Question #54035

Consider a quantum particle confined in a well of width a. If the particle is in its ground
state calculate the quantity DxDp where ( ) 2 2 2
Dx = x − x and ( ) . 2 2 2
Dp = p − p

Expert's answer

Answer on Question #54035, Physics Quantum Mechanics

Consider a quantum particle confined in a well of width aa. If the particle is in its ground state calculate the quantity DxDpDxDp where ()222Dx=x−x() 222Dx = x - x and ()().


222Dp=p−p222Dp = p - p

Solution:

The average value of PXP_{X}

⟨PX⟩=∫0aψ1∗(x)P^Xψ1(x)dx=∫0a2asin⁡(πxa)(−iℏ∂∂x)2asin⁡(πxa)dx=−2iℏaπna∫0asin⁡(πxa)cos⁡(πxa)dx=−iℏπa2∫0asin⁡(2πxa)dx=0\begin{aligned} \langle P_{X} \rangle &= \int_{0}^{a} \psi_{1}^{*}(x) \hat{P}_{X} \psi_{1}(x) dx = \int_{0}^{a} \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) \left(-i\hbar \frac{\partial}{\partial x}\right) \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) dx = -\frac{2i\hbar}{a} \frac{\pi n}{a} \int_{0}^{a} \sin\left(\frac{\pi x}{a}\right) \cos\left(\frac{\pi x}{a}\right) dx = \\ &- \frac{i\hbar \pi}{a^{2}} \int_{0}^{a} \sin\left(\frac{2\pi x}{a}\right) dx = 0 \end{aligned}


where aa is the width of the potential well; ψ1=2asin⁡(πxa)\psi_{1} = \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) is the wave function of a one-dimensional potential well in ground state.

The average value of PX2P_{X}^{2}

⟨PX2⟩=∫0aψ1∗(x)P^X2ψ1(x)dx=∫0a2asin⁡(πxa)(−iℏ∂∂x)22asin⁡(πxa)dx=2ℏ2a(πna)2∫0asin⁡(πxa)sin⁡(πxa)dx=ℏ2π2a2\begin{aligned} \langle P_{X}^{2} \rangle &= \int_{0}^{a} \psi_{1}^{*}(x) \hat{P}_{X}^{2} \psi_{1}(x) dx = \int_{0}^{a} \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) \left(-i\hbar \frac{\partial}{\partial x}\right)^{2} \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) dx = \\ &\frac{2\hbar^{2}}{a} \left(\frac{\pi n}{a}\right)^{2} \int_{0}^{a} \sin\left(\frac{\pi x}{a}\right) \sin\left(\frac{\pi x}{a}\right) dx = \frac{\hbar^{2} \pi^{2}}{a^{2}} \end{aligned}


The average value of x^\hat{x}

⟨x⟩=∫0aψ1∗(x)x^ψ1(x)dx=∫0a2asin⁡(πxL)x2Lsin⁡(πxL)dx=2a∫0asin⁡(πxa)xsin⁡(πxa)dx=a/2\langle x \rangle = \int_{0}^{a} \psi_{1}^{*}(x) \hat{x} \psi_{1}(x) dx = \int_{0}^{a} \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{L}\right) x \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right) dx = \frac{2}{a} \int_{0}^{a} \sin\left(\frac{\pi x}{a}\right) x \sin\left(\frac{\pi x}{a}\right) dx = a / 2


The average value of x2x^{2}

⟨x2⟩=∫0aψ1∗(x)x2ψ1(x)dx=∫0a2asin⁡(πxa)x22asin⁡(πxa)dx=2a∫0asin⁡2(πxa)x2dx=a26(2−3/π2)\left\langle x ^ {2} \right\rangle = \int_ {0} ^ {a} \psi_ {1} ^ {*} (x) x ^ {2} \psi_ {1} (x) d x = \int_ {0} ^ {a} \sqrt {\frac {2}{a}} \sin \left(\frac {\pi x}{a}\right) x ^ {2} \sqrt {\frac {2}{a}} \sin \left(\frac {\pi x}{a}\right) d x = \frac {2}{a} \int_ {0} ^ {a} \sin^ {2} \left(\frac {\pi x}{a}\right) x ^ {2} d x = \frac {a ^ {2}}{6} \left(2 - 3 / \pi^ {2}\right)


Then


ΔxΔp=(ℏ2π2a2−0)(a26(2−3/π2)−a24)=ℏ23π2−6\Delta x \Delta p = \sqrt {\left(\frac {\hbar^ {2} \pi^ {2}}{a ^ {2}} - 0\right) \left(\frac {a ^ {2}}{6} \left(2 - 3 / \pi^ {2}\right) - \frac {a ^ {2}}{4}\right)} = \frac {\hbar}{2 \sqrt {3}} \sqrt {\pi^ {2} - 6}


Answer: ΔxΔp=ℏ23π2−6\Delta x\Delta p = \frac{\hbar}{2\sqrt{3}}\sqrt{\pi^2 - 6}

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