Let's say we have a particle in an infinite well, and let's also say it is in the ground state. Now we make the well bigger by very quickly moving one of the boundaries of the well.   How do we compute the probability of the particle being in the ground state of the new well? Also, what is the prob of the particle being in the first excited state of the new well?  Assume for concreteness that the inital (1D)well is from x=0 to x=L and the new well is from x=0 to x=2L. We simply pull the right wall of the well to x=2L.
120491
I am currently studying for an exam in Quantum Mechanics and came across a solution to a problem I have trouble with understanding.  The Problem:  A Particle sits in an infinite potential well described by  \begin{align} V(x) &amp;= 0, &amp; 0 \leq x \leq L \\ V(x) &amp;= \infty, &amp; \text{otherwise} \end{align}  We know that the energies are given by $E_n = \dfrac{n^2 \pi^2 \hbar^2}{2 m L^2}$ and $\Psi(x) = A_n \sin(n \pi x /L)$.  At time $t_0$ the potential well is suddenly doubled in size, such that the potential is now  \begin{align} V(x) &amp;= 0, &amp; 0 \leq x \leq 2L \\ V(x) &amp;= \infty, &amp; \text{otherwise} \end{align}  So the energies are now given by $\tilde{E}_n = \dfrac{n^2 \pi^2 \hbar^2}{2 \cdot 4 m L^2}$ and $\tilde{\Psi}(x) = \tilde{A}_n \sin(n \pi x /2L)$.   If the particle is in the ground state of the potential well before the change, what is the probability to find the particle in the ground state of the new potential after the change?   This is absolutely clear to me. We find a non vanishing probability as a result. But now it gets tricky:   What is the expectation value of the energy of the particle directly after the change? How does the expectation value of the energy evolve in time?   The solution suggests that the expectation value of the energy does not evolve in time, which is clear to me, since the Hamiltonian is time independent and thus energy is conserved. But it also suggests that the expectation value does not change after we double the width of the potential wall which I understand from the argument of energy conservation but not in terms of quantum mechanics. If the probability that the particle is in the state $\tilde{\Psi}$ does not vanish the particle could have the energy $\tilde{E}_n$ which is lower than $E_n$ and this would mean that the expectation value of energy could change (with a given probability).  What am I missing here, where is my mistake? Any help is appreciated!
76579
Is there a way to resolve probability of an event, given another event that never happens? Mathematically speaking the problem is:  Given that $P(B) = 0$,  $$P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{0}{0}$$  Is this probability vacuously $0$ of $1$? Can we show that it's one or the other?
109341
Negative probabilities are naturally found in the Wigner function (both the original and its discrete variants), the Klein paradox (where it is an artifact of using a one-particle theory) and the Klein-Gordon equation.  Is a general treatment of such quasi-probability distributions, besides naively using 'legit' probabilistic formulas? For example, is there a theory saying which measurements are allowed, so to screen negative probabilities? Additionally, is there an intuition behind negative probabilities?
55906
So assume we have a probability space $(\Omega, \mathcal{F}, P)$ and a random variable $X : \Omega \rightarrow \mathbb{R}^*$. We can derive from this a distribution $P_X$, and a distribution function $F$. My book says that when working with distribution function directly we have no need to know the underlying probability space since everything is well determined. We just need to show that we can find one. So here is my attempt :   Here we start with the distribution function $F$. increasing and right continuous, st. $F(\infty) = 1$ and $F(-\infty) = 0$. We define :   \begin{align} P_I : I &amp;\longmapsto [0,1] \quad I \in \mathbb{R}^2 ( interval) \\ [a,b] &amp;\longmapsto F(b) - F(a) \end{align}   We can define $P_X$ on the borel sets by decomposing the set into disjoint intervals :    \begin{align} P_X : \mathcal{B}(\mathbb{R}^*) &amp;\longmapsto [0,1] \quad B = \dot \bigcup B_i \quad B_i \in \mathbb{R}^2 \\ B &amp;\longmapsto \sum_i P_I[a_i,b_i]  \end{align}   By taking $X$ as the identity function we have $\Omega = \mathbb{R}^*$ and $P = (P_X)_{\Omega}$   I suspect that I can use the fact that F is increasing and right continuous to say that we can find a measure $\mu$ s.t $\mu(a,b) = F(b)-F(a)$ and then use this measure as probability measure. But I'm scared to miss any vital step along this way.  I would like to know if there is any flaw in my reasoning. Or anything missing. Thanks for any input !
