# Find all values of E for a transcendental equation

• Mathematica
• happyparticle
In summary, the potential well has a lot of bound states, and the mass of the particle is around 7 GeV.
happyparticle
Hi,
I'm trying to solve a transcendental equation. I would like all the values of E that solve this equation.
##k = -l \cdot Cot(la)##
However, using Nsolve or FindRoot, they give me a precision error. Hence, I'm trying this form.
##\sqrt{-e /(e+v)} = -Cot(la)##
FindRoot only give me an imaginary value and NSolve just keep running.
Here is my code
Code:
k = Sqrt[-2*m*e] /h
l = Sqrt[2*m*(e+v)] /h
m = (2.66*10^-26)(2.32*10^-26) / ((2.66*10^-26) + (2.32*10^-26))
h = 6.58-10^-16
v = 0.05
a = 3.2*10^-10

Nsolve[Sqrt[-e / (e+v)] == -(Cot[l-a]), e]

I'm trying to find all the energies corresponding to the bound states in a potential well.
https://quantummechanics.ucsd.edu/ph130a/130_notes/node151.html

Thank you

Last edited:
It is not Cot[l-a], but Cot[l a].

However, if you try and plot the two sides of the equality with the correct term in the cotangent, you will see that the two curves never cross in the range of interest, ##[-V_0,0]##.

I find the mass is about 7 GeV. In an infinite square well of width 0.32 nm, the ground state energy is around 0.00027 eV, so there should be plenty of bound states in the finite well of depth 0.05 eV.

DrClaude
happyparticle said:
Here is my code
Code:
k = Sqrt[-2*m*e] /h
l = Sqrt[2*m*(e+v)] /h
m = (2.66*10^-26)(2.32*10^-26) / ((2.66*10^-26) + (2.32*10^-26))
h = 6.58-10^-16
v = 0.05
a = 3.2*10^-10

Nsolve[Sqrt[-e / (e+v)] == -(Cot[l-a]), e]
In addition to the other errors @DrClaude has pointed out, you have a typo in defining ##\hbar## as well.

Here's a little trick to simplifying the calculations and avoid a bunch of unit conversions.
$$k = \sqrt{-\frac{2mE}{\hbar^2}} = \sqrt{-\frac{2(mc^2)E}{(\hbar c)^2}}.$$ Express the mass ##m## in units of ##{\rm eV}/c^2## and the energy ##E## in eV, and use ##\hbar c = 197.3~\rm eV\cdot nm##. Then the calculated value of ##k## will have units of ##\rm nm^{-1}##.

DrClaude

## 1. What is a transcendental equation?

A transcendental equation is an equation that contains a transcendental function, such as trigonometric, exponential, or logarithmic functions, and an unknown variable. These equations cannot be solved algebraically and require numerical methods to find the values of the variable.

## 2. How do you find all values of E for a transcendental equation?

To find all values of E for a transcendental equation, you can use numerical methods such as the bisection method, Newton's method, or the secant method. These methods involve iteratively guessing and refining the value of E until a desired level of accuracy is achieved.

## 3. Can a transcendental equation have infinite solutions?

Yes, it is possible for a transcendental equation to have infinite solutions. This is because transcendental functions, such as trigonometric and exponential functions, are periodic and can produce an infinite number of solutions for certain values of the variable.

## 4. How do you know if a solution for a transcendental equation is accurate?

The accuracy of a solution for a transcendental equation can be determined by checking if the solution satisfies the original equation. Additionally, the solution can be compared to known values or approximations to determine its accuracy.

## 5. Are there any limitations to using numerical methods for finding solutions to transcendental equations?

Yes, there are some limitations to using numerical methods for finding solutions to transcendental equations. These methods may not always converge to a solution, and they may also be computationally expensive and time-consuming for certain equations. Additionally, there may be multiple solutions or no solutions at all for certain transcendental equations.

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