a) The basic principles of quantum mechanics are:

Question

a) The basic principles of quantum mechanics are:

Asked on March 27, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-27T15:12:36.786Z

a) The basic principles of quantum mechanics are: 1. The state of a quantum mechanical system is completely described by a wave function $\Psi(\vec{r}, t)$, which contains all possible information about the system. 2. Every observable physical quantity (like position, momentum, energy) corresponds to a linear, Hermitian operator in quantum mechanics. 3. The only possible results of a measurement of an observable are the eigenvalues of the corresponding operator. 4. The time evolution of the wave function is governed by the time-dependent Schrödinger equation. b) For a particle in a 1D box of length $L$ (from $x=0$ to $x=L$), the normalized wave functions are given by: $$ \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) $$ The position operator is $\hat{x} = x$. The expectation value of the position operator is given by: $$ \langle x \rangle = \int_0^L \psi_n^*(x) x \psi_n(x) \, dx $$ Since $\psi_n(x)$ is real, $\psi_n^*(x) = \psi_n(x)$. Step 1: Substitute the wave function into the expectation value integral. $$ \langle x \rangle = \int_0^L \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right) x \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right) \, dx $$ $$ \langle x \rangle = \frac{2}{L} \int_0^L x \sin^2\left(\frac{n\pi x}{L}\right) \, dx $$ Step 2: Use the trigonometric identity $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$. $$ \langle x \rangle = \frac{2}{L} \int_0^L x \left(\frac{1 - \cos\left(\frac{2n\pi x}{L}\right)}{2}\right) \, dx $$ $$ \langle x \rangle = \frac{1}{L} \int_0^L \left(x - x \cos\left(\frac{2n\pi x}{L}\right)\right) \, dx $$ Step 3: Split the integral into two parts. $$ \langle x \rangle = \frac{1}{L} \left[ \int_0^L x \, dx - \int_0^L x \cos\left(\frac{2n\pi x}{L}\right) \, dx \right] $$ Step 4: Evaluate the first integral. $$ \int_0^L x \, dx = \left[\frac{x^2}{2}\right]_0^L = \frac{L^2}{2} - 0 = \frac{L^2}{2} $$ Step 5: Evaluate the second integral using integration by parts, $\int u \, dv = uv - \int v \, du$. Let $u = x$ and $dv = \cos\left(\frac{2n\pi x}{L}\right) \, dx$. Then $du = dx$ and $v = \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right)$. $$ \int_0^L x \cos\left(\frac{2n\pi x}{L}\right) \, dx = \left[x \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right)\right]_0^L - \int_0^L \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right) \, dx $$ Evaluate the first term: $$ \left[x \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right)\right]_0^L = \left(L \frac{L}{2n\pi} \sin(2n\pi)\right) - \left(0 \cdot \frac{L}{2n\pi} \sin(0)\right) = 0 - 0 = 0 $$ Evaluate the second term: $$ - \int_0^L \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right) \, dx = - \frac{L}{2n\pi} \left[-\frac{L}{2n\pi} \cos\left(\frac{2n\pi x}{L}\right)\right]_0^L $$ $$ = \frac{L^2}{4n^2\pi^2} \left[\cos\left(\frac{2n\pi x}{L}\right)\right]_0^L = \frac{L^2}{4n^2\pi^2} [\cos(2n\pi) - \cos(0)] $$ Since $\cos(2n\pi) = 1$ for integer $n$ and $\cos(0) = 1$: $$ = \frac{L^2}{4n^2\pi^2} [1 - 1] = 0 $$ So, the second integral $\int_0^L x \cos\left(\frac{2n\pi x}{L}\right) \, dx = 0$. Step 6: Substitute the results back into the expression for $\langle x \rangle$. $$ \langle x \rangle = \frac{1}{L} \left[ \frac{L^2}{2} - 0 \right] = \frac{1}{L} \frac{L^2}{2} = \frac{L}{2} $$ The expectation value of the position operator for a particle in a 1D box is $\boxed{\frac{L}{2}}$. c) At least three assumptions associated with the problem of a particle in a 1D box are: 1. The potential energy of the particle is zero inside the box ($0 < x < L$). 2. The potential energy is infinite outside the box ($x \le 0$ or $x \ge L$), meaning the particle is perfectly confined within the box. 3. The particle is treated as a point particle, meaning it has no internal structure or spin.