101363
I am starting to learn about quantum mechanics and was wondering about the solutions to a question about a particle in an infinite potential well of length $L$.  The solutions are: $$\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi n x}{L} \right)$$  And I was wondering what is the meaning of $n$. Isn't there supposed to be only one wave function? In that case, $\Psi = \sum_{n=1} ^\infty a_n \psi_n$, but what are the $a_n$? Or maybe it is probabilistic, and we don't know which $\psi_n$ the particle is currently in?
97321
Recently I read a  about solving radial Schrodinger equation with inverse power law potential.  Consider the radial Schrodinger equation(simply set $\mu=\hbar=1$):  $$\left(-\frac{1}{2}\Delta+V(\mathbf{r})\right)\psi(\mathbf{r})=E\psi(\mathbf{r}).$$  A  gives a one-dimension equation:  $$-\frac{1}{2}D^2\phi(r)+\left(V(r)+\frac{1}{2}\frac{l(l+1)}{r^2}\right)\phi(r)=E\phi(r),$$  where $D=\dfrac{d}{dr}$, and $l$ is the azimuthal quantum number.  If we only consider the ground state, then $l=0$, so  $$-\frac{1}{2}D^2 \phi(r)+V(r)\phi(r)=E\phi(r).$$  We want to find eigenvalue $E$ such that $\phi(0)=\phi(+\infty)=0$.  The central potential discussed in the paper is of this form:  $$V(r)=\alpha r^{-\beta}.$$  It states (see page 4) that if $\beta&gt;2$ then the potential is repulsive (i.e. $\alpha&gt;0$).  My questions are:    Is this conclusion (i.e. If $\beta&gt;2$ then we must have $\alpha&gt;0$) generally valid in physics? What would happen if $\alpha&lt;0$ and $\beta&gt;2$? Is there 'ground state' in this condition? What about the condition $\alpha&gt;0$ and $\beta=1,2$? I have tried to solve the equation numerically with $\alpha=1,\beta=1,2$, and the ground state energy in this two conditions seem to be $0$, is my result correct?   P.S. When I try to find ground state when  $\alpha=-1,\beta=2$, the energy seem to be $-\infty$, which is qualitatively different from $\alpha=-1,\beta=1$.
16822
In section 134 of Vol. 3 (Quantum Mechanics), Landau and Lifshitz make the energy complex in order to describe a particle that can decay:  $$ E = E_0 - \frac{1}{2}i \Gamma. $$  The propagator $U(t) = \exp(-i H t)$ then makes the wavefunction die exponentially with time.  But also, $H$ is non-Hermitian.  My question: Do we have to modify the basic postulates of quantum mechanics (as described by Shankar, say, or the earlier sections of Landau &amp; Lifshitz) to describe unstable particles?
66015
Sorry if this is really naive, but we learned in Newtonian physics that the total energy of a system is only defined up to an additive constant, since you can always add a constant to the potential energy function without changing the equation of motion (since force is negative the gradient of the potential energy).   Then in Quantum Mechanics we showed how the ground state of a system with potential energy $V(x) = \frac{1}{2} m \omega^{2} x^{2}$ has an energy $E_{0}=\frac{1}{2} \hbar \omega $.   But if we add a constant to $V(x)$ won't that just shift the ground state energy by the same constant? So in what sense can we actually say that the ground state energy has an absolute value (as opposed to just a relative value)? Is there some way to measure it?   I ask this in part because I have heard that Dark Energy might be the ground state energy of quantum fields, but if this energy is only defined up to a constant, how can we say what it's value is?
