The Schrodinger equation for a single particle in one dimension: $$i \hbar \frac{\partial \Psi}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2\Psi}{\partial x^2} + V\Psi$$ where $m$ is the mass of the particle, $\hbar$ is the reduced planck constant, $V$ is the potential the particle is in and $\Psi$ is the wavefunction of the particle describes where the particle is; i.e: $$\int_a^b |\Psi(x,t)|^2 dx$$ describes the probability of finding particle between $a$ and $b$ at time t. Evidently, the particle must be somewhere, so we have that: $$\int_{-\infty}^{+\infty} |\Psi(x,t)|^2 dx = 1$$ Notice that any such $\Psi$ a solution to the Shrodinger equation gives rise to an infinite number of solutions $a\Psi$ where $a$ is a constant, so we must normalize the solution by making sure that the probability of the particle being somewhere is 1. It is true that if we normalise the wavefunction at $t = 0$ and let the function evolve dynamically by the Shrodinger equation, that the wavefunction stays normalised.

Problem 1.5
Consider the wavefunction $$\Psi(x,t) = A e^{-\lambda |x|}e^{-i\omega t}$$ where $A,\lambda,$ and $\omega$ are positive real constants.

Normalise $\Psi$.
Determine the expectation values of $x$ and $x^2$.
Find the standard deviation of $x$. Sketch the graph of $|\Psi|^2$, as a function of $x$, and mark the points $ (\langle x \rangle + \sigma) $ and $ (\langle x \rangle - \sigma) $, to illustrate the sense in which $\sigma$ represents the "spread" in $x$. What is the probability that the particle would be found outside this range?

Solution:

$$1 = \int_{-\infty}^{+\infty} |\Psi (x,t)|^2 dx= \int_{-\infty}^{+\infty} A^2 e^{-2\lambda |x|}dx = \frac{A^2}{\lambda}$$ Therefore: $$A = \sqrt{\lambda}$$
$$\langle x \rangle = \int_{-\infty}^{+\infty} x|\Psi (x,t)|^2 dx$$ But $x$ is an odd function and $|\Psi(x,t)|^2$ is an even function (as it depends in some way on $|x|$), an odd function times an even function is an odd function and the integral of an odd function over a symmetric domain is 0, hence $\langle x \rangle = 0$ $$\langle x^2 \rangle = \int_{-\infty}^{+\infty} x^2 |Psi (x,t)|^2 dx = \frac{1}{2\lambda^2}$$ By repeated usage of the product rule for integration.
$$\sigma = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} = \sqrt{\frac{1}{2\lambda^2}} = \frac{1}{\lambda\sqrt{2}}$$

This is the graph for the case where $\lambda = 1$. The probability that the particle is found outside one sigma range of the average is given by: $$P = 2\int_{\frac{1}{\lambda\sqrt{2}}}^{\infty} \lambda e^{-2\lambda x} dx = e^{-2} \simeq 24.3 \%$$

$$\langle x \rangle = \int_{-\infty}^{+\infty} \Psi^*(x) \Psi dx$$ $$\langle p \rangle = m \frac{d\langle x \rangle}{dt} = \int \Psi^* \bigg(\frac{\hbar}{i} \frac{\partial}{\partial x}\bigg) \Psi dx$$ We say that the operator $x$ "represents" position, and the operator $(\hbar/i)(\partial/\partial x)$ "represents" momentum, in quantum mechanics; to calculate expectation values we "sandwich" the appropriate operator between $\Psi^*$ and $\Psi$, and integrate. The nice thing is, all classical dynamical variables can be expressed in terms of position and momentum. Kinetic energy, for example, is: $$\langle T \rangle = \frac{p^2}{2m}$$ To calculate the expectation value of any such quantity, $Q(x,p)$, we simply replace every $p$ by it's associated operator, sandwich the resulting operator between the conjugate of the wavefunction and the wavefunction, and integrate: $$\langle Q(x,p) \rangle = \int \Psi^* Q\bigg(x,\frac{\hbar}{i}\frac{\partial}{\partial x} \bigg) \Psi dx$$ For example, the expectation value of kinetic energy is $$\langle T \rangle = -\frac{\hbar^2}{2m} \int \Psi^* \frac{\partial^2}{\partial x^2} \Psi dx$$

Solution:

