How Does a Particle Behave in a One-Dimensional Delta Potential?

[skrat]

Homework Statement

A particle is moving in a one-dimensional potential $$V(x)=-\lambda \delta (x)$$ where $\lambda >0$. Calculate
a) Expected value and uncertainty of particle energy.
b) At $t=0$ we measure the position of the particle. What is the probability that the particle has $|x|>\frac{\hbar ^2}{m\lambda }$ right after the measurement?
c) At $t=0$ we measure the energy of the particle. What are possible results and their probabilities?
d) What if we do all those measurements at $t>0$?

Homework Equations

The Attempt at a Solution

Let us firstly find that wavefunction:
Before the Delta function we expect something like $\psi _1=A e^{ikx}$ and after $\psi _2=Ae^{-ikx}$.

Boundary condition $$\psi (-0)=\psi (+0)$$ eliminates all odd functions while the second boundary condition $${\psi (+0)}'-{\psi (-0)}'=-2\frac{m\lambda }{\hbar ^2}\psi (0)$$ leaves me with $k=-\frac{im\lambda}{\hbar ^2}$.

With given $k$ and after normalization of the function, we finally get to the desired wavefunction $$\psi (x)=\sqrt{\frac{m\lambda }{\hbar ^2}}e^{-\frac{m\lambda }{\hbar ^2}|x|}$$

a) I seriously hope there is an easier way than $$<E>=\int _{-infty}^{\infty} \psi(x)[-\frac{\hbar ^2}{2m}\frac{\partial^2 }{\partial x^2}-\lambda \delta (x)]\psi (x)dx=\frac{5m\lambda}{2\hbar ^2}$$ Doing the same horrible integral for $<E^2>=<\frac{p^4}{4m^2}-\lambda \frac{p^2}{2m}\delta(x)+\lambda ^2 \delta (x)>$ will eventually bring me to $\delta _E=\sqrt{<E^2>-<E>^2}$

b) I am quite unsure if this is correct: $$P=\frac{<x>_{part}}{<x>}$$ where $<x>_{part}=2\int_{0}^{\frac{\hbar^2}{2m}}\psi (x)^2 dx$ and $<x>=2\int_{0}^{\infty }\psi (x)^2 dx$. Or is it?

Why I am confused, is that if we for example had Harmonic oscillator $|\psi,0>=\frac{1}{\sqrt 2}(|0>+|1>)$. Than we only have to calculate $|\psi, t>=p_1|0>+p_2|1>$ because $p_1^2$ and $p_2^2$ represent the probability that the particle is in given state. However, I am bit lost in my case. :/

Ok and no idea about c) and d) :/

 

[mfb]

The problem statement is misleading - in general, a particle in this potential can be free, and have whatever energy it likes to have.
In a bound state (=what you calculated - it also helps for c and d), what do you expect for $\delta_E$? And I would expect <E> to be negative.

(b) I would not call it <x> because it is not the expectation value of the position. The formulas look good.

 

[skrat]

Yeah I can see that. That is because I left out the most important part.
At time $t=0$ the particle is in ground state.

a) Energy is simply $E=\frac{\hbar^2}{2m}k^2=-\frac{m\lambda ^2}{2\hbar ^2}$. I was also suspicious because energy wasn't negative, but this seems to be a better solution now. No integration needed.

b) I was worried the notation may be a bit strange, but as long as we both understand it, it's fine. =)

Also, I got some answers for c) and d) and I would like to check if they make any sense at all:

c) Since the particle is in ground state, it can only move to higher energy levels, but for that it needs to somehow get that extra energy. Since the problem doesn't state anything about putting extra energy to the system, I assume it is safe to say that the particle will only (and also always - meaning for $t>0$) be in ground state.
According to this, the answer to c) is the energy calculated in b) with probability 1. And the answer to part d) is that nothing changes.

 

[mfb]

I would answer the same for (c) and (d). They just look a bit odd.

 

[paranormal]

I would suggest using the Schrödinger equation to solve for the particle's energy in this potential. This will give a more rigorous and general solution than the approach of finding the wavefunction and then calculating the energy.

For part a), the expected value and uncertainty of the particle's energy can be found by solving the time-independent Schrödinger equation and using the resulting wavefunction to calculate the expectation value and uncertainty using the appropriate operators for energy.

For part b), the probability can be found by integrating the wavefunction squared over the region of interest. This will give the probability of finding the particle in that region after the measurement.

For part c), measuring the energy at t=0 will give a definite result, as the particle is in an eigenstate of energy. The possible results will be the eigenvalues of the energy operator, and the probabilities will be the coefficients of the wavefunction squared.

For part d), the measurements at t>0 will give the same results as at t=0, but the wavefunction will evolve according to the time-dependent Schrödinger equation. This means that the probabilities will change as the wavefunction evolves over time.

In summary, to fully answer this problem, it would be best to solve the Schrödinger equation and use the resulting wavefunction to calculate the expected value, uncertainty, and probabilities for the particle's energy and position.