112907
I am preparing for the exam. And I need to know the answer to one question which I can't understand.  "Give an example of non-Hamiltonian systems: in case of infinite number of particles; for a finite number of particles".  I hope somebody can help me.
82078
A particle is subject to an infinite square well potential with  $$V(x)= \begin{cases}  0  &amp;   −a \lt x \lt a\\   \infty &amp;  \,\,\,\,\text{otherwise} \end{cases}$$  At a time $t=0$ its wavefunction is given by $$\psi(x, t=0)=\frac{1}{\sqrt{5a}}\cos\left(\frac{\pi x}{2a}\right)+\frac{2}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right)$$       (a) What are the possible results of an energy measurement at $t = 0$, and with what probabilities?     Re-writing the terms in $\psi(x,t=0)$ as $$u_1(x)=a^{-1/2}\cos\left(\frac{\pi x}{2a}\right)\qquad \text{&amp;} \qquad u_2(x)=a^{-1/2}\sin\left(\frac{\pi x}{a}\right)$$ such that $$\psi(x,t=0)= \frac{1}{\sqrt{5}}u_1(x)+\frac{2}{\sqrt{5}}u_2(x)$$ with energies $$E_1=\frac{\hbar^2 \pi^2}{8ma^2}\qquad \text{&amp;} \qquad E_2=\frac{\hbar^2 \pi^2}{2ma^2}$$ and probabilities $$P(E_1)=\frac15 \qquad \text{&amp;} \qquad P(E_2)=\frac45$$       (b) If no measurement is performed, what is the wavefunction $\psi(x, t)$ at all times $t$? What are the possible energies and their probabilities if a measurement is first performed at time $t$?     At a time $t$ the wavefunction is given by $$\psi(x, t=0)=\frac{1}{\sqrt{5a}}\cos\left(\frac{\pi x}{2a}\right)\exp\left(\frac{-i E_1 t}{\hbar}\right)+\frac{2}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right)\exp\left(\frac{-i E_2 t}{\hbar}\right)$$    The problem is that I cannot answer the second part of the question (b).    The answer states that:      Because the two terms are energy eigenstates, the relative probabilities do not change and are as in the previous part.     I have some questions about the above statement: why does an "energy eigenstate" mean that the probablities don't change with time? Or, put in another way, does a "momentum eigenstate" (for example) have constant relative probabilities when measured at any time?   The answer seems to be implying there is something special about energy eigenstates when it comes to time evolution. The eigenvalues of energy measurements don't seem to change regardless of how many times you measure them, and no matter how long you wait until you measure them. I'm not sure if this is the case, but could someone please explain why all quantum operators (except Hamiltonian) do not exhibit this behaviour.
25399
I'm studying for a test in quantum mechanics and I'm having a hard time understanding how to use . There are no examples in my text book, only definitions that I can't understand how to use, so I hope you can help me instead.  So, there's an assignment where you're supposed to calculate first order correction to the ground state using ladder operators. In the right answer they say that  $$E_0^1 = \langle 0|H'|0\rangle$$  where does the zeros come from? I understand that they're coming from the wave function $\psi$. But I don't really understand the theory behind this.  Then, after some calculations it's written in the right answer  $$ \langle 0|aa^\dagger a^\dagger|0\rangle +\langle0|aaa^\dagger a^\dagger|0\rangle =\langle0|aa^\dagger a|1\rangle +\langle0|aaa^\dagger|1\rangle$$  What are the mathematical rules of these ladder operators? How did these zeros become ones?  I would really appreciate if someone could help me understand this a little bit better.
