mathrm{i}}{2\alpha}\,M_{0} = \frac{\mathrm{i}\hbar\,\delta t}{m}\,M_{0}, \]

which is exactly the result quoted in \(M_0 = \sqrt{2\pi\mathrm{i}\hbar\delta t/m},\; M_1=0,\;M_2 = (\mathrm{i}\hbar\delta t/m)\,M_0\). Note the physically important feature: \(M_{2}/M_{0} = \mathrm{i}\hbar\,\delta t/m\) is linear in \(\delta t\). After multiplication by the normalization \(A = 1/M_{0}\), the second moment contributes a term of order \(\delta t\) — exactly the order that survives to build the kinetic term of the Schrödinger equation.

3. Verifying the slice normalization. For the closed-form free slice propagator \(K_{\delta t}^{\mathrm{free}}(x,x') = \sqrt{m/(2\pi\mathrm{i}\hbar\,\delta t)}\,\exp\!\bigl[\mathrm{i}\,m(x-x')^2/(2\hbar\delta t)\bigr]\), evaluate \(\int K_{\delta t}^{\text{free}}(x,x')\,\mathrm{d}x'\) by Gaussian integration. Show that the result equals \(1\), independent of \(x\) and \(\delta t\). What does this say about how a constant initial wavefunction evolves over one slice?

Solution.

The free slice propagator is

\[ K_{\delta t}^{\text{free}}(x,x') = \sqrt{\frac{m}{2\pi\mathrm{i}\hbar\,\delta t}}\;\exp\!\left[\frac{\mathrm{i}\,m\,(x-x')^{2}}{2\hbar\,\delta t}\right]. \]

Integrate over the source point \(x'\). Substitute \(u = x'-x\), so \(\mathrm{d}x' = \mathrm{d}u\) and \((x-x')^{2} = u^{2}\). The kinetic exponent becomes \(\mathrm{i}\alpha u^{2}\) with \(\alpha = m/(2\hbar\,\delta t)\):

\[ \int_{-\infty}^{\infty}K_{\delta t}^{\text{free}}(x,x')\,\mathrm{d}x' = \sqrt{\frac{m}{2\pi\mathrm{i}\hbar\,\delta t}}\int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}\alpha u^{2}}\,\mathrm{d}u = \sqrt{\frac{m}{2\pi\mathrm{i}\hbar\,\delta t}}\;M_{0}. \]

The remaining integral is precisely the zeroth Gaussian moment from Problem 1,

\[ M_{0} = \sqrt{\frac{\mathrm{i}\pi}{\alpha}} = \sqrt{\frac{2\pi\mathrm{i}\hbar\,\delta t}{m}}. \]

Multiplying the two square roots, the entire \(\delta t\)- and \(m\)-dependence cancels:

\[ \int_{-\infty}^{\infty}K_{\delta t}^{\text{free}}(x,x')\,\mathrm{d}x' = \sqrt{\frac{m}{2\pi\mathrm{i}\hbar\,\delta t}}\cdot\sqrt{\frac{2\pi\mathrm{i}\hbar\,\delta t}{m}} = 1. \]

The result is \(1\) for every \(x\) and every \(\delta t\) — the integral does not depend on either.

Physical interpretation. Consider a spatially uniform initial wavefunction \(\psi(x',t) = c\) (a constant). One slice later,

\[ \psi(x,t+\delta t) = \int_{-\infty}^{\infty}K_{\delta t}^{\text{free}}(x,x')\,\psi(x',t)\,\mathrm{d}x' = c\int_{-\infty}^{\infty}K_{\delta t}^{\text{free}}(x,x')\,\mathrm{d}x' = c. \]

A constant wavefunction is left exactly unchanged by the free slice propagator, for any slice duration. This is precisely the self-consistency condition \(A(\delta t)\,M_{0} = 1\) that was used in the lecture to fix the normalization \(A(\delta t) = 1/M_{0}\). The condition was imposed only in the limit \(\delta t\to 0\) to fix one constant; this computation shows it actually holds at finite \(\delta t\) as well. Consistency check against the free Schrödinger equation \(\mathrm{i}\hbar\,\partial_{t}\psi = -(\hbar^{2}/2m)\,\partial_{x}^{2}\psi\): a constant \(\psi\) has \(\partial_{x}^{2}\psi = 0\), hence \(\partial_{t}\psi = 0\) — a uniform wave is stationary, exactly as the slice integral confirms.

4. Slice propagator approaches a delta function. As the slice duration \(\delta t \to 0\), the slice propagator \(K_{\delta t}^{\mathrm{free}}(x, x')\) should reduce to the identity: \(\psi(x, t + \delta t) \to \psi(x, t)\). Make this precise.

(a) Show that \(\int_{-\infty}^{\infty} K_{\delta t}^{\mathrm{free}}(x, x')\, f(x')\, \mathrm{d}x' \to f(x)\) as \(\delta t \to 0\) for any smooth, bounded test function \(f\).

(b) Compute \(\int_{-\infty}^{\infty} K_{\delta t}^{\mathrm{free}}(x, x')\, x'\, \mathrm{d}x'\) exactly. Show it equals \(x\) — the slice propagator preserves the mean position.

