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mgal95
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The problem stated below is from Liboff "Introductory Quantum Mechanics" (2nd Edition), exercise 5.4.
A pulse ## 1m ## long contains ##1000 \alpha ## particles. At ## t=0## each ##\alpha## particle is in the state:
<br /> \psi (x,0)=\frac{1}{10}\exp (ik_ox)<br />
for |x|\leq 50cm and zero elsewhere. It is given that k_o=\pi/50
(a) At t=0, how many \alpha particles have momentum in the interval 0\leq\hbar k\leq \hbar k_o ?
(b) At which values of momentum will ##\alpha## particles not be found at ##t=0##?
(c) Describe an experiment to "prepare" such a state.
(d) Construct ##\Delta x## and ##\Delta p## for this state, formally. What is ##\Delta x\Delta p##? (Hint: To calculate ##\Delta p##, use ##|b(k)|^2##)
As I get it, we are in a Hilbert Space spanned by the eigenfunctions of the momentum operator, which constitute a continuous set. Thus, the state wavefunction should be written as:
<br /> \psi (x,0)=\int_{-\infty}^{+\infty} b(k)\phi_k dx<br />
where
<br /> \phi_k =\frac{1}{\sqrt{2\pi}}e^{ikx}<br />
Thus, we have (let L=100cm): ##\displaystyle b(k)=\int_{-\infty}^{+\infty}\psi (x,0)\phi^{\star}_k dx=...=\sqrt{\frac{2}{\pi L}}\frac{\sin[(k_o-k)L/2]}{k_o-k}##
(a) That being said, the number of particles in the given interval is just the integral:
<br /> N_{interval}=N_{total}\int_0^{k_o}|b(k)|^2dk<br />
(b) We will not find particles in the states whom ##b(k)## equals to zero, that is: ##\sin[(k_o-k)L/2]=0\rightarrow k=\frac{2n\pi}{L}-k_o\rightarrow p_k=\hbar k=\cdots## with ##n\in\mathbb Z##
(c) I suppose that we could get a laser beam of definite wavelength, which means of definitive momentum, and just send a pulse of ##100cm## in lenght. The state of such a laser beam (thought in the whole x axis) should be an eigenstate of the momentum (the one with ##k_o## should be chosen obviously). When we send the pulse (turn on and off the laser), we should get the given wavefunction. Is that any close to being right?
(d) I have encountered many problems here. I used the definition for the uncertainty, that is:
<br /> (\Delta p)^2=\left<\hat{p}^2\right>-\left<\hat{p}\right>^2=\left<\psi \right|\hat{p}^2\left|\psi\right>-\left<\psi\right|\hat{p}\left|\psi\right>^2<br />
Computing the above, I get:
##\displaystyle \left<\psi \right|\hat{p}^2\left|\psi\right>=\int_{-\infty}^{+\infty}\psi^{\star}\hat{p}^2\psi dx=(-i\hbar)^2(ik_o)^2\int_{-L/2}^{L/2}\psi^{\star}\psi dx=k_o^2\hbar^2##
And the other one: ##\displaystyle \left<\psi \right|\hat{p}\left|\psi\right>=\int_{-\infty}^{+\infty}\psi^{\star}\hat{p}\psi dx=(-i\hbar)(ik_o)\int_{-L/2}^{L/2}\psi^{\star}\psi dx=k_o\hbar##
But this gives ##\Delta p =0## ! This could be true if the uncertainty in ##x## was infinite. However, it is easily seen that this is not the case because: ##\displaystyle \left<x\right>=\int_{-L/2}^{L/2} x|\psi|^2dx=0## and ##\displaystyle \left<x^2\right>=\int_{-L/2}^{L/2} x^2|\psi|^2dx=\frac{L^2}{12}##
My first issue is that Heisenberg's uncertainty principle is violated, which means that somewhere lies a mistake. I see that the uncertainty in the momentum cannot be zero because I prooved in (a) that there is a continuous spectrum of the momentum eigenstates which constitute our wavefunction. In addition to that, I tried to compute the integrals in the case where the exponential is all over the x axis, namely when we have as state function an eigenstate of the momentum (where I should get zero uncertainty in the momentum). Both integrals (obviously) go to infinity and cannot be substracted. Last but not least, we are supposed to have ##C^{\infty}## functions as wavefunctions (at least ##C^2##). This postulate gives us the boundary conditions in problems like the infinite well. However ##e^z## does not equal zero for any ##z\in\mathbb C##. How can we have, then, such a wavefunction?
I want to know, first of all, which of my ideas are right and which are false. I study quantum mechanics on my own and there is no one else to ask for help. I am supposed to take the course at the University this Fall and I am trying to prepare myself a bit. I have a lot of other questions as well, but this one has got on my nerves for 2 days now and I cannot sleep, because I think I am deprived of basic knowledge.
