The time-independent Schrodinger equation for a particle moving along the x-axis is: $$H\psi (x) + V(x) \psi (x) = E \psi(x)$$  I know that the physically acceptable solutions to this equation can only be found for a specific value of E (so there are $n$ solutions for each $E_1$, $E_2$, ... , $E_n$). Why in general (I am aware that this is the case for the infinite square well for example), mathematically, is that the case?
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?
108850
I am new to QM but I don't really understand when to use time dependent or time independent schrodingers equation? For example, for a free particle in a box, how do we know we should use TISE. more specifically, how do you determine if a wave function should be time dependent?
78523
For $a$ being positive what are the quantisation conditions for an exponential potential?  $$ - \frac{d^{2}}{dx^{2}}y(x)+ ae^{|x|}y(x)=E_{n}y(x) $$ with boundary conditions  $$ y(0)=0=y(\infty) $$ I believe that the energies $ E_{n} $ will be positive and real   I have read a similar paper:     P. Amore, F. M. Fernández. Accurate calculation of the complex eigenvalues of the Schrödinger equation with an exponential potential. ,  [math-ph].   However, I get this strange quantisation condition          $$ J_{2i\sqrt{E_{n}}}(\sqrt{-a})=0 $$  How can I handle this in the case $ a &gt;0 $?
8845
Srednicki's Quantum Field Theory mentions the following at the end of the unit on path integrals in non-relativistic quantum mechanics:   Assume that the total Hamiltonian is of the form,     $$ H = H_0 + H_1, $$   where $H_0$ is the exactly solvable part and $H_1$ is treated as a perturbation.  Without the presence of the perturbation, the ground-state t0 ground-state transition amplitude in the presence of external forces/classical sources is:     $$ \langle 0 | 0 \rangle_{f,h} = \int \mathcal{D}p \; \mathcal{D}q \; \exp \Bigg [i \int_{-\infty}^{\infty} dt \; (p\dot{q} - (1- i\epsilon)H_{0} + fq + h p \Bigg ]. \tag{6.21} $$   In the presence of the perurbation, the amplitude is now given by:    Questions:  1) To get equation $(6.22)$, how can we take the exponential inside the integral, make the necessary manipulations and then take the exponential outside the integral. This seems very ad-hoc to me. How and when do we know that the these steps are valid? Doesn't this required justification from mathematical analysis?  2) More importantly, I don't get the first term involving the perturbing Hamiltonian. Why do we have functional derivatives in its argument? How did it come out to be? Why don't we similarly have functional derivatives in the argument of $H_0$ as well? What's the distinction?  Note: The factor of $(1 - i \epsilon)$ has been suppressed in $(6.22)$.
116863
I am trying to solve the wave equation:  $u_{tt}$  = $u_{xx}$   With initial values:  $u(x,0) =\begin{cases} x^3 - x, &amp;\text{for }|x|\le 1,\ \\0, &amp;\text{for }|x|\ge1\end{cases}$  $u_t(x,0) =\begin{cases} 1 - x^2, \text{for } |x|\le 1, \\ 0, \text{for } |x|\ge1    \end{cases}$  In the domain $D= \{(x,t) | - \infty &lt; x&lt;\infty, \space\space t&gt;0\}  $  I am using D'Alembert's Formula and after inputting the values in I have:  $u(x,t) = \frac{1}{2}[\frac{2}{3}(x+t)^3 + \frac{4}{3}(x-t)^3  - 2(x-t)] $  I am not sure where to go from here.  I think because the initial conditions are not differentiable at some points  then there will be points in the solution that are not differentiable   So I'd need to consider the different cases of $|x+t|\le 1$?  I could be going down the complete wrong path so any help is appreciated.  Thanks
97189
I'm having a discussion with someone. I said that it is -even theoretically- impossible to reach $0$ K, because that would imply that all molecules in the substance would stand perfectly still.  He said that this isn't true, because my theory violates energy-time uncertainty principle. He also told me to look up the Schrödinger equation and solve it for an oscillator approximating a molecule. See that it's lowest energy state is still non-zero.  Is he right in saying this and if so, can you explain me a bit better what he is talking about.
80084
This is inspired by Griffiths QM section 2.2, on the infinite square well, which is about how far I've gotten (so, sorry if this is addressed later in the book).  For any given starting wavefunction, you can express it as a sum over the solutions of the time-independent Schrödinger equation. The coefficients in the sum are constant in time, therefore the expectation value of energy is constant in time. The chapter says "this is a manifestation of conservation of energy in quantum mechanics." Ok.  Now in problem 2.5, we have a wavefunction that is an even mixture of the first two stationary states. Part D has us compute the expectation value of momentum, and the solution is sinusoidal in time. So momentum is not conserved.  How can momentum be not conserved? Why is energy conserved but not momentum?