(c) Compute \(\int_{-\infty}^{\infty} K_{\delta t}^{\mathrm{free}}(x, x')\, x'^{2}\, \mathrm{d}x'\). Show that it equals \(x^{2} + \mathrm{i}\hbar\,\delta t/m\) — a complex number, signalling that the slice propagator is a phase-distribution kernel rather than a probability distribution. Interpret physically.

Solution.

(a) The free slice propagator is

\[ K_{\delta t}^{\mathrm{free}}(x, x') = \sqrt{\frac{m}{2\pi\mathrm{i}\hbar\,\delta t}}\,\exp\!\left[\frac{\mathrm{i}m(x-x')^{2}}{2\hbar\,\delta t}\right]. \]

Substitute \(u = x - x'\). The exponent’s coefficient is \(\alpha = m/(2\hbar\,\delta t)\), which diverges as \(\delta t \to 0\). Writing the kernel as \(K = \sqrt{\alpha/(\mathrm{i}\pi)}\,\mathrm{e}^{\mathrm{i}\alpha u^{2}}\), this is a Fresnel-Gaussian peak: the oscillations become arbitrarily rapid as \(\alpha \to \infty\), and only the immediate neighborhood \(\vert u\vert \lesssim 1/\sqrt\alpha = \sqrt{\hbar\,\delta t/m}\) contributes coherently to the integral. The width of the coherent region shrinks as \(\sqrt{\delta t} \to 0\), while Problem 3 showed the normalization stays \(1\). These are the defining properties of a delta function:

\[ \int K_{\delta t}^{\mathrm{free}}(x, x') f(x')\,\mathrm{d}x' \xrightarrow{\delta t \to 0} f(x). \]

Explicit verification by Taylor-expanding \(f\): with \(f(x') = f(x - u) = \sum_n \frac{(-u)^n}{n!} \partial_x^n f(x)\), the moment expansion (from Problem 6) gives \(\sum_k \frac{A M_{2k}}{(2k)!}\partial_x^{2k}f(x)\). As \(\delta t \to 0\), only the \(k=0\) term (\(A M_0 = 1\)) survives; every higher \(k\) vanishes as \(O(\delta t^k)\). So the integral collapses to \(f(x)\).

(b) Compute the first moment using \(u = x - x'\), so \(x' = x - u\):

\[ \int K_{\delta t}^{\mathrm{free}}(x, x')\, x'\, \mathrm{d}x' = \int K(x, x - u)\,(x - u)\,\mathrm{d}u = x\int K\,\mathrm{d}u - \int K\,u\,\mathrm{d}u. \]

The first integral is \(x \cdot A\,M_0 = x \cdot 1 = x\) (Problem 3). The second integral has integrand \(u\,\mathrm{e}^{\mathrm{i}\alpha u^2}\) which is odd in \(u\), so it vanishes: \(A\,M_1 = 0\). Therefore

\[ \int K_{\delta t}^{\mathrm{free}}(x, x')\, x'\, \mathrm{d}x' = x. \quad\checkmark \]

The slice propagator preserves the mean position — consistent with a free particle (no force, no drift).

\[ \int K\,x'^{2}\,\mathrm{d}x' = \int K(x-u)^{2}\,\mathrm{d}u = \int K\,(x^{2} - 2xu + u^{2})\,\mathrm{d}u = x^{2} \cdot 1 - 2x\cdot 0 + A\,M_{2}. \]

From Problem 2, \(A\,M_{2} = M_{2}/M_{0} = \mathrm{i}\hbar\,\delta t/m\). Hence

\[ \int K_{\delta t}^{\mathrm{free}}(x, x')\, x'^{2}\, \mathrm{d}x' = x^{2} + \frac{\mathrm{i}\hbar\,\delta t}{m}. \quad\checkmark \]

This is a complex number. A genuine probability distribution always has a real mean and a real (non-negative) variance, so the slice kernel cannot be a probability distribution — it is the amplitude kernel that defines the propagation of wavefunctions, not the diffusion kernel of a stochastic process.

Physical interpretation. The imaginary \(\mathrm{i}\hbar\,\delta t/m\) in the second moment is the seed of the Schrödinger equation’s kinetic term. Compared with a real diffusion kernel — where \(\langle (\Delta x)^{2}\rangle = 2D\,\delta t\) generates the heat equation \(\partial_{t}\rho = D\,\partial_{x}^{2}\rho\) — the substitution \(D \to \mathrm{i}\hbar/(2m)\) converts the heat equation into the free-particle Schrödinger equation \(\mathrm{i}\hbar\,\partial_{t}\psi = -(\hbar^{2}/2m)\,\partial_{x}^{2}\psi\). Quantum evolution is “imaginary-time” diffusion. The factor of \(\mathrm{i}\) is not cosmetic: it flips real exponential decay (broadening of a heat distribution) into pure phase rotation (oscillations of a wavefunction at fixed amplitude), which is why \(\vert\psi\vert^{2}\) stays normalized while a heat distribution spreads. The imaginary-time substitution \(t \to -\mathrm{i}\tau\) would convert Schrödinger evolution to diffusion and vice versa, and this duality underlies many computational techniques (imaginary-time path integral, ground-state Monte Carlo, etc.) that map quantum problems to statistical ones.

