In  (on Scattering states and the step potential) by Dr. Barton Zwiebach, at $2:28$, the professor says:  &quot;The energy of any energy eigenstate here has to be bigger than the lowest point of the potential...&quot;  The reason provided seems a little vague - I am unable to understand how the above statement is being concluded. Could someone kindly help?
50221
Suppose we have a particle of mass $m$ that is in an eigenstate $|\psi\rangle$ of the Hamiltonian $\hat{H}=\hat{T}+\hat{V}$, where $\hat{T}$ is the kinetic energy operator and $\hat{V}=V(\hat{r})$ is the potential energy operator. If the potential has lower bound $V_0$, then is it necessary for the energy eigenvalue $E$ of $|\psi\rangle$ to be greater than $V_0$? Classically this is true, since we regard negative kinetic energy as physically unrealizable/meaningless. However, I don't know if I can necessarily say the same in the quantum case. For example, what I'm tempted to do is write  $$\langle\psi | \hat{T}|\psi \rangle = \langle\psi | E-\hat{V}|\psi\rangle$$  and say "If $E&lt;V_0$, then the RHS is necessarily negative, implying the LHS is as well, which we will regard as physically meaningless. If $E=V_0$, then $|\psi\rangle$ is trivial.", but I'm not sure if that's right.   Confused by that, I then wanted to show that, if $E&lt; V_0$, then a non-trivial $|\psi\rangle$ is non-normalizable. However, I'm not entirely sure how to do this.
91422
I am learning about various signal processing methods in my university course, and I can't seem to grasp what 'energy' in signals represent. I mean, I know that it is the integral of the absolute value of energy squared (side question: why is absolute value needed?), but I don't know the meaning of what this energy is. Is there an intuitive way to think about these energies? Thanks a lot.
101913
I was wondering if the good old quadratic potential was the only potential with equally spaced eigenvalues. Obviously you can construct others, such as a potential that is infinite in some places and quadratic in others, but that's only trivially different. I am not referring to equally spaced as a limiting behavior either, I mean truly integer spaced.  Any ideas? If not, is there a proof for its uniqueness?  If there are other potentials with equally spaced eigenvalues, can one use them as starting points for a free-field QFT? It would be interesting to know if there is a deeper mathematical relation between all of these potentials and whether they could be used to study interacting systems.
86342
I am studying Perturbation Theory from J.J. Sakurai's textbook Modern Quantum Mechanics. I am having trouble proving formulas on page 299 (5.2.5) and (5.2.6) from the previous ones [mainly (5.2.4)]. Can anyone help?  This is a problem of working with perturbation theory when the non-perturbed state has degeneracy. To elaborate I do not get why $(E-H_0-\lambda P_1VP_1)$ is not singular.  More importantly is the equation below correct and if so why:  $$(E-\lambda P_1VP_1)|l \rangle = E_D^{(0)}|l \rangle $$  That is the core of my problem. If the above equation - in the framework of Sakurai's statement of the problem - is correct I can work out all the rest on my own. However I do not get why this equation would hold true.
99922
In Griffith's QM book, he introduces scattering matrices as an end-of-the-chapter Problem 2.52.  For a Dirac-Delta potential $V(x) = \alpha \delta (x - x_0)$, I've derived the scattering matrix and observed that it is unitary $S^{-1} = S^{\dagger}$.  I'm trying to explain why this is intuitively, but I don't really have an intuitive picture of what hermitian conjugation $S^{\dagger}$ is doing here. Thoughts?
30946
I understand that if a particle approaches a finite potential barrier of height $V_0$ with energy $E &lt; V_0$, there is still a finite probability of finding the particle on the other side of the barrier due to quantum tunneling.  My question is, since the wavefunction is nonzero inside the barrier region, is it possible to actually make a position measurement and locate a particle inside the barrier?  I mean, if we can say that "there is a nonzero probability that the particle is inside the barrier", surely this would suggest we can do so?  If not, why not? Am I understanding the whole wavefunction/probability distribution thing right?
53447
My apologies if this question has already been answered. Using the notation of Sakurai, we have an energy eigenket N and an eigenvalue n.  $$N | n \rangle = n | n \rangle$$  For the number operator $N = a^\dagger a $. My question is precisely why does n have to be an integer. I do understand that it must be positive definite due to the norm condition on $ a | n \rangle $:  $$\langle n | a^\dagger a | n \rangle \geq 0$$  I can see that we can show:  $$ a | n \rangle = \sqrt{n-1} | n-1 \rangle $$ and similarly for $a^\dagger | n \rangle$  Sakurai argues that the sequential operation of $a$ operators leads to the form which he does not write, but I will write here for clarity:  $$ a^k | n \rangle = \left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right) | n - k + 1 \rangle $$   Since  $$ a|0 \rangle = 0 $$  the sequence must terminate, but can only terminate if n is an integer. How can we make that jump? Is there a formal mathematical proof that the sequence $\left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right)$ will terminate if and only if n is an integer?
50158
Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$. The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle?
66530
Let $\langle\Omega|$ be the ground state of an interacting theory, just as Peskin &amp; Schroeder(PS) describes on page 82 and page 213. On page 213 PS do the following   $$\tag{1}\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle = \langle\Omega|\exp(+\mathrm{i}Px)\phi(0)\exp(-\mathrm{i}Px)|\lambda_{\vec{p}}\rangle \\ =\langle\Omega|\phi(0)|\lambda_{0}\rangle\exp(-\mathrm{i}px)\bigg|_{p^0=E_{\vec{p}}} $$  In the above $|\lambda_0\rangle$ is an eigenstate of the full interacting hamiltonian $H$. The ket $|\lambda_{\vec{p}}\rangle$ is the boost of $|\lambda_0\rangle$ with momentum $\vec{p}.$ Furthermore $E_{\vec{p}} = \sqrt{p^2+m_\lambda^2}$ and $\vec{P}|\lambda_0\rangle = 0.$  I assume PS used $$\tag{2}\langle\Omega|\mathrm{e}^{\mathrm{i}Px}=\langle\Omega|\mathrm{e}^{\mathrm{i}0x}$$ in $(1)$.   Is $(2)$ true?   I thought $\langle0|\mathrm{e}^{\mathrm{i}Px}=\langle0|\mathrm{e}^{\mathrm{i}0x}$ where $\langle0|$ is the ground state of the free theory.   I hope I have provided enough info for the question to make sense.
