The title explains itself. If the potential is an even function then prove that wave function is either odd or even. I set $-x$ in Schrodinger equation and find out that $\psi(-x)$ is also a solution for the equation therefore any linear combination of $\psi(x)$ and $\psi(-x)$ is also a solution but I couldn't go any further from that. Any help is appreciated.
24421
I have done the Problem 2.1 in Griffiths' quantum mechanics,  and it seems not making sense to me.  What if the wave function isn't symmetric at all?  Then obviously the proof doesn't work. The solution confuses me.      If $V(x)=V(-x)$ then the "position" wave function can always be taken to be either even or odd.
56655
How to determine whether a wave is travelling or standing? I have been told that wave function of the form $f(k_1x \pm k_2t)$ is a travelling wave. But then even a standing wave equation $2A \cos \omega t \sin kx$ can be written as superposition of two waves $$A\sin\frac{kx-\omega t}{2} + A\sin\frac{kx+\omega t}{2}$$ This is of the form $f(k_1x \pm k_2t)$ but still not a travelling wave.  In particular is $y=\cos x \sin t \ + \cos 2x \sin 2t$ a travelling wave?
92741
I have a question concerning delta function potentials inside an infinite well, like this one  to which we add the following delta function potential: $V_2(x)=\frac{\hbar^2}{2m}\frac{\beta}{a}\delta (x)$. I found the bound states equations for even solutions: $\tan\frac{y}{2}=-\frac{2y}{\beta}$ and odd solutions: $y=2n\pi$, where $y=\sqrt{\frac{2mE}{\hbar^2}}$. Before adding the delta function potential, the ground state plot was:   and after adding the potential:   and this is what I don't understand. Why is the function "pulled" down by the delta potential?  Would this potential do the same thing if it was put in another place? How can I explain the effects of the delta potential in the ground state, or any other state wave function?
38405
I have the one dimensional free particle Schrödinger equation  $$i\hbar \frac{\partial}{\partial t} \Psi (x,t) =  -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi (x,t), \tag{1}$$  with general solution  $$\psi(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}. \tag{2}$$  I'd expect that the solution is normalized:  $$ \int_{-\infty}^\infty |\psi(x,t)|^2 dx = 1. \tag{3}$$  But   $$|\psi(x,t)|^2 = \psi(x,t)\psi^* (x,t) = A^2 + B^2 + AB ( e^{2ikx} + e^{-2ikx} ), \tag{4}$$  and the integral diverges:  $$ \int_{-\infty}^\infty |\psi(x,t)|^2 dx = \frac{AB}{2ik} (e^{2ikx} - e^{-2ikx})\biggl|_{-\infty}^\infty + (A^2+B^2) x\biggl|_{-\infty}^\infty. \tag{5}$$  What is the reason for this? Can it be corrected?
100089
For simple potentials like square wells and harmonic oscillators, one can explicitly calculate the product $\Delta x \Delta p$ for stationary states.  When you do this, it turns out that higher energy levels have higher values of   $\Delta x \Delta p$.    Is this true for all time-independent potentials?  Certainly, it is possible to find two states $\mid \Psi_1 \rangle$ and $\mid \Psi_2 \rangle$ with $\langle \Psi_1 \mid H \mid \Psi_1 \rangle &gt; \langle \Psi_2 \mid H \mid \Psi_2 \rangle$ and also $\Delta x_1 \Delta p_1 &lt; \Delta x_2 \Delta p_2$.  For example, choose a quadratic potential, let $\mid \Psi_2 \rangle$ be the first state and let $\mid \Psi_1 \rangle$ be a Gaussian coherent state (thus with minimum uncertainty) and fairly high energy.  So I'm asking here just about the stationary states.  As Ron pointed out in the comments, this question is most interesting if we consider potentials with only a single local minimum, and increasing potential to the right of it and decreasing to the left.