★ 5. Probability conservation. Starting from \(\mathrm{i}\hbar\,\partial_t\psi = \hat{H}\psi,\quad\hat{H} = -(\hbar^2/2m)\partial_x^2 + V(x)\) with real \(V(x)\), derive the continuity equation \(\partial_{t}\rho + \partial_{x}j = 0\) with \(\rho = \vert\psi\vert^{2}\) and \(j = (\hbar/m)\operatorname{Im}(\psi^{*}\partial_{x}\psi)\).

(a) Compute \(\partial_{t}\vert\psi\vert^{2}\) using the Schrödinger equation and its complex conjugate.

(b) Identify the probability current \(j\) and verify \(\partial_{t}\rho + \partial_{x}j = 0\).

(c) Integrate over all \(x\), assuming \(\psi\to 0\) at infinity, to conclude that \(\frac{\mathrm{d}}{\mathrm{d}t}\int\vert\psi\vert^{2}\,\mathrm{d}x = 0\).

(d) Continuity on the Gaussian wavepacket. Take the free-particle limit \(V = 0\). The Gaussian wavepacket of HW 3.2.1.1 admits the closed-form wave function at time \(t > 0\),

\[ \psi(x,t) = \frac{1}{(\pi\sigma^{2})^{1/4}}\,\frac{1}{\sqrt{1 + \mathrm{i}t/\tau}}\,\exp\!\left[-\frac{x^{2}}{2\sigma^{2}\,(1 + \mathrm{i}t/\tau)}\right], \qquad \tau = \frac{m\sigma^{2}}{\hbar}. \]

Compute the probability density \(\rho(x,t) = \vert\psi(x,t)\vert^{2}\) and the probability current \(j(x,t)\) identified in (b). Verify that the continuity equation \(\partial_{t}\rho + \partial_{x}j = 0\) derived in (a)–(b) is satisfied identically at all \((x,t)\).

Solution.

The Schrödinger equation and its complex conjugate (with \(V(x)\) real, so \(V^{*} = V\)) are

\[ \mathrm{i}\hbar\,\partial_{t}\psi = -\frac{\hbar^{2}}{2m}\,\partial_{x}^{2}\psi + V\psi, \qquad -\mathrm{i}\hbar\,\partial_{t}\psi^{*} = -\frac{\hbar^{2}}{2m}\,\partial_{x}^{2}\psi^{*} + V\psi^{*}. \]

Solve each for the time derivative:

\[ \partial_{t}\psi = \frac{\mathrm{i}\hbar}{2m}\,\partial_{x}^{2}\psi - \frac{\mathrm{i}}{\hbar}\,V\psi, \qquad \partial_{t}\psi^{*} = -\frac{\mathrm{i}\hbar}{2m}\,\partial_{x}^{2}\psi^{*} + \frac{\mathrm{i}}{\hbar}\,V\psi^{*}. \]

(a) Time derivative of the density. With \(\rho = \vert\psi\vert^{2} = \psi^{*}\psi\), the product rule gives

\[ \partial_{t}\rho = \psi^{*}\,\partial_{t}\psi + \psi\,\partial_{t}\psi^{*}. \]

Substitute the two expressions above:

\[\begin{split} \begin{split} \partial_{t}\rho &= \psi^{*}\!\left(\frac{\mathrm{i}\hbar}{2m}\,\partial_{x}^{2}\psi - \frac{\mathrm{i}}{\hbar}V\psi\right) + \psi\!\left(-\frac{\mathrm{i}\hbar}{2m}\,\partial_{x}^{2}\psi^{*} + \frac{\mathrm{i}}{\hbar}V\psi^{*}\right)\\ &= \frac{\mathrm{i}\hbar}{2m}\left(\psi^{*}\,\partial_{x}^{2}\psi - \psi\,\partial_{x}^{2}\psi^{*}\right) - \frac{\mathrm{i}}{\hbar}V\,\psi^{*}\psi + \frac{\mathrm{i}}{\hbar}V\,\psi\psi^{*}. \end{split} \end{split}\]

The two potential terms are equal and opposite — they cancel exactly. This cancellation is why \(V\) must be real: a real potential contributes no source or sink of probability. Hence

\[ \partial_{t}\rho = \frac{\mathrm{i}\hbar}{2m}\left(\psi^{*}\,\partial_{x}^{2}\psi - \psi\,\partial_{x}^{2}\psi^{*}\right). \]

