Spin-3/2 particle and degeneracy in excited state

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SUMMARY

The discussion focuses on calculating the minimum energy of a particle in a one-dimensional potential defined by V(x) = mgx for x > 0 and V(x) = ∞ for x ≤ 0. The uncertainty principle, expressed as ΔxΔp ≥ ħ/2, is utilized to derive the energy expression E₀ = (1/8α²)(mħ²g²)^(1/3), where α is a dimensionless constant. The challenge lies in determining an appropriate length scale for Δx, with suggestions made to reference similar problems, such as estimating the ground state energy of the hydrogen atom.

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Students and researchers in quantum mechanics, particularly those interested in energy calculations and the application of the uncertainty principle in one-dimensional potentials.

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Homework Statement



Consider a particle with mass m in the following 1D potential:
<br /> V(x)=\left\{ \begin{array}{lr} mgx \ \ \ x&gt;0 \\ \infty \ \ \ \ \ \ \ x\leq 0 \end{array} \right.<br />
What is its minimum energy calculated using the uncertainty relation?

Homework Equations



<br /> \Delta x \Delta p \geq \frac{\hbar}{2}<br />

The Attempt at a Solution



My problem is, I don't know what to use for \Delta x! I see no length scale I can use. The only thing that came into my mind was making a constant with the dimension of length using the available constants and so I got \alpha(\frac{\hbar^2}{gm^2})^{\frac 1 3}(where \alpha is a dimensionless constant) and this gives \Delta p \geq \frac{1}{2\alpha} (m^2 \hbar g)^{\frac 1 3} \Rightarrow E_0=\frac{1}{8\alpha^2} (m \hbar^2 g^2)^{\frac 1 3}.
But the problem is, this method can give any multiple of (m \hbar^2 g^2)^{\frac 1 3}!
What should I do?
Thanks
 
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