56020
This question is about non-normalisable states in quantum mechanics, e.g. the eigenstates of the position operator $|x\rangle$ which are defined by the eigenvalue equation  $$\tag{1} \hat{X}|x\rangle = x|x\rangle.$$  Since the spectrum of the operator is continuous, the states are not square normalisable. Instead one has an equation such as  $$\tag{2} \langle x'|x\rangle = \delta(x-x')$$  which gives the overlap meaning as a . However it is clear that there is still at least an ambiguity in choosing a constant, i.e. why no have  $$\langle x'|x\rangle = a \times \delta(x-x')$$  where $a$ is $a \in \mathbb{R} $.    So I went to see how Equation (2) is derived. In Schleich's textbook on quantum optics in phase space I found the following, very short and in my opinion not clear/complete derivation.  From Equation (1) which can be taken to be the definition of the position operator and the assumption that the operator is hermitian which is a fundamental postulate of quantum mechanics one obtains:  $$\tag{3} \left( x - x' \right) \langle x'|x\rangle = 0.$$  That was easy so far, but then Schleich jumps immediately to equation (2). But the solutions for $\langle x'|x\rangle$ of (3) are all functions/generalised functions that are non-zero everywhere except at $x=x'$. So my question is: Why is (2) the unique solution? Where does the additional information come from?  My suspicion is that we impose another requirement, which is probably the completeness of the states, which would make (2) clear since it is the identity element in the position representation.
58203
I stumbled on this question while answering the following question:  Given a general wavefunction $\psi(x)$ and an operator $\hat Q$, with eigenvalue spectrum $\{q_r\}$ and corresponding eigenfunctions $\{u_r\left(x\right)\}$, show that the expectation value of $\hat Q$ is given by $$\langle\hat Q\rangle = \int_{-\infty}^\infty{\psi^*\hat Q \psi \,dx}$$  I started of with the ususal definition of the expected value and inserted the identity operator in position representation: $$\langle\hat Q\rangle = \langle \psi|\hat Q|\psi\rangle = \langle \psi|\left(\int dx|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\right)|\psi\rangle$$  Taking $\langle\psi|$ and $|\psi\rangle$ into the integral gives: $$\left(\int dx\langle \psi|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\psi\rangle\right)$$  Pulling the 2nd integral into the first:$$\int dx\int  {dx}'\langle \psi|x\rangle\langle x'|\psi\rangle\langle x| \hat Q|x'\rangle$$  This is the point where I'm stuck. If $\langle x|\hat Q|x'\rangle = \hat Q \delta(x-x')$, then the last expression would evaluate to the answer. However I  don't know why this should be true and even if it is true I wouldn't know how to prove this equality.
100771
The wavefunction $\Psi(x,t)$ for a free particle is given by  $$\Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)}$$  This wavefunction is non-normalisable. Does this mean that free particles do not exist in nature?  Why then do we use free particles $\psi(\vec{x}) = e^{ikz}$, for example, in scattering theory?
25116
The question is to show that for any $x,y \in\Re^n$  $||x+y||\cdot||x-y||\le||x||^2+||y||^2$  with equality if and only if $\langle x,y \rangle= 0.$  I have that $||x+y||^2=||x||^2+2 \langle x,y \rangle+||y||^2$ , which helps me prove equality but I'm lost on how to show this and show that the inequality part.
27861
Hm, this just occurred to me while answering another question:   If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of $\delta$-functions  $\psi_x = \delta(x)$, and that indicates that the size of my Hilbert space is that of $\mathbb{R}$.  On the other hand, we all know that we can diagonalize $H$ by going to the occupation number states, so the Hilbert space would be $|n\rangle, n \in \mathbb{N}_0$, so now the size of my Hilbert space is that of $\mathbb{N}$ instead.  Clearly they can't both be right, so where is the flaw in my logic?
61693
Let me remind you about the following classical  examples in quantum mechanics.     Example 1. Bound states in 1-dim potential V(x).   Let $V(x)$ be a symmetric potential i.e.   $$V(x) = V(-x)$$   Let us introduce the parity operator $\hat\Pi$ in the following way:   $$\hat\Pi f(x) = f(-x).$$   It is obvious that    $$[\hat H,\hat\Pi] = 0.$$   Therefore, for any eigenfunction of $\hat H$ we have:   $$\hat H|\psi_E(x)\rangle = E|\psi_E(x)\rangle = E\hat\Pi|\psi_E(x)\rangle,$$   i.e. state $\hat\Pi|\psi_E(x)\rangle$ is eigenfunction with the same eigenvalue. Is $E$ a degenerate level? No, because of linear dependence of $|\psi\rangle$ and $\hat\Pi|\psi\rangle.$   Consider the second example.     Example 2. Bound states in 3-dim a potential $V(r)$. Where $V(r)$ possesses central symmetry, i.e. depends only on distance to center.   In that potential we can choose eigenfunction of     angular momentum $\hat L^2$ for basis    $$|l,m\rangle,$$   where $l$ is total angular momentum and $m$ - its projection on chosen axis (usually $z$). Because of isotropy eigenfunction with different $m$ but the same $l$ correspond to one energy level and linearly independent. Therefore, $E_l$ is a degenerate level.      My question is if there is some connection between symmetries and degeneracy of energy levels. Two cases are possible at the first sight:   Existence of symmetry $\Rightarrow$ Existence of degeneracy Existence of degeneracy $\Rightarrow$ Existence of symmetry     It seems like the first case is not always fulfilled as shown in the first example. I think case 1 may be fulfilled if there is continuous symmetry. I think the second case is always true.