(b) The probability current. The bracket is a total spatial derivative. Indeed,

\[ \partial_{x}\!\left(\psi^{*}\,\partial_{x}\psi - \psi\,\partial_{x}\psi^{*}\right) = \underbrace{\partial_{x}\psi^{*}\,\partial_{x}\psi + \psi^{*}\,\partial_{x}^{2}\psi}_{\partial_{x}(\psi^{*}\partial_{x}\psi)} - \underbrace{\partial_{x}\psi\,\partial_{x}\psi^{*} - \psi\,\partial_{x}^{2}\psi^{*}}_{\partial_{x}(\psi\partial_{x}\psi^{*})} = \psi^{*}\,\partial_{x}^{2}\psi - \psi\,\partial_{x}^{2}\psi^{*}, \]

since the cross terms \(\partial_{x}\psi^{*}\,\partial_{x}\psi\) cancel. Therefore

\[ \partial_{t}\rho = \frac{\mathrm{i}\hbar}{2m}\,\partial_{x}\!\left(\psi^{*}\,\partial_{x}\psi - \psi\,\partial_{x}\psi^{*}\right) = -\,\partial_{x}\!\left[\,\frac{\mathrm{i}\hbar}{2m}\!\left(\psi\,\partial_{x}\psi^{*} - \psi^{*}\,\partial_{x}\psi\right)\right]. \]

Define the probability current as the quantity inside the spatial derivative,

\[ j \;\equiv\; -\frac{\mathrm{i}\hbar}{2m}\left(\psi^{*}\,\partial_{x}\psi - \psi\,\partial_{x}\psi^{*}\right), \]

so that \(\partial_{t}\rho = -\partial_{x}j\), i.e. the continuity equation

\[ \partial_{t}\rho + \partial_{x}j = 0. \]

To bring \(j\) to the stated form, note that \(\psi\,\partial_{x}\psi^{*} = (\psi^{*}\,\partial_{x}\psi)^{*}\), so the bracket is a number minus its own complex conjugate:

\[ \psi^{*}\,\partial_{x}\psi - \psi\,\partial_{x}\psi^{*} = \psi^{*}\,\partial_{x}\psi - (\psi^{*}\,\partial_{x}\psi)^{*} = 2\mathrm{i}\,\operatorname{Im}\!\left(\psi^{*}\,\partial_{x}\psi\right). \]

Hence

\[ j = -\frac{\mathrm{i}\hbar}{2m}\cdot 2\mathrm{i}\,\operatorname{Im}\!\left(\psi^{*}\,\partial_{x}\psi\right) = \frac{\hbar}{m}\,\operatorname{Im}\!\left(\psi^{*}\,\partial_{x}\psi\right), \]

exactly as stated.

(c) Global conservation. Integrate the continuity equation over all space and exchange the order of \(\partial_{t}\) and \(\int\mathrm{d}x\):

\[ \frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}\vert\psi\vert^{2}\,\mathrm{d}x = \int_{-\infty}^{\infty}\partial_{t}\rho\,\mathrm{d}x = -\int_{-\infty}^{\infty}\partial_{x}j\,\mathrm{d}x = -\,\bigl[\,j(x,t)\,\bigr]_{x=-\infty}^{x=+\infty}. \]

For a normalizable state \(\psi\to 0\) as \(x\to\pm\infty\) (and its derivative is bounded), so the current \(j = (\hbar/m)\operatorname{Im}(\psi^{*}\partial_{x}\psi)\) vanishes at both endpoints. Therefore

\[ \frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}\vert\psi\vert^{2}\,\mathrm{d}x = 0. \]

The total probability is constant in time: if \(\psi\) is normalized once, it stays normalized under Schrödinger evolution. Probability is locally conserved (continuity equation) and globally conserved (constant norm) — this is the statement that Schrödinger time evolution is unitary.

(d) Continuity on the Gaussian wavepacket. Squaring the magnitude of the given \(\psi(x,t)\) and using \(\vert 1 + \mathrm{i}t/\tau\vert = \sqrt{1 + (t/\tau)^{2}}\) together with \(\operatorname{Re}\!\left[1/(1 + \mathrm{i}t/\tau)\right] = 1/(1 + (t/\tau)^{2})\),

\[ \rho(x,t) = \frac{1}{\sqrt\pi\,\sigma_{t}}\,\exp\!\left(-\frac{x^{2}}{\sigma_{t}^{2}}\right), \qquad \sigma_{t} = \sigma\sqrt{1 + (t/\tau)^{2}}. \]

For the current, differentiate the wave function,

\[ \partial_{x}\psi = \psi\cdot\!\left[-\frac{x}{\sigma^{2}\,(1 + \mathrm{i}t/\tau)}\right], \qquad \psi^{*}\partial_{x}\psi = \rho\cdot\!\left[-\frac{x}{\sigma^{2}\,(1 + \mathrm{i}t/\tau)}\right]. \]

The bracket equals \(-x\,(1 - \mathrm{i}t/\tau)/[\sigma^{2}(1 + (t/\tau)^{2})]\), whose imaginary part is \(x\,t/(\tau\,\sigma_{t}^{2})\). Hence

\[ j(x,t) = \frac{\hbar}{m}\,\operatorname{Im}(\psi^{*}\partial_{x}\psi) = \frac{\hbar\,t\,x}{m\,\tau\,\sigma_{t}^{2}}\,\rho(x,t). \]

Using \(\hbar/(m\tau) = \hbar^{2}/(m^{2}\sigma^{2})\) and \(\sigma_{t}^{2} = \sigma^{2}(\tau^{2} + t^{2})/\tau^{2}\), this simplifies to the compact form

\[ j(x,t) = \frac{x\,t}{\tau^{2} + t^{2}}\,\rho(x,t). \]