121390
I'm trying to learn how to apply the WKB approximation. Given the following problem:      An electron, say, in the nuclear potential   $$U(r)=\begin{cases}  &amp; -U_{0} \;\;\;\;\;\;\text{ if } r &lt; r_{0} \\   &amp; k/r \;\;\;\;\;\;\;\;\text{ if } r &gt; r_{0} \end{cases}$$   1. What is the radial Schrödinger equation for the $\ell=0$ state?      2. Assuming the energy of the barrier (i.e. $k/r_{0}$) to be high, how do you use the WKB approximation to estimate the bound state energies inside the well?     For the first question, I thought the radial part of the equation of motion was the following  $$\left \{ - {\hbar^2 \over 2m r^2} {d\over dr}\left(r^2{d\over dr}\right) +{\hbar^2 \ell(\ell+1)\over 2mr^2}+V(r) \right \} R(r)=ER(r)$$  Do I simply just let $\ell=0$ and obtain the following? Which potential do I use?  $$\left \{ - {\hbar^2 \over 2m r^2} {d\over dr}\left(r^2{d\over dr}\right) +V(r) \right \} R(r)=ER(r)$$  For the other question, do I use $\int \sqrt{2m(E-V(r))}=(n+1/2)\hbar π$, where $n=0,1,2,...$ ? If so, what are the turning points? And again, which of the two potentials do I use?
47714
Just to get the thoughts rolling...  Consider a two state system with discrete energy levels $E_1$ and $E_2$ where $E_2 &gt; E_1$ which contains $N$ particles.  We can easily deduce that the state of maximum entropy is when energy level $E_1$ and $E_2$ each contain $\frac{N}2$ particles.  Based on the (flipped-upside-down) thermodynamic definition of temperature, $$(\frac{\partial E}{\partial S})=T$$ we see that when energy as a function of entropy increases, temperature also increases.  This means, for our two state system, every time a particle populates level $E_2$, there is a corresponding increase in temperature because that population increases entropy.  What about when we go beyond the point of maximum entropy where there are $\frac N2$ particles in $E_1$ and $E_2$?  Well, then we are adding energy to the system, but the entropy is decreasing.  Based on the Boltzmann factor, this can't be possible unless the temperature of the system is negative. That is, $$\frac{P_1}{P_2}=e^{-\frac{(E_1-E_2)}{kT}}$$ which approaches infinity unless one says the system has a negative absolute temperature.  Now we're getting to the point: this increase in energy which is connected to a decrease in entropy is what has been called a . This is most often connected to .  Forgive me for saying things you already know.  All of this raises a lot of questions though, the most striking to me is that there does not seem to me to be a good physical explanation for how this can be possible. For instance, when people say that the negative temperature system is "hotter" I get what that means as far as saying that heat will flow from a negative temp. system to a pos. temp system.  I'm fine with that except that it invokes the idea of a fairly macroscopic quantity--hotness--for a purely quantum phenomenon. So, is there any macroscopic physical meaning to this statement that the negative temperature is "hotter"?  Concerning population inversion in lasers: doesn't this mean that the negative temperature required for that laser to exist ought to be so "hot" no mere mortal can handle such a device.  After all, negative absolute temperatures are often explained as being "hotter than infinite temperature." (I've heard this other places, but also see the neg. temp. Wikipedia article for quote.) I understand that again from a mathematical perspective, but what is the physical connection? A statement like that must surely manifest itself in some way that is noticeable beyond the quantum level.  I guess I really want to understand how it could be this "hotter than infinite temperature" system isn't perceptible in the macroscopic realm? I mean, lasers are cool, but they aren't just melting through everything as if they have infinite temperature.    My bad for the obvious misuse of the word "physical" when I probably meant something more like "macroscopic". I hope the point is clear anyways.  Thoughts?
