Nature of Eigenvalue Spectrum for Potential Problems in One Dimension

This units contains problems on finding nature of energy eigenvalues and eigenfunctions for potential problems in one dimension.

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### Objecives and Roadmap

Objevtives

For energy eigenvalues for potential problems in one dimension there are only

following three possibilities.

- The energy levels can be discrete and non degenerate. These correspond to the bound states.
- The energy eigenvalues can be continuous and non degenerate.
- These correspond to a particle being able to escape to $+\infty$ or $-\infty$ but not both.
- The energy eigenvalues can be continuous and doubly degenerate.These correspond to a particle being able to escape to both $+\infty$ and $-\infty$

The last two possibilities correspond to states in which particle is not bounded. In this unit you will solve problems on about nature of spectrum of

energy eigenvalues for potential problems in one dimension. Specifically, given a potential one can ask the following questions

- Do the bound states exists for the potential? If yes what is the range of bound state energies?
- Can the energy be continuous? If continuous energy solutions are possible what is the range of energy for which continuous energy spectrum is likely to be possible?
- What can one say about the degeneracy of energy eigenvalues? In which range of energy the eigenvalues are non degenerate and in which range the energy eigenvalues have degeneracy two?

RoadMap

The thumb rules, given below, will be useful to decide about existence of bound states and degeneracy of energy eigenvalues in one dimension.*These rules, stated below **without a proof, hold for commonly encountered potential problems, and are not **theorems valid for all cases. The exceptions to the rules are well **known$^*$.*

The nature of energy eigenvalues, discrete or continuous, degenerate or non-degenerate, is generally given by the following rules. It may be added that the rules give us an idea what to expect for given potential and that exceptions to some of these rules below are known to exist. Let $V(x)$ be a potential whose absolute minimum value is $V_\text{min}$ and which tends to $V_\pm$ as $x\to \pm \infty$.

- It can be proved that the energy eigenvalues must always be greater than or equal to $V_{\mbox{min}}$. This means that the no acceptable solutions of the Schr\"{o}dinger equation exists if $E < V_\text{min}$. \item Bound states exist for energy greater than $V_{\mbox{min}}$ and $E$ below both $V_+$ and $V_-$: $$ E > V\text{min}\text{ and } E < V_-, \text{ and } E < V_+ .$$ The corresponding energy eigenvalues are discrete and non degenerate.
- For $E$ between $V_+, V_-$, the eigenvalues are continuous and non-degenerate.
- For $E$ greater than both $V_+$ and $V_-$ ( and of course one must have $E > V_\text{min}$, the energies eigenvalues are continuous and doubly degenerate.

For the above purpose the values of the potential at $\pm \infty$ and its absolute minimum value $V_0$ alone are required.

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***Terms and conditions apply

These rules are illustrated for a particular potential sketched below. The behaviour of the potential at $\pm \infty$ and having $V_0$ a minimum alone are important, behaviour of the potential at other places otherwise is unimportant except that it should be continuous.

### Examples

Questions for You

You should think how the nature of energy spectrum will change if, for example,

- $V_\pm$ both go to infinity;
- $V_-=V_+$;
- $V\pm$ both are less than or equal to $V_0$.
One of the two, $V_-$ or $V_+$, is less than or equal to zero.

### Exercise

For each of the potentials drawn below, \Figref{pots1} and \Figref{pots2} find the ranges of energy values as specified in statements (a),(b),(c),(d). Give short reasons for your answers and write your answers in a tabular form given below.

- The range of allowed energy eigenvalues
- The range of energies between which the bound states can be found.
- The range of energies for which the energy eigenvalue will be continuous and non-degenerate.
- Range for which the energy eigenvalues are continuous and doubly degenerate.

Potential | (a) | (b) | (c) | (d) |

1 | ||||

2 | ||||

3 | ||||

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### Problems

Plot the following potentials listed in the table below. In the third column indicate if quantum bound states are possible or not? If YES, give the range of energies for allowed bound states energies.

SN | Potential | Do Bound States Exist? | Range of Energies for Bound States in QM |

1 | $V_0\tanh \alpha x$ | ||

2 | $\begin{cases} \infty & x < 0 \\ - \dfrac{k}{x} + \dfrac{M^2}{x^2} & x > 0, \end{cases}\\ k >0$.\\ | $\cdots$ | |

3 | $\begin{cases} V_0 (x/L)^4 & |x| < L\\ V_0 & |x| > L \end{cases} \\ V_0>0$ | $\cdots$ | $\cdots$ |

4 | $\displaystyle V(x) = V_0 \tanh^2 \alpha x$ | ||

5 | $\frac{1}{2}m \omega^2 x^2 + \dfrac{\lambda}{x^2}, \lambda >0, x\ge$ | ||

6 | $V(x) =\begin{cases} V_0 & x < 0 \\ 0 & x >0 \text{ and } x < L\\ 2 V_0 & x > L \end{cases}$ | ||

7 | $V(x) = \begin{cases} V_0[(x/L)^4 -1] & |x|> L \\ 0 & |x| < L \end{cases} V_0 >0$ | ||

8 | $V(x) = a x^2 + b x^3 , a>0, b<0.$ | ||

9 | $V(x) = \begin{cases} \infty & x < 0 \\ \dfrac{k}{x} & x > 0, k >0 \end{cases}$ | ||

10 | $V(x) = V_0 [\exp(-2\alpha x) - 2 \exp(-\alpha x)], \alpha >0 V_0>0$ |

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