Verifying \(\partial_{t}\rho + \partial_{x}j = 0\). Differentiate \(\rho\) in time, using \(\sigma_{t}^{2} = \sigma^{2}(\tau^{2} + t^{2})/\tau^{2}\) so that \((\sigma_{t}^{2})'/\sigma_{t}^{2} = 2t/(\tau^{2} + t^{2})\):

\[ \partial_{t}\rho = \rho\left[-\frac{t}{\tau^{2} + t^{2}} + \frac{2\tau^{2}\,t\,x^{2}}{\sigma^{2}\,(\tau^{2} + t^{2})^{2}}\right], \]

where the two terms come from differentiating the prefactor \(1/\sigma_{t}\) and the Gaussian exponent, respectively. For the current,

\[ \partial_{x}j = \frac{t}{\tau^{2} + t^{2}}\,\rho + \frac{x\,t}{\tau^{2} + t^{2}}\,\partial_{x}\rho, \qquad \partial_{x}\rho = -\frac{2\tau^{2}\,x}{\sigma^{2}(\tau^{2} + t^{2})}\,\rho, \]

\[ \partial_{x}j = \rho\left[\frac{t}{\tau^{2} + t^{2}} - \frac{2\tau^{2}\,t\,x^{2}}{\sigma^{2}\,(\tau^{2} + t^{2})^{2}}\right]. \]

Adding the two,

\[ \partial_{t}\rho + \partial_{x}j = 0 \]

identically — the prefactor pieces cancel and the \(x^{2}\) pieces cancel separately. The free Gaussian realises probability conservation explicitly: the spreading governed by \(\sigma_{t}\) and the local flow \(j(x,t) = x\,t\,\rho/(\tau^{2} + t^{2})\) are exactly matched, so no probability leaks anywhere. The “outflow velocity” \(j/\rho = x\,t/(\tau^{2} + t^{2})\) approaches \(x/t\) at late times, exactly the free-expansion rate \(\partial_{t}\langle x^{2}\rangle^{1/2}\) — the cloud spreads like a distribution of classical free particles labelled by their asymptotic velocity \(x/t\).

6. Higher moments do not matter. The Taylor expansion \(\psi(x-\delta x,t) = \psi(x,t) - \delta x\,\partial_x\psi + \tfrac{1}{2}(\delta x)^2\partial_x^2\psi + O((\delta x)^3)\) produces moments \(M_{n}\) for \(n = 0, 1, 2, 3, 4, \ldots\).

(a) Use the derivative trick repeatedly to show that \(M_{2k} = c_{k}\,M_{0}\,\alpha^{-k}\) for some dimensionless constants \(c_{k}\).

(b) Substituting \(\alpha = m/(2\hbar\,\delta t)\), conclude that \(A\,M_{2k} = O(\delta t^{\,k})\). Why does the small-\(\delta t\) limit therefore truncate exactly at the second moment?

Solution.

(a) All even moments from repeated differentiation. Each \(\alpha\)-derivative of a moment lowers an extra \((\delta x)^{2}\) from the exponent:

\[ \partial_{\alpha}M_{n} = \int\mathrm{i}(\delta x)^{2}\,(\delta x)^{n}\,\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\mathrm{d}(\delta x) = \mathrm{i}\,M_{n+2}. \]

Applying this \(k\) times starting from \(M_{0}\),

\[ M_{2k} = \frac{1}{\mathrm{i}^{k}}\,\partial_{\alpha}^{k}M_{0}. \]

(Odd moments need not be tracked: \(M_{2k+1} = 0\) by parity, since the integrand \((\delta x)^{2k+1}\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\) is odd.) With \(M_{0} = \sqrt{\mathrm{i}\pi}\;\alpha^{-1/2}\), the repeated power-rule derivative is

\[ \partial_{\alpha}^{k}\,\alpha^{-1/2} = \left(-\tfrac{1}{2}\right)\!\left(-\tfrac{3}{2}\right)\cdots\!\left(-\tfrac{2k-1}{2}\right)\alpha^{-1/2-k} = (-1)^{k}\,\frac{(2k-1)!!}{2^{k}}\;\alpha^{-1/2-k}, \]

where \((2k-1)!! = 1\cdot 3\cdot 5\cdots(2k-1)\) and \((-1)!! \equiv 1\). Hence

\[ \partial_{\alpha}^{k}M_{0} = (-1)^{k}\,\frac{(2k-1)!!}{2^{k}}\;M_{0}\,\alpha^{-k}, \]

and therefore

\[ M_{2k} = \frac{1}{\mathrm{i}^{k}}\,(-1)^{k}\,\frac{(2k-1)!!}{2^{k}}\;M_{0}\,\alpha^{-k} = \left(\frac{-1}{\mathrm{i}}\right)^{\!k}\frac{(2k-1)!!}{2^{k}}\;M_{0}\,\alpha^{-k}. \]

Using \(-1/\mathrm{i} = \mathrm{i}\), this is

\[ \boxed{\;M_{2k} = c_{k}\,M_{0}\,\alpha^{-k},\qquad c_{k} = \left(\frac{\mathrm{i}}{2}\right)^{\!k}(2k-1)!!\;} \]

The constants \(c_{k}\) are pure numbers — dimensionless — independent of \(m\), \(\hbar\), \(\delta t\). Check: \(c_{0} = 1\) (so \(M_{0} = M_{0}\)); \(c_{1} = \mathrm{i}/2\), reproducing \(M_{2} = (\mathrm{i}/2\alpha)M_{0}\) from Problem 2; \(c_{2} = (\mathrm{i}/2)^{2}\cdot 3 = -3/4\).