Accepted Answer · 2026-03-27T15:12:36.786Z

a) The basic principles of quantum mechanics are: 1. The state of a quantum mechanical system is completely described by a wave function $\Psi(\vec{r}, t)$, which contains all possible information about the system. 2. Every observable physical quantity (like position, momentum, energy) corresponds to a linear, Hermitian operator in quantum mechanics. 3. The only possible results of a measurement of an observable are the eigenvalues of the corresponding operator. 4. The time evolution of the wave function is governed by the time-dependent Schrödinger equation. b) For a particle in a 1D box of length $L$ (from $x=0$ to $x=L$), the normalized wave functions are given by: $$ \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) $$ The position operator is $\hat{x} = x$. The expectation value of the position operator is given by: $$ \langle x \rangle = \int_0^L \psi_n^*(x) x \psi_n(x) \, dx $$ Since $\psi_n(x)$ is real, $\psi_n^*(x) = \psi_n(x)$. Step 1: Substitute the wave function into the expectation value integral. $$ \langle x \rangle = \int_0^L \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right) x \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right) \, dx $$ $$ \langle x \rangle = \frac{2}{L} \int_0^L x \sin^2\left(\frac{n\pi x}{L}\right) \, dx $$ Step 2: Use the trigonometric identity $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$. $$ \langle x \rangle = \frac{2}{L} \int_0^L x \left(\frac{1 - \cos\left(\frac{2n\pi x}{L}\right)}{2}\right) \, dx $$ $$ \langle x \rangle = \frac{1}{L} \int_0^L \left(x - x \cos\left(\frac{2n\pi x}{L}\right)\right) \, dx $$ Step 3: Split the integral into two parts. $$ \langle x \rangle = \frac{1}{L} \left[ \int_0^L x \, dx - \int_0^L x \cos\left(\frac{2n\pi x}{L}\right) \, dx \right] $$ Step 4: Evaluate the first integral. $$ \int_0^L x \, dx = \left[\frac{x^2}{2}\right]_0^L = \frac{L^2}{2} - 0 = \frac{L^2}{2} $$ Step 5: Evaluate the second integral using integration by parts, $\int u \, dv = uv - \int v \, du$. Let $u = x$ and $dv = \cos\left(\frac{2n\pi x}{L}\right) \, dx$. Then $du = dx$ and $v = \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right)$. $$ \int_0^L x \cos\left(\frac{2n\pi x}{L}\right) \, dx = \left[x \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right)\right]_0^L - \int_0^L \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right) \, dx $$ Evaluate the first term: $$ \left[x \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right)\right]_0^L = \left(L \frac{L}{2n\pi} \sin(2n\pi)\right) - \left(0 \cdot \frac{L}{2n\pi} \sin(0)\right) = 0 - 0 = 0 $$ Evaluate the second term: $$ - \int_0^L \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right) \, dx = - \frac{L}{2n\pi} \left[-\frac{L}{2n\pi} \cos\left(\frac{2n\pi x}{L}\right)\right]_0^L $$ $$ = \frac{L^2}{4n^2\pi^2} \left[\cos\left(\frac{2n\pi x}{L}\right)\right]_0^L = \frac{L^2}{4n^2\pi^2} [\cos(2n\pi) - \cos(0)] $$ Since $\cos(2n\pi) = 1$ for integer $n$ and $\cos(0) = 1$: $$ = \frac{L^2}{4n^2\pi^2} [1 - 1] = 0 $$ So, the second integral $\int_0^L x \cos\left(\frac{2n\pi x}{L}\right) \, dx = 0$. Step 6: Substitute the results back into the expression for $\langle x \rangle$. $$ \langle x \rangle = \frac{1}{L} \left[ \frac{L^2}{2} - 0 \right] = \frac{1}{L} \frac{L^2}{2} = \frac{L}{2} $$ The expectation value of the position operator for a particle in a 1D box is $\boxed{\frac{L}{2}}$. c) At least three assumptions associated with the problem of a particle in a 1D box are: 1. The potential energy of the particle is zero inside the box ($0 < x < L$). 2. The potential energy is infinite outside the box ($x \le 0$ or $x \ge L$), meaning the particle is perfectly confined within the box. 3. The particle is treated as a point particle, meaning it has no internal structure or spin.