23488
In the particle in a box, harmonic oscillator and in Hydrogen Atom, we can safely assume $$\Psi(x,t) = \psi(x)e^{-i\omega t}.$$ So why not make it a postulate to consider the wave function to be always in the form $$\psi(x)e^{-i\omega t}$$ We can still explain all the foundational experiments, so my question is why not make this amendment to the theory so that we can avoid a lot of nightmares and some peace of mind. Why unnecessarily consider so general thing when we can do away with simple things most of the time?  Oh my actual question is where do we run into problems (practical physical situations/experiments) if we consider such a thing? If we run into problem anywhere, I hope to make any other simple changes to fix them but without going back on this one.
112674
Why can't the Klein-Gordon equation with a Couloumb potential describe the hydrogen atom?  Why can the first order Dirac equation explain it? What are the failures?
55118
I'm reading  and I'm trying to understand the proof under "Degeneracy in One Dimension".  Here's what it says:     Considering a one-dimensional quantum system in a potential $V(x)$ with degenerate states $|\psi_1\rangle$ and $|\psi_2\rangle$ corresponding to the same energy eigenvalue $E$, writing the time-independent Schrödinger equation for the system:       $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_1}{ \partial x^2} + V\psi_1 =E\psi_1$$       $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_2}{ \partial x^2} + V\psi_2 =E\psi_2$$   Multiplying the first equation by $\psi_2$ and the second by $\psi_1$ and subtracting one from the other, we get:       $$\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$$   Integrating both sides       $$\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$$   In case of well-defined and normalizable wave functions, the above constant vanishes, provided both the wave functions vanish at at least one point, and we find: $\psi_1(x)=c\psi_2(x)$ where $c$ is, in general, a complex constant. So, the two degenerate wave functions are not linearly independent. For bound state eigenfunctions, the wave function approaches zero in the limit $x\to\infty$ or $x\to-\infty$, so that the above constant is zero and we have no degeneracy.   Here are the parts I don't understand:   How does integrating $\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$ yield $\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$.  Without having explicit formulas for $\psi_1(x)$ and $\psi_2(x)$, how do we just apparently pull them both out of the integral? I don't understand how the constant in the last equation vanishes just because $\psi_1$ and $\psi_2$ vanish at infinity. Even if it does vanish, how does that imply $\psi_1=c\psi_2$?
47934
Wave equations take the form:  $$\frac{ \partial^2 f}  {\partial t^2} = c^2 \nabla ^2f$$  But the Schroedinger equation takes the form:  $$i \hbar \frac{ \partial f}  {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$  The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation?  And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions?
32722
I'm having trouble finding a simple answer to this question (maybe because there isn't one), but I'm just confused about how the Schrödinger Equation would look for two electrons. I understand that it would exist in 6 dimensional configuration space, but how does the potential look? It's confusing that the potential would be different for each electron depending on where the other one is.
114639
I am reading "The undivided universe: an ontological interpretation of quantum theory" and cannot understand this derivation.  From the Schrödinger equation: $$ i\hbar \frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V\psi $$ They derive the quantum Hamilton-Jacobi equation $$ \frac{\partial S}{\partial t} + \frac{(\nabla S)^2}{2m} + V + Q = 0 $$  using the polar form $\psi = Re^{iS/\hbar}$, but I don't understand what they do afterwards.
95226
If a particle is totally localized at $x=0$, its wave function $\Psi(x,t)$ should be a Dirac delta function $\delta(x)$. Accordingly, its Fourier transform $\Phi(p,t)$ would be a constant for all $p$, thus the particle's momentum is totally uncertain. I guess this is what the Uncertainty Principle told us.  But in another hand, since the particle is totally localized at $x=0$, it is not moving and therefore static. Its velocity should be 0, so is its momentum, which contradicts the conclusion above.  So, what's wrong with my thinking?  Is it that there shouldn't be a TOTALLY STATIC particle, or it's in a non-normalizable state, so the uncertainty principle is not applicable?
63304
Consider the time-dependent  (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{\mathrm T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm  dt^{\prime}\right]\Psi (0), \qquad t&gt;0, $$ where $\hat{\mathrm T}$ is the time-ordering operator.  It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake?
2369
Let's consider an infinite square well. In the first exited state there is a node at the middle of the well (i.e. wave function and thus probability of finding the particle is zero there). If I measure the position right now, I may find the particle on the left side of the node, and if I measure the particle after some time, I may find the particle on the right side of the node. But must  a particle on the left side to go to right side not pass through the node point? Thus the particle will spend some time at that point, so there should be some probability of finding the particle there, but the wave function is zero there, so what's wrong with the arguement? Does the particle move here to there, without passing through the space in between? But that will just violate continuity equation?
93385
There's an imaginary $i$ in the Schrödinger equation, which I guess is to define the position of the particle in a space-time involving a complex function. But what is the real physical significance of $i$ in the equation?