(b) The order in \(\delta t\) and the exact truncation. The kinetic-phase coefficient is \(\alpha = m/(2\hbar\,\delta t)\), so

\[ \alpha^{-k} = \left(\frac{2\hbar\,\delta t}{m}\right)^{\!k} = \left(\frac{2\hbar}{m}\right)^{\!k}\delta t^{\,k}. \]

Each inverse power of \(\alpha\) carries one power of \(\delta t\). The normalization fixed in the lecture is \(A = 1/M_{0}\), so

\[ A\,M_{2k} = \frac{M_{2k}}{M_{0}} = c_{k}\,\alpha^{-k} = c_{k}\left(\frac{2\hbar}{m}\right)^{\!k}\delta t^{\,k} = O(\delta t^{\,k}). \]

The factor \(M_{0}\) — the only piece with awkward \(\delta t\)-dependence — cancels against \(A\), leaving a clean power \(\delta t^{k}\).

Now recall the moment expansion of one slice. Taylor-expanding the source wave \(\psi(x-\delta x,t) = \psi(x,t) - \delta x\,\partial_x\psi + \tfrac{1}{2}(\delta x)^2\partial_x^2\psi + O((\delta x)^3)\) and integrating term by term gives

\[ \psi(x,t+\delta t) = A\sum_{n\ge 0}\frac{(-1)^{n}}{n!}\,M_{n}\,\partial_{x}^{n}\psi(x,t) = \sum_{k\ge 0}\frac{A\,M_{2k}}{(2k)!}\,\partial_{x}^{2k}\psi(x,t), \]

where odd terms dropped out (\(M_{2k+1}=0\)). Substituting \(A\,M_{2k} = O(\delta t^{k})\), the \(k\)-th term is \(O(\delta t^{k})\,\partial_{x}^{2k}\psi\). To extract the Schrödinger equation we form

\[ \partial_{t}\psi = \lim_{\delta t\to 0}\frac{\psi(x,t+\delta t) - \psi(x,t)}{\delta t}. \]

Examine each \(k\):

\(k=0\): the term is \(A\,M_{0}\,\psi = \psi\), which cancels the \(-\psi(x,t)\) in the numerator. It contributes nothing to the difference.
\(k=1\): the term is \(O(\delta t^{1})\); divided by \(\delta t\) it gives a finite, nonzero limit — this is the kinetic term \(\tfrac{\mathrm{i}\hbar}{2m}\partial_{x}^{2}\psi\).
\(k\ge 2\): the term is \(O(\delta t^{k})\) with \(k\ge 2\); divided by \(\delta t\) it is \(O(\delta t^{\,k-1})\to 0\).

So in the limit \(\delta t\to 0\) only \(k=0\) (trivial) and \(k=1\) (the second moment \(M_{2}\)) survive; every higher even moment is suppressed by at least one extra power of \(\delta t\) and vanishes. The expansion truncates exactly at the second moment — not approximately. This is the deep reason the Schrödinger equation is first order in \(\partial_{t}\) and exactly second order in \(\partial_{x}\): the slice kernel’s Gaussian width scales as \(\delta x\sim\alpha^{-1/2}\sim\sqrt{\delta t}\), so the \(2k\)-th moment scales as \((\delta x)^{2k}\sim\delta t^{k}\), and only \(k=1\) matches the single power of \(\delta t\) removed by the time derivative.

7. Phase-space spreading: loss of minimum uncertainty. A free particle starts in the minimum-uncertainty Gaussian state \(\psi(x, 0) = (\pi\sigma^{2})^{-1/4}\exp(-x^{2}/(2\sigma^{2}))\), for which \(\sigma_{x}(0)\,\sigma_{p}(0) = \hbar/2\) exactly.

(a) Compute \(\sigma_{x}(t)\) from the propagator (or quote the result of HW 3.2.1.1).

(b) Argue from momentum conservation that \(\sigma_{p}(t) = \sigma_{p}(0) = \hbar/(2\sigma)\) for all \(t > 0\).

\[ \sigma_{x}(t)\,\sigma_{p}(t) = \frac{\hbar}{2}\sqrt{1 + (t/\tau)^{2}}, \qquad \tau = \frac{m\sigma^{2}}{\hbar}. \]

The product grows monotonically with time: the state begins at minimum uncertainty and loses this status under free evolution.