46558
A typical problem in quantum mechanics is to calculate the spectrum that corresponds to a given potential.    Is there a one to one correspondence between the potential and its spectrum?  If the answer to the previous question is yes, then given the spectrum, is there a systematic way to calculate the corresponding potential?
17594
In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being   $$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$  (cf. last formula on p. 101). He then says that these eigenfunctions are not square integrable because   $$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx ~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx ~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right) ~\rightarrow~\infty$$  (cf. second formula on p. 102). My question is, how does he arrive at the final term, more specifically, where does the $\delta\left(\lambda-\lambda\right)$ bit come from?   My total knowledge of the Dirac delta function was gleaned earlier on in Griffiths and extends to just about understanding   $$\tag{2.95}\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right)$$  (cf. second formula on p. 53).  References:   D.J. Griffiths, Introduction to Quantum Mechanics, (1995) p. 101-102.
21409
Given some potential $V$, we have the eigenvalue problem   $$ -\frac{\hbar^2}{2m}\Delta \psi + V\psi = E\psi $$   with the boundary condition   $$ \lim_{|x|\rightarrow \infty} \psi(x) = 0 $$   If we wish to seek solutions for   $$ E &lt; \lim_{|x|\rightarrow \infty}V(x), $$   I find textbooks like Sakurai and others assert that (without proof) set of all such eigenvalues $E$ that admits nontivial solutions is discrete, which gives us liberty to index them with natural numbers (principal quantum numbers). Can anyone help me with a little elaboration of this mathematical fact ? As a student of mathematics, I know that compact operators have discrete spectrums, so for the operator $ T : L^2 \rightarrow L^2 $given as  $$ T(\psi) = -\frac{\hbar^2}{2m}\Delta \psi + V\psi. $$   Are there methods to show that $T$ is compact? Or possibly other ways to establish discreteness of its eigenspectrum?
100859
Motivation: In odd dimensions, solutions to the wave equation: $$u_{tt}(x,t)=\nabla u(x,t), \qquad u_t(x,0)=0, \qquad u(x,0)=f(x)$$ where $t \geq 0$ and $x \in \mathbb{R}^n$, have the nice property that the value of $u(x,t)$ only depends on the values $f(y)$ with $|y-x|=t$. For even dimensions, the value $u(x,t)$ depend on all the values $f(y)$ with $|y-x|\leq t$. A consequence of this is that when you switch a light bulb on and then off (in 3D), there will be a light wave traveling with the speed on light and behind the wave, there will be total darkness. But when you throw a rock into a pond (with a 2D surface), there will be lots of waves traveling outwards from where the rock hit the water and, in theory, the water will never a still again.  Question: Can anyone give an intuitive explanation of this difference between odd and even dimensions?
57696
I am having trouble proving this directly. I know that it is easy to prove by contradiction by assuming both $a$ and $b$ to be odd, but how should I start to try to prove this? This is a homework problem and I am just looking for help getting started, not the full answer.
3628
I am reading and following along the appendices of , and trying to learn how he derives Schrödinger's Equation from his Matrix Mechanics, but I have run into a bit of trouble. It seems like for his derivation to work, it must be necessary for the integral of a function times the Dirac Delta Function's derivative be: $$\int^{\infty}_{-\infty}f(\xi)\delta'(a-\xi)d\xi=f'(a). \tag{36}$$ But the actual identity is $$\int^{\infty}_{-\infty}f(\xi)\delta'(\xi-a)d\xi=-f'(a).$$ Does anybody care to explain why it is like this in Heisenberg's book, or provide a derivation along the same vein, but with the correct identity for the delta function?