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0

No, you can't simply substitute. What you can do is change $x$ with a linear shift: $$x \rightarrow x+\frac{a}{2}.$$ But if this is homework, you'll need to show your instructor WHY this is the proper way to handle the situation. Or you could solve the problem with the new boundary conditions, but be aware that you have two solutions which are possible. ...


1

$\newcommand{\d}{\;\mathrm{d}}$ Elaborating on the answer of zeldredge, I want to say why the following expression works: $$P(E_0) = \left| \int_{-\infty}^{\infty} \ \Phi^* \psi \d x \right|^2 \tag{1}$$ Notice that the eigenfunctions of the Hamiltonian spans the space. That is you can write any function $\psi$ as a linear combination of eigenfunctions ...


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First you have to find the eigenstate of the Hamiltonian, in other words, you have to solve the equation: $$\hat H \psi_{E}(x)=E \psi_{E}(x),$$ then take the inner product of $$\psi_{E}(x)\quad and \quad \psi(x)$$,it is the probability you want.


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You have to know the energy eigenstates. Suppose you know that $\Phi (x)$ is an energy eigenstate with $E = E_0$. Then the probability that your state $\psi(x)$ will be measured to have energy $E_0$ is: $$ P(E_0) = \int_{-\infty}^{\infty} \mathrm{d} x \ \Phi^*(x) \psi(x) $$ You can do this for the full set of eigenstates $\Phi_i$ to get the wavefunction in ...


3

There is no need, nor mathematically nor physically, for the wavefunction or its (spatial) derivative to be continuous. In fact, the space of wavefunctions is usually considered to be an Hilbert space (and there are very poignant physical and mathematical motivations, the first is that any algebra of physical observable that satisfies reasonable assumptions ...


2

For the time-dependent Schrödinger equation there might be one. The spatial derivative of the wave function is connected to a "flow of probability" associated with the squared absolute value of the wave function which gives a probability density. You can view this probability density as a fluid with mass conservation (probability conservation). A ...


-3

It started out smooth, and it can't go from smooth to not smooth.


0

I'd say that a good physical reason for the first derivative being continuous is for the probability current to be continuous too, since it's constructed from $\psi(x,t)$ and $ \nabla \psi(x,t)$.


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To find the solution to your problem: Area 1: U1,E>U1 The Schrodinger equation and it's solution is: $$ψ'' + {2m \over \hbar ^2}(E-U1)ψ=0 ---> ψ_A=e^{ik_1 x} + c_1e^{-i k_1 x} $$ where $k_1 ^2 = {2m \over \hbar ^2}(E - U1) $. Area 2: U2, E As in the above: $$ψ'' - {2m \over \hbar ^2}(U2-E)ψ=0--> Ψ_Β = c_2 e^{k_2 x} + c_3 e^{-k_2 x} $$ where $k_2 ^2 ...


2

The solution you give is unphysical, because it cannot be normalized (as you noticed). This happens a lot in physics (e.g. try solving the wave equation in cylindrical coordinates: you'll get Bessel functions that rise to infinity at the origin, which you'll also discard if the origin is part of the solution domain). Unphysical solutions are discarded, and ...


1

First off, there are some very misleading answers given above. Introductory quantum courses fail to properly discuss "time." It is a parameter, not an observable. E(operator)=ih(bar) d/dt has no meaning. That operator simply discribes the time evolution of a wavefuntion that is complex. So it does not describe a physical observable. I know this might be hard ...


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It is easiest if you consider a rectangular box in 3D, with length $L_x$ in the $x-y$ plane, and length $L_z \ll L_x$ in the $z$ direction. Now, $L_x$ is very large, and you are only interested in the physics deep within the bulk, far from the boundaries. Therefore it is of little importance exactly which boundary conditions you choose. It is often easier to ...


2

A state $\psi$ corresponds to an energy $E$ if: $$H\psi = E\psi$$ Clearly, if there is a state $\psi = \sum_i c_i \psi_i$ where $H\psi_i = E_0\psi_i\ \forall i$, then $$H\psi = \sum_i c_i H\psi_i = \sum_i c_i E_0\psi_i = E_0\psi$$ A linear combination of states with the same energy value again has the same energy value. Now consider $\psi = c_1\psi_1 + ...


3

I) Well, $r=0$ is the boundary of a $d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but $r=0$ does not correspond to a physical boundary per se, other than what we can deduce from the TISE. The mantra is that a boundary condition (for finite $r$) should only be imposed if it is a consequence of the TISE. See ...


0

Assuming you've done the algebra correctly, these equations can be solved for a relationship between $k$ and $K$, which should lead to the quantization of energy levels in terms of $a$, $b$, and $V_o$. Then you solve for $C$ in terms of $A$ from either equation (you MUST get the same result with either) and then normalize.


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The wave function must be symmetrical because you aren't "allowed" to know which of the two bosons is in which level. If so, you were capable of distinguish one particle from the other. Indeed the problem is not that you can't know in which level they are, they aren't just in any precise level, taken alone. It's they're superposition which has a precise ...


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Observe that if $\psi(x_1,x_2)$ is not symmetric, $\psi(x_1,x_2) + \psi(x_2,x_1)$ certainly is. This corresponds to us really not knowing whether it is the "first" particle in the ground state or the "second" particle, since they're indistinguishable.


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First I think that the hamiltonian is $\hat{H}=\frac{1}{2m}\left(p+\frac{e}{c}A\right)^2$, not $\hat{H}=\frac{1}{2m}\left(p+\frac{e}{c}A\right)$. Then the remaining is just calculation. Using $[\hat{p},A]=0$, expanding hamiltonian, we get $$\hat{H}=\frac{1}{2m}(-\hbar^2 \frac{\partial^2}{\partial x^2}+\frac{e^2B^2}{c^2}x^2+\frac{2 \hbar e ...



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