(d) Identify the physical reason. The free Hamiltonian \(\hat H = \hat p^{2}/(2m)\) is a function of \(\hat p\) alone, so it commutes with \(\hat p\) but not with \(\hat{x}\). Use this to explain why \(\sigma_{p}\) is preserved while \(\sigma_{x}\) is not. Comment on what initial state would maintain minimum uncertainty under free evolution (hint: there isn’t one — minimum-uncertainty states are not stable under free flight).

Solution.

(a) From HW 3.2.1.1, the free Gaussian wavepacket has

\[ \sigma_{x}(t) = \sigma\sqrt{1 + (t/\tau)^{2}}, \qquad \tau = \frac{m\sigma^{2}}{\hbar}. \]

(b) The free Hamiltonian \(\hat H = \hat p^{2}/(2m)\) is a function of \(\hat p\), so \([\hat H, \hat p] = 0\). By 1.3.1 Problem 1 (conservation theorem for observables commuting with \(\hat H\)), all moments of \(\hat p\) are time-independent:

\[ \langle\hat p\rangle(t) = \langle\hat p\rangle(0), \qquad \langle\hat p^{2}\rangle(t) = \langle\hat p^{2}\rangle(0). \]

Therefore \(\sigma_{p}(t) = \sigma_{p}(0)\). For the initial Gaussian, the Fourier transform is also Gaussian, with width \(\sigma_{p}(0) = \hbar/(2\sigma)\) — saturating the Heisenberg bound at \(t = 0\).

\[ \sigma_{x}(t)\,\sigma_{p}(t) = \sigma\sqrt{1 + (t/\tau)^{2}}\cdot\frac{\hbar}{2\sigma} = \frac{\hbar}{2}\sqrt{1 + (t/\tau)^{2}}. \]

At \(t = 0\): \(\sigma_{x}\sigma_{p} = \hbar/2\), the minimum uncertainty. For \(t > 0\) the square root exceeds \(1\) and the product strictly grows. For \(t \gg \tau\) the product grows linearly with \(t\), signalling unbounded loss of phase-space locality.

(d) Why \(\sigma_{p}\) is conserved. Momentum is a constant of free motion: \(\hat p\) generates translations, and a translationally-invariant Hamiltonian (no \(\hat{x}\) dependence) preserves \(\hat p\). So the momentum distribution is frozen.

Why \(\sigma_{x}\) grows. Different momentum components travel at different velocities \(v = p/m\). A wavepacket that started narrowly localized in \(x\) contained a wide spread of momenta (Heisenberg), and those momenta carry their portions of the wavefunction apart. The position uncertainty therefore grows; the rate is set by the velocity spread \(\sigma_{p}/m = \hbar/(2m\sigma)\), giving the linear-in-\(t\) growth \(\sigma_{x}(t \gg \tau) \approx \sigma_{p}t/m\).

No minimum-uncertainty steady state. A stationary minimum-uncertainty state would require both \(\sigma_{x}\) and \(\sigma_{p}\) to be time-independent. For a free particle this is impossible: \(\sigma_{p}\) is fixed by initial conditions, but the position spread must grow because of the velocity dispersion just argued. The minimum-uncertainty Gaussian is a fleeting initial condition, not a steady state.

The situation is qualitatively different for the harmonic oscillator (2.1.3 P3): there, coherent states maintain minimum uncertainty forever, because the oscillator periodically refocuses what free flight would spread. The harmonic restoring force is precisely what keeps the wavepacket gathered — a free particle has no such mechanism, and minimum-uncertainty is irreversibly lost.

This is the operator-level statement that under any non-trivial unitary evolution, the phase-space volume explored by an initially localized state can only grow, never shrink. Quantum free flight is one explicit, exactly-solvable example.

8. Potential placement freedom. Repeat the derivation of the full Schrödinger equation, but evaluate the slice potential at the source point \(x' = x - \delta x\) instead of the endpoint \(x\).

(a) Show that the additional slice phase becomes \(\exp[-\mathrm{i}V(x-\delta x)\,\delta t/\hbar]\).

(b) Expand \(V(x-\delta x) = V(x) - \delta x\,V'(x) + O((\delta x)^{2})\) and use the Gaussian moments to show that the difference relative to evaluating \(V\) at \(x\) contributes only at \(O(\delta t^{\,2})\).

(c) Conclude that the Schrödinger equation is independent of where \(V\) is sampled within the slice, and explain in one sentence why this freedom matches the analogous freedom for the slice action (HW 3.2.1.4).

Solution.

(a) The slice phase with \(V\) at the source. For the Lagrangian \(L = \tfrac{1}{2}m\dot x^{2} - V(x)\), the slice action over a step of duration \(\delta t\) is the kinetic part plus a potential contribution \(-V\,\delta t\), where \(V\) is evaluated at some point of the slice. If we choose the source point \(x' = x-\delta x\) rather than the endpoint \(x\), the potential contributes action \(\Delta S_{V} = -V(x-\delta x)\,\delta t\). By the path-integral rule, action enters the amplitude as a phase \(\mathrm{e}^{\mathrm{i}\Delta S/\hbar}\), so the slice propagator picks up the extra factor

\[ \exp\!\left[\frac{\mathrm{i}\,\Delta S_{V}}{\hbar}\right] = \exp\!\left[-\frac{\mathrm{i}\,V(x-\delta x)\,\delta t}{\hbar}\right], \]

multiplying the free kinetic kernel \(K_{\delta t}^{\text{free}}\). This is the analogue of \(K_{\delta t}(x,x') = K_{\delta t}^{\mathrm{free}}(x,x')\,\exp[-\mathrm{i}V(x)\delta t/\hbar]\) with \(V(x)\to V(x-\delta x)\).

