# Is the time independent Schrodinger equation commutative?

• etotheipi
Thanks for pointing that out!In summary, the conversation discusses the time independent Schrodinger equation and a potential mistake in its derivation. The mistake is corrected and further calculations are discussed, including the use of an integrating factor. The conversation concludes with a clarification on a term that was initially overlooked.
etotheipi
Homework Statement
A particle moves in the potential ##V(x) = -\frac{\hbar^2}{m} \text{sech}^2{x}##. Show that ##A^{\dagger}A \psi = (\mathcal{E} + 1)\psi## where ##\mathcal{E} = \frac{2mE}{\hbar^2}## and ##A = \frac{d}{dx} + \tanh{x}##, ##A^{\dagger} = -\frac{d}{dx} + \tanh{x}##.

Hence show that the ground state has ##\mathcal{E} \geq -1## and that a wavefunction ##\psi_0(x)## has an energy eigenstate with ##\mathcal{E} = -1## iff ##\frac{d\psi_0}{dx} + \tanh{(x)} \psi_0 = 0## and find ##\psi_0(x)##.
Relevant Equations
N/A
I'm falling at the first hurdle here; the time independent Schrodinger becomes $$-\frac{\hbar^2}{2m} \psi''(x) - \frac{\hbar^2}{m}\text{sech}^2(x) \psi(x) = E\psi(x)$$ $$\left(-\frac{d^2}{dx^2} - 2\text{sech}^2(x) \right)\psi(x) = \frac{2mE}{\hbar^2}\psi(x) = \mathcal{E} \psi(x)$$ $$\left(-\frac{d^2}{dx^2} + 2\tanh^2(x) \right)\psi(x) = (\mathcal{E}+2) \psi(x)$$ But this can't be right since ##A^{\dagger}A = \tanh^2(x) - \frac{d^2}{dx^2}## and I've also got a 2 on the other side.

I wondered if anyone could point out my mistake? Thanks!

Applying coefficient 1/2 on V(x) seems to satisfy the given relation.

Last edited:
etotheipi
anuttarasammyak said:
Applying coefficient 1/2 on V(x) seems to satisfy the given relation.

I noticed that but the source of these questions is usually very reliable so I thought there's no way it could be a mistake! But if you say so also, then I am happy to correct it and work from that. Thanks!

A further question; with an integrating factor and then with some normalisation I can find $$\psi(x) = \frac{1}{\sqrt{2}\cosh{x}}$$ for the equation ##\frac{d\psi}{dx} + \tanh{x} \psi(x) = 0##. But how was it possible to obtain this differential equation from the second order one above? We know that ##A^{\dagger}A## has eigenvalues greater than or equal to zero.

I found my post #2 was wrong. From OP
$$(-\frac{d^2}{dx^2}+2\tanh^2 x -1)\psi(x)=(\epsilon+1) \psi(x)$$
$$(-\frac{d}{dx}+\tanh x)(\frac{d}{dx}+\tanh x)\psi(x)=(\epsilon+1) \psi(x)$$
using
$$-\frac{d}{dx}(\tanh x\ \psi(x))= (-1 + tanh^2x)\psi(x) - \tanh x \frac{d}{dx}\psi(x)$$

Last edited:
etotheipi
anuttarasammyak said:
I found my post #2 was wrong.
How so? It gives the correct result: \begin{align*}-\frac{\hbar^2}{2m} \psi''(x) - \frac{\hbar^2}{2m}\text{sech}^2(x) \psi(x) &= E\psi(x) \\ \left(-\frac{d^2}{dx^2} - \text{sech}^2(x) \right)\psi(x) &= \frac{2mE}{\hbar^2}\psi(x) = \mathcal{E} \psi(x) \\ A^{\dagger}A \psi(x) = \left(-\frac{d^2}{dx^2} + \tanh^2(x) \right)\psi(x) &= (\mathcal{E}+1) \psi(x) \end{align*}

We have forgotten the term
$$-\frac{d \tanh x }{dx} \psi(x)$$

anuttarasammyak said:
We have forgotten about the term
$$-\frac{d \tanh x }{dx} \psi(x)$$

I'm sorry, I don't follow. Isn't it just a difference of two squares? That along with ##\tanh^2(x) + \text{sech}^2(x) = 1##.

$$(-\frac{d}{dx}+\tanh x)(\frac{d}{dx}+\tanh x)\psi(x)=-\frac{d^2 \psi(x)}{dx^2}+\tanh x\frac{d \psi(x)}{dx}+\tanh^2 x\ \psi(x)-\frac{d}{dx}[\tanh x\ \psi(x)]$$

etotheipi
Ah okay, yes you're right it's not commutative.

## 1. What is a particle in a potential?

A particle in a potential refers to a particle, such as an electron, that is subjected to a force in a specific area or field, known as a potential. This potential can be either attractive or repulsive and can affect the motion and behavior of the particle.

## 2. What is the significance of studying particles in a potential?

Studying particles in a potential is important in understanding the behavior and properties of particles in various fields, such as quantum mechanics and electromagnetism. It also helps in predicting and explaining the behavior of particles in real-world scenarios, such as in chemical reactions and electronic devices.

## 3. How is the motion of a particle in a potential described?

The motion of a particle in a potential is described by the Schrödinger equation in quantum mechanics, which takes into account the potential energy, kinetic energy, and mass of the particle. This equation can be solved to determine the probability of finding the particle at a certain position in the potential.

## 4. What factors affect the behavior of a particle in a potential?

The behavior of a particle in a potential is affected by various factors, such as the strength and shape of the potential, the mass and energy of the particle, and external forces acting on the particle. These factors can alter the trajectory and properties of the particle, leading to different outcomes.

## 5. Can a particle in a potential have multiple energy levels?

Yes, a particle in a potential can have multiple energy levels, also known as energy states. These energy states correspond to different levels of energy that the particle can possess while being confined in the potential. The number of energy levels depends on the specific potential and the properties of the particle.

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