78529
what do it mean by "A real nxn matrix always has a real eigenvalue when n is odd?" Shouldnt it be that regardless of odd or even there will always have a real root?  hope to hear from you guys soon:)
30841
So I'm going through my notes and I think I've confused myself. We often imply  $$ |\psi\rangle \to \psi(x)\\ \langle\psi| \to \psi(x)^* $$ for instance when we talk about eigenvalue equations we interpret  $$ \hat{H}|\psi\rangle =E|\psi\rangle $$ as simply  $$ \hat{H}\psi(x)=E\psi(x) $$ but I don't understand why we say $|\psi\rangle \to \psi(x)$ because if that was the case then  $$ \langle\psi|\psi\rangle=|\psi|^2 $$ when clearly $$ \langle\psi|\psi\rangle=\int_{-\infty}^\infty |\psi|^2 dx = 1 $$ I'm obviously missing something simple, could anyone point out where I'm going wrong?
49357
I'd like some suggestions for good reading materials on the Keldysh formalism in the condensed matter physics community. I'm familiar with the imaginary time, coherent state, and path integral formalisms, but lately, I've been seeing Keldysh more and more in papers. My understanding is that it is superior to the imaginary time formalism at least in that one can evaluate non-equilibrium expectations.
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?
117186
The $\delta(x)$ Dirac delta is not the only "point-supported" potential that we can integrate; in principle all their derivatives $\delta', \delta'', ...$ exist also, do they?  If yes, can we look for bound states in any of these $\delta'^{(n)}(x)$ potentials? Are there explicit formulae for them (and for the scattering states)?  To be more precise, I am asking for explicit solutions of the 1D Schroedinger equation with point potential,  $$- {\hbar^2 \over 2m} \Psi_n''(x) + a \  \delta'^{(n)}(x) \Psi(x) \ = E_n \Psi_n(x) $$  I should add that I have read at least of three set of boundary conditions that are said to be particular solutions:   $\Psi'(0^+)-\Psi'(0^-)= A \Psi(0)$ with $\Psi(0)$ continuous, is the zero-th derivative case, the "delta potential". $\Psi(0^+)-\Psi(0^-)= B \Psi'(0)$ with $\Psi'(0)$ continuous, was called "the delta prime potential" by Holden. $\lambda \Psi'(0^+)=\Psi'(0^-)$ and $\Psi(0^+)=\lambda\Psi(0^-)$ simultaneusly, was called "the delta prime potential" by Kurasov   The zero-th derivative case, $V(x)=a \delta(x)$ is a typical textbook example, pretty nice because it only has a bound state, for negative $a$, and it acts as a kind of barrier for positive $a$. So it is interesting to ask for other values of $n$ and of course for the general case and if it offers more bound states or other properties. Is it even possible to consider $n$ beyond the first derivative?  Related questions  (If you recall a related question, feel free to suggest it in the comments, or even to edit directly if you have privileges for it)  For the delta prime, including velocity-dependent potentials, the question has been asked in   In the halfline $r&gt;0$, the delta is called "Fermi Pseudopotential". As of today I can not see questions about it, but  seems to be the same potential.  A general way of taking with boundaring conditions is via the theory of self-adjoint extensions of hermitian operators. This case is not very different of the "particle in 1D box", question  A general take was the question  A related but very exotic question is  because obviosly the point-supported interacctions have a peculiar scaling  Comments  Of course upgrading distributions to look as operators in $L^2$ is delicate, and it goes worse for derivatives of distributions when you consider its evaluation $&lt;\phi | \rho(x) \psi&gt;$. Consider the case $\rho(x) = \delta'(x) = \delta(x) {d\over dx}$. Should the derivative apply to $\psi$ only, or to the product $\phi^*\psi$?
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?
81219
Consider a particle in a potential well. Let’s assume it’s a simple harmonic oscillator potential and the particle is in its ground state with energy E0 = (1/2) &#8463;&omega;0. We measure its position (measurement-1) with a high degree of accuracy which localises the particle, corresponding to a superposition of momentum (and therefore energy) states.  Now we measure the particle’s energy (measurement-2) and happen to find that it’s E10 = (21/2) &#8463;&omega;0. Where did the extra energy come from?  In the textbooks it’s claimed that the extra energy comes from the act of observation but I wonder how that could work. Measurement-1 which probed the position of the particle can’t have delivered to it a precise amount of energy, while measurement-2 might just have been passive. No doubt there is entanglement here between the particle state and the measuring device but where, and which measurement?
13181
Just beginning to learn quantum mechanics and my textbook says that all pure eigenstates of the Schrodinger equation ( with a specified potential function) have a zero standard deviation for the expectation values of energy.So every measurement of the energy yields a fixed value. I've also read that pure eigenstates are physically realisable solutions in some situations ( like the infinite potential well). But wouldn't that violate the uncertainty principle? the standard deviation in expectation of energy is zero and hence $$ΔE~Δt=0.$$ Where am I going wrong? Also could someone also give me an insight as to what uncertainty in time even means? I haven't been able to grasp that concept clearly yet.