(b) The difference is \(O(\delta t^{2})\). Because the slice integral always carries one explicit \(\delta t\) in the potential phase, expand it to first order:

\[ \exp\!\left[-\frac{\mathrm{i}V(x-\delta x)\,\delta t}{\hbar}\right] = 1 - \frac{\mathrm{i}\,\delta t}{\hbar}\,V(x-\delta x) + O(\delta t^{2}). \]

Taylor-expand the potential about the endpoint \(x\):

\[ V(x-\delta x) = V(x) - \delta x\,V'(x) + \tfrac{1}{2}(\delta x)^{2}V''(x) + O((\delta x)^{3}). \]

The difference between sampling \(V\) at the source and at the endpoint is therefore, inside the slice integral,

\[ \Delta\bigl[\text{phase}\bigr] = -\frac{\mathrm{i}\,\delta t}{\hbar}\Bigl[\,-\delta x\,V'(x) + \tfrac{1}{2}(\delta x)^{2}V''(x)\,\Bigr] + O(\delta t^{2}). \]

This correction multiplies \(\psi_{\text{kin}}\) and must be integrated against the kinetic Gaussian, \(\psi(x,t+\delta t)\supset A\!\int\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\Delta[\text{phase}]\,\psi(x-\delta x,t)\,\mathrm{d}(\delta x)\). Take the two pieces in turn, using \(\psi(x-\delta x)\approx\psi(x) - \delta x\,\partial_{x}\psi + \cdots\) and the moment values \(A\,M_{0} = 1\), \(M_{1}=0\), \(A\,M_{2} = O(\delta t)\):

Linear piece \(\propto\delta x\,V'(x)\). The leading term pairs \(\delta x\) with \(\psi(x)\), giving a factor \(A\,M_{1} = 0\) by parity. The first nonzero term pairs \(\delta x\) with the \(-\delta x\,\partial_{x}\psi\) piece of the Taylor expansion of \(\psi\), producing \(\int(\delta x)^{2}\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\mathrm{d}(\delta x) = M_{2}\). Its contribution is

\[ \frac{\mathrm{i}\,\delta t}{\hbar}\,V'(x)\,\bigl(-\partial_{x}\psi\bigr)\,A\,M_{2} = \underbrace{\frac{\mathrm{i}\,\delta t}{\hbar}}_{O(\delta t)}\times\underbrace{A\,M_{2}}_{O(\delta t)} \times(\cdots) = O(\delta t^{2}). \]

Quadratic piece \(\propto(\delta x)^{2}V''(x)\). The leading term pairs \((\delta x)^{2}\) with \(\psi(x)\), giving \(A\,M_{2} = O(\delta t)\):

\[ -\frac{\mathrm{i}\,\delta t}{\hbar}\cdot\tfrac{1}{2}V''(x)\,\psi(x)\,A\,M_{2} = \underbrace{\frac{\mathrm{i}\,\delta t}{\hbar}}_{O(\delta t)}\times\underbrace{A\,M_{2}}_{O(\delta t)}\times(\cdots) = O(\delta t^{2}). \]

In both cases the explicit \(\delta t\) from the potential phase multiplies a quantity that is itself at least \(O(\delta t)\) — because every surviving even moment obeys \(A\,M_{2k} = O(\delta t^{k})\) and the odd moment \(M_{1}\) vanishes. Hence the entire difference between evaluating \(V\) at \(x-\delta x\) and at \(x\) is

\[ \psi^{(x')}(x,t+\delta t) - \psi^{(x)}(x,t+\delta t) = O(\delta t^{2}). \]

(c) Conclusion. The Schrödinger equation is built from the \(O(\delta t)\) part of the slice evolution: one divides \(\psi(x,t+\delta t)-\psi(x,t)\) by \(\delta t\) and lets \(\delta t\to 0\). Any difference that enters only at \(O(\delta t^{2})\) becomes \(O(\delta t)\) after the division and vanishes in the limit. Since sampling \(V\) at the source point instead of the endpoint changes the result only at \(O(\delta t^{2})\) — and the same estimate covers the midpoint or any interior point — the resulting equation

\[ \mathrm{i}\hbar\,\partial_{t}\psi = -\frac{\hbar^{2}}{2m}\,\partial_{x}^{2}\psi + V(x)\,\psi \]

is identical regardless of where \(V\) is sampled within the slice. This matches the freedom found for the slice action in HW 3.2.1.4: the coherent paths have displacements only of order \(\delta x\sim\sqrt{\hbar\,\delta t/m}\), so across one slice \(V\) is effectively constant and the choice of reference point shifts the slice action by merely \(O(\delta t^{2})\) — negligible in the continuum limit.