Thank you for your time and forgive my english. It is not my mother tongue.
Homework Statement
A pulse ## 1m ## long contains ##1000 \alpha ## particles. At ## t=0## each ##\alpha## particle is in the state:
<br /> \psi (x,0)=\frac{1}{10}\exp (ik_ox)<br />
for |x|\leq 50cm and zero elsewhere. It is given that k_o=\pi/50
(a) At t=0, how many \alpha particles have momentum in the interval 0\leq\hbar k\leq \hbar k_o ?
(b) At which values of momentum will ##\alpha## particles not be found at ##t=0##?
(c) Describe an experiment to "prepare" such a state.
(d) Construct ##\Delta x## and ##\Delta p## for this state, formally. What is ##\Delta x\Delta p##? (Hint: To calculate ##\Delta p##, use ##|b(k)|^2##)
Homework Equations
The Attempt at a Solution
As I get it, we are in a Hilbert Space spanned by the eigenfunctions of the momentum operator, which constitute a continuous set. Thus, the state wavefunction should be written as:
<br /> \psi (x,0)=\int_{-\infty}^{+\infty} b(k)\phi_k dx<br />
where
<br /> \phi_k =\frac{1}{\sqrt{2\pi}}e^{ikx}<br />
Thus, we have (let L=100cm): ##\displaystyle b(k)=\int_{-\infty}^{+\infty}\psi (x,0)\phi^{\star}_k dx=...=\sqrt{\frac{2}{\pi L}}\frac{\sin[(k_o-k)L/2]}{k_o-k}##
(a) That being said, the number of particles in the given interval is just the integral:
<br /> N_{interval}=N_{total}\int_0^{k_o}|b(k)|^2dk<br />
(b) We will not find particles in the states whom ##b(k)## equals to zero, that is: ##\sin[(k_o-k)L/2]=0\rightarrow k=\frac{2n\pi}{L}-k_o\rightarrow p_k=\hbar k=\cdots## with ##n\in\mathbb Z##
(c) I suppose that we could get a laser beam of definite wavelength, which means of definitive momentum, and just send a pulse of ##100cm## in lenght. The state of such a laser beam (thought in the whole x axis) should be an eigenstate of the momentum (the one with ##k_o## should be chosen obviously). When we send the pulse (turn on and off the laser), we should get the given wavefunction. Is that any close to being right?
(d) I have encountered many problems here. I used the definition for the uncertainty, that is:
<br /> (\Delta p)^2=\left<\hat{p}^2\right>-\left<\hat{p}\right>^2=\left<\psi \right|\hat{p}^2\left|\psi\right>-\left<\psi\right|\hat{p}\left|\psi\right>^2<br />
Computing the above, I get:
##\displaystyle \left<\psi \right|\hat{p}^2\left|\psi\right>=\int_{-\infty}^{+\infty}\psi^{\star}\hat{p}^2\psi dx=(-i\hbar)^2(ik_o)^2\int_{-L/2}^{L/2}\psi^{\star}\psi dx=k_o^2\hbar^2##
And the other one: ##\displaystyle \left<\psi \right|\hat{p}\left|\psi\right>=\int_{-\infty}^{+\infty}\psi^{\star}\hat{p}\psi dx=(-i\hbar)(ik_o)\int_{-L/2}^{L/2}\psi^{\star}\psi dx=k_o\hbar##
But this gives ##\Delta p =0## ! This could be true if the uncertainty in ##x## was infinite. However, it is easily seen that this is not the case because: ##\displaystyle \left<x\right>=\int_{-L/2}^{L/2} x|\psi|^2dx=0## and ##\displaystyle \left<x^2\right>=\int_{-L/2}^{L/2} x^2|\psi|^2dx=\frac{L^2}{12}##
My first issue is that Heisenberg's uncertainty principle is violated, which means that somewhere lies a mistake. I see that the uncertainty in the momentum cannot be zero because I prooved in (a) that there is a continuous spectrum of the momentum eigenstates which constitute our wavefunction. In addition to that, I tried to compute the integrals in the case where the exponential is all over the x axis, namely when we have as state function an eigenstate of the momentum (where I should get zero uncertainty in the momentum). Both integrals (obviously) go to infinity and cannot be substracted. Last but not least, we are supposed to have ##C^{\infty}## functions as wavefunctions (at least ##C^2##). This postulate gives us the boundary conditions in problems like the infinite well. However ##e^z## does not equal zero for any ##z\in\mathbb C##. How can we have, then, such a wavefunction?
I want to know, first of all, which of my ideas are right and which are false. I study quantum mechanics on my own and there is no one else to ask for help. I am supposed to take the course at the University this Fall and I am trying to prepare myself a bit. I have a lot of other questions as well, but this one has got on my nerves for 2 days now and I cannot sleep, because I think I am deprived of basic knowledge.
Thank you for your time and forgive my english. It is not my mother tongue.