78507
While Solving the TISE for a particle an infinite square well with potential given by: $$ U(x) = \left\{         \begin{array}{ll}             0 &amp; \quad -L/2 \leq x \leq L/2 \\             \infty &amp; \quad otherwise         \end{array}     \right. $$  we get two sets of solutions: $$ \psi(x) = \left\{         \begin{array}{ll}             A\sin(\frac{n\pi x}{L}) &amp; \quad n = 2,4,6,... \\             B\cos(\frac{n \pi x}{L}) &amp; \quad n = 1,2,3,...         \end{array}     \right. $$  But when we solve the TISE for potential $ U(x) = 0 \quad for \quad 0 \leq x \leq L $ we get only one solution i.e. $\phi(x) = A\sin(\frac{n\pi x}{L}) \space where \space A=\sqrt{\frac{2}{L}}$.  I want to know what difference does the change of coordinates makes on the system so that for one there are two solutions and for other there is only one?
103970
$\bullet$ 1. For the one-particle states, the completeness relation is given in Peskin and Schroeder, $$(\mathbb{1})_{1-particle}=\int\frac{d^3\textbf{p}}{(2\pi)^{3}}|\textbf{p}\rangle\frac{1}{2E_\textbf{p}}\langle\textbf{p}|$$ How does one derive this?   $\bullet$ 2. In expanding a general Fock state $|\Psi\rangle$ as $|\Psi\rangle=\mathbb{1}|\Psi\rangle$ can we approximate the identity operator as $$\mathbb{1}\approx(\mathbb{1})_{1-particle}?$$ If yes, when?  $\bullet$ 3. If not, how does one derive similar completeness relations for 2-particle states $|\textbf{p}_1,\textbf{p}_2\rangle$ (or in general, for $N$-particle states $|\textbf{p}_1,\textbf{p}_2...|\textbf{p}_N\rangle$)?
54908
Is it logically correct to assert that the expectation of the momentum $$\langle \hat p \rangle=0$$ for any bound state because it is bound to some finite region? What is the physical interpretation of the fact that $$\langle \hat p \rangle=0$$ in an energy eigenstate $\psi_n(x,t)$  but $$\langle \hat p \rangle\neq0$$ in some superposition state $$\psi(x,t)=c_m\psi_m(x,t)+c_n\psi_n(x,t)~?$$ Here $\psi_n(x,t)$ the eigenstates of the Hamiltonian, for example, in the problem of particle in a box (say).
32148
I'm having a hard time understanding the following evaluation of the infinite product  $$ \prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right).$$ In particular, I don't understand how you go from line 2 to line 3. Here $\omega = -\frac{1}{2}+ i \frac{\sqrt{3}}{2}$, which is a primitive third root of unity.  $$\begin{align}\prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)  &amp;=\prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3} \\  &amp;=\lim_{m\to\infty}\frac{1}{m}\prod_{n=2}^m\frac{(n-\omega)(n-\omega^2)}{n^2} \\  &amp;=\lim_{m\to\infty}\frac{\Gamma(m+1-\omega)\Gamma(m+1-\omega^2)}{m(m!)^2\Gamma(-\omega)\Gamma(-\omega^2)(1-\omega)(1-\omega^2)(-\omega)(-\omega^2)} \\  &amp;=\frac{1}{3\Gamma(-\omega)\Gamma(-\omega^2)} \\ &amp;=\frac{\sin{\pi(-\omega)}}{3\pi} \\  &amp;=\frac{\cosh (\frac{\sqrt{3}\pi}{2})}{3\pi } \end{align}$$ EDIT: I think I'm starting to make sense out of this by writing out the terms. $$ \begin{align} \prod_{n=2}^{m} \frac{(n- \omega)(n-\omega^{2})}{n^{2}} &amp;= \frac{(2-\omega)(3-\omega) \cdots (m- \omega)(2-\omega^{2})(3-\omega^{2})\cdots (m- \omega^{2})}{2^{2} \cdot 3^{2} \cdot \cdots  \cdot m^{2}}  \\ &amp;= \frac{1}{(m!)^{2}} \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})} \end{align}$$ Now I have to figure out how to take the limit as $m \to \infty$.