By Shrodinger's equation and it's conjugate: $$i \hbar \frac{\partial \Psi}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V\Psi$$ $$-i\hbar \frac{\partial \Psi^*}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi^*}{\partial x^2} + V\Psi^*$$ (The negative sign comes from the conjugation of $i$ on the left hand side) Now: $$ \frac{d}{dt} \langle p \rangle = \frac{d}{dt} \int \psi^* \bigg( \frac{\hbar}{i} \frac{d}{dx} \bigg) \psi dx$$ Interchanging derivative and integration, do derivative by parts, we end up with: $$ \frac{d}{dt} \langle p \rangle = \frac{\hbar}{i} \int \frac{d\Psi^*}{dt} \frac{d\Psi}{dx} + \Psi^* \frac{d^2\Psi}{dtdx} dx$$ and then substitute in the time derivatives of $\Psi$ and $\Psi^*$ that we obtained from Shrodinger's equation, $$ \frac{d}{dt} \langle p \rangle = \frac{\hbar}{i} \int \frac{\hbar i}{2m} \bigg(\Psi^* \frac{d\Psi}{dx^3} - \frac{d^2 \Psi^*}{dx^2} \frac{d\Psi}{dx}\bigg) - \Psi^* \bigg(\frac{i}{h} \frac{dV}{dx}\bigg) \Psi dx$$ Applying the product rule to the first term, we find that everything cancels (under the normal observation that $\Psi$ tends to $0$ as $x$ tends to positive or negative infinity). Hence we have our desired relation: $$\frac{d}{dt} \langle p \rangle = - \int \Psi^* \bigg(\frac{dV}{dx}\bigg) \Psi dx = -\bigg\langle \frac{dV}{dx}\bigg\rangle$$

The de Broglie formula tells us that a particle's momentum and wavelength are related by: $$p = \frac{h}{\lambda} = \frac{2\pi \hbar}{\lambda}$$ Heisenberg's uncertainty principle tells us: $$\sigma_x \sigma_p \geq \frac{\hbar}{2}$$ where $\sigma_x$ is the standard deviation of $x$, and $\sigma_p$ is the standard deviation of $p$.

Problem 1.9

A particle of mass $m$ is in the state $$\Psi(x,t) = A e^{-a[(mx^2/\hbar)+it]}$$ where $A,a$ are positive real constants.

Find $A$.
For what potential energy function $V(x)$ does $\Psi$ satisfy the Shrodinger equation?
Calculate the expectation values of $x$,$x^2$,$p$, and $p^2$.
Find $\sigma_x$ and $\sigma_p$. Is their product consistent with the uncertainty principle?

Solution:

Normalising the wavefunction (and integrating the Gaussian integral by doing the square trick) we find that $$A = \bigg(\frac{2am}{\hbar \pi}\bigg)^{1/4}$$
Substituting $\Psi$ into the Shrodinger equation we find that $$V = 2ma^2x^2$$
$$\langle x \rangle = \langle p \rangle = 0$$ by symmetry arguments. $$\langle x^2 \rangle = \frac{\hbar^2}{4am}$$ $$\langle p^2 \rangle = \hbar ma$$ by integration.
$$\sigma_x = \sqrt{\frac{\hbar}{4am}}$$ $$\sigma_p = \sqrt{\hbar ma}$$ $$\sigma_x \sigma_p = \frac{\hbar}{2}$$ So this is in some sense an 'optimal' situation (in that 'maximal' amount of information retention)

Problem 1.15

Suppose we want to describe an unstable particle, that spontaneously disintegrates with 'lifetime' $\tau$; i.e. $$P(t) = \int \Psi^* \Psi dx = e^{-t/\tau}$$ A crude way of approximating this situation would be to throw away the assumption that $V$ (our potential) must be real, and instead replace it with $V = V_0 + i\Gamma$ where $V_0, \Gamma$ are positive real constants. Show that instead of $P(t) = 1$, we have: $$\frac{dP}{dt} = - \frac{2\Gamma}{\hbar} P$$ Solve for $P(t)$, and find the lifetime of the particle $\tau$.

Solution: $$\frac{dP}{dt} = \int \frac{d\Psi^*}{dt}\Psi + \Psi^* \frac{d\Psi}{dt}$$ Now, using the modified Shrodinger's equations: $$i\hbar \frac{d\Psi}{dt} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V_0 \Psi - i \Gamma \Psi$$ $$-i\hbar\frac{d\Psi^*}{dt}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi^*}{\partial x^2} + V_0 \Psi + i \Gamma \Psi^*$$ Replace the time derivitives of the wavefunction with spatial derivatives. Most of the terms drop out by the usual arguments (that $\Psi$ vanishes far away from the origin) and we are left with our desired equality $$\frac{dP}{dt} = -\frac{2\Gamma}{\hbar}P$$ Solving, we find that $$P = e^{-t/\tau}$$ where $\tau = \frac{\hbar}{2\Gamma}$ is the lifetime of the particle.