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A) The basic principles of quantum mechanics are:

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Answer

b

L2\frac{L}{2}

Step-by-step solution

6 steps

Part a

The basic principles of quantum mechanics are

  1. The state of a quantum mechanical system is completely described by a wave function Ψ(r⃗,t)\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.

Part b

L2\frac{L}{2}

For a particle in a 1D box of length LL (from x=0x=0 to x=Lx=L), the normalized wave functions are given by

ψn(x)=2Lsin⁡(nπxL)\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)

The position operator is x^=x\hat{x} = x. The expectation value of the position operator is given by:

⟨x⟩=∫0Lψn∗(x)xψn(x) dx\langle x \rangle = \int_0^L \psi_n^*(x) x \psi_n(x) \, dx

Since ψn(x)\psi_n(x) is real, ψn∗(x)=ψn(x)\psi_n^*(x) = \psi_n(x).

  1. Substitute the wave function into the expectation value integral.

    ⟨x⟩=∫0L(2Lsin⁡(nπxL))x(2Lsin⁡(nπxL)) dx\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 ⟨x⟩=2L∫0Lxsin⁡2(nπxL) dx\langle x \rangle = \frac{2}{L} \int_0^L x \sin^2\left(\frac{n\pi x}{L}\right) \, dx

  2. Use the trigonometric identity sin⁡2θ=1−cos⁡(2θ)2\sin^2\theta = \frac{1 - \cos(2\theta)}{2}.

    ⟨x⟩=2L∫0Lx(1−cos⁡(2nπxL)2) dx\langle x \rangle = \frac{2}{L} \int_0^L x \left(\frac{1 - \cos\left(\frac{2n\pi x}{L}\right)}{2}\right) \, dx ⟨x⟩=1L∫0L(x−xcos⁡(2nπxL)) dx\langle x \rangle = \frac{1}{L} \int_0^L \left(x - x \cos\left(\frac{2n\pi x}{L}\right)\right) \, dx

  3. Split the integral into two parts.

    ⟨x⟩=1L[∫0Lx dx−∫0Lxcos⁡(2nπxL) dx]\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]

  4. Evaluate the first integral.

    ∫0Lx dx=[x22]0L=L22−0=L22\int_0^L x \, dx = \left[\frac{x^2}{2}\right]_0^L = \frac{L^2}{2} - 0 = \frac{L^2}{2}

  5. Evaluate the second integral using integration by parts, ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

    Let u=xu = x and dv=cos⁡(2nπxL) dxdv = \cos\left(\frac{2n\pi x}{L}\right) \, dx.
    Then du=dxdu = dx and v=L2nπsin⁡(2nπxL)v = \frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right).

    ∫0Lxcos⁡(2nπxL) dx=[xL2nπsin⁡(2nπxL)]0L−∫0LL2nπsin⁡(2nπxL) dx\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:

    [xL2nπsin⁡(2nπxL)]0L=(LL2nπsin⁡(2nπ))−(0⋅L2nπsin⁡(0))=0−0=0\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:

    −∫0LL2nπsin⁡(2nπxL) dx=−L2nπ[−L2nπcos⁡(2nπxL)]0L- \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 =L24n2π2[cos⁡(2nπxL)]0L=L24n2π2[cos⁡(2nπ)−cos⁡(0)]= \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⁡(2nπ)=1\cos(2n\pi) = 1 for integer nn and cos⁡(0)=1\cos(0) = 1:

    =L24n2π2[1−1]=0= \frac{L^2}{4n^2\pi^2} [1 - 1] = 0

    So, the second integral ∫0Lxcos⁡(2nπxL) dx=0\int_0^L x \cos\left(\frac{2n\pi x}{L}\right) \, dx = 0.

  6. Substitute the results back into the expression for ⟨x⟩\langle x \rangle.

    ⟨x⟩=1L[L22−0]=1LL22=L2\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 L2\boxed{\frac{L}{2}}.

Part 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<L0 < x < L).
  2. The potential energy is infinite outside the box (x≤0x \le 0 or x≥Lx \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.
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