If the dot behaves as a perfectly cubical 3 -D square well, the ground state is a. non degenerate. b. twofold degenerate. c. threefold degenerate. d. You can't tell without knowing the energy.

Short Answer

Expert verified
The answer is a. The ground state of a perfectly cubical 3D square well is non-degenerate.

Step by step solution

01

Understanding the concept of a degenerated state

In quantum mechanics, degeneracy of a system refers to the property where the system can have the same energy state in more than one configuration. In a quantum well, due to its bounded nature, there are distinct levels that are allowed for energy, known as quantum energy levels.
02

Connecting the dimension of the well to degeneracy

For any given quantum state, there are three quantum numbers \(n\), \(l\) and \(m\) that can take multiple values, and each unique set of these numbers represents a unique state. For a 3-d square well \(n\), \(l\) and \(m\) are all valid, and they can each take multiple values which indicate the number of nodal planes in each of the three directions. The ground state corresponds to \(n=1\), \(l=1\), and \(m=3\).
03

Evaluating the degeneracy of the ground state of a 3D well

For a 3D well in the ground state (\(n=1\), \(1=1\), \(m=3\)), there is only one possible state which means it is non-degenerate. Thus the ground state of a 3D square well is non-degenerate.

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