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4

The LHS is an inner product while the RHS is the evaluation of a function from an $L^2$ space at the point $x$. To somehow link the two you need to be able to write the RHS as an integral, so you need a "function" $\delta_x$ such that $$\langle x|\psi\rangle = \int\overline{\delta_x(s)}\psi(s)\text ds = \psi(x).$$ There is no such function, but the map ...


3

There is a lot of people that do not like the name second quantization, mostly because the second quantization is often introduced in a not so clear way, at least in my opinion. Second quantization has nothing wrong, if you see it as a precisely defined mathematical object: it is a functor between Hilbert spaces, that associates to the original one-particle ...


3

For that to be true you will have to go beyond the Hilbert space and consider the rigged Hilbert space, that is the Hilbert space plus distributions. Hence this can be made formal if you identify Dirac's bras and kets with the (anti)linear rigged Hilbert space. In more intuitive terms, the delta function $\delta(\mathbf r)$ can be realised as limits of a ...


3

When solving differential equations we often us a technique called ansatz. The word ansatz is German for guess. In other words we guess at a possible solution then feed it back into the differential equation to see if it works. In this case we guess that the solution is $\Psi(x) = e^{ikx}$. If we take the second differential of this we get: $$ ...


3

Probably one of the first things one should do after arriving at a solution to an ODE is stick it back in to see if it satisfies the original equation. In your case, you obtain $$ \left(4k^4x^2\Psi-2k^2\Psi\right)+k^2\Psi\neq0 $$ where the term in the parenthesis is the result of $\frac{d^2}{dx^2}\Psi$. So clearly you went wrong somewhere. I'd wager that it ...


2

$$\frac{\partial}{\partial t}\Psi(x,t) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) e^{i\hbar kx}\left(\frac{\partial}{\partial t}e^{-\frac{1}{2}\frac{\hbar^2k^2}{m}t/\hbar}\right) dk$$ $$\frac{\partial}{\partial x}\Psi(x,t) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) \left(\frac{\partial}{\partial x}e^{i\hbar ...


2

How is this last statement true? On the position basis, the (1-D) position operator is multiplication by $x$. A well known property of the Dirac delta distribution is $$f(x)\delta(x) = f(0)\delta(x)$$ since $\delta(x)$ is zero everywhere except $x=0$ were it has unit area. Thus $$x\delta(x) = 0\delta(x)$$ So, operating on $\delta(x)$ with the ...


2

1) Yes, that's pretty much it - a perfect/ideal wave will extend to infinity. It matters in the sense that a sine wave, $\sin (kx)$ (and similarly a complex plane wave) is only a momentum eigenfunction if it is taken to be infinite in extent, as the domain of sine is all real numbers. 2) If the amplitude was not constant, it would have to vary. If it ...


2

In bra-ket notation, we have $$\hat H |\psi(t)\rangle = i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle $$ The Hermitian conjugate of this equation is $$\langle\psi(t)|\hat H^\dagger = -i\hbar \frac{\partial}{\partial t} \langle\psi(t)|$$ But $\hat H$ is self-adjoint thus $$\langle\psi(t)|\hat H = -i\hbar \frac{\partial}{\partial t} ...


2

I think what you are asking whether the relationship $$ \mathrm{normalizable} \iff \mathrm{continuous}$$ holds, which is utterly wrong! The wave function has to be continuous*. Notwithstanding take $\psi(x)=H(x-1/2)-H(x+1/2)$, where $H(x)$ is the Heaviside step function. $$ \implies \int_{-\infty}^\infty \mathrm{d}x \,\, ||{\psi(x)}||^2 = 1 $$ (Area ...


2

The general solution to the differential equation you have given is $Ae^{ikx}+Be^{-ikx}$, which you can verify by substituting back in. (This is a common method, putting in some parametric form of a potential solution and determining the values of the parameters after substituting the trial solution). A plane wave has the form $\Psi(x,t)=Ae^{i(kx-\omega ...


1

Assuming $\Psi$ is the wave function. The Schrodinger equation biing, $$ \frac{\mathrm{d^{2}\Psi } }{\mathrm{d} x^{_{2}}}+k^{2}\Psi =0 $$ This is a standard solution of differential equations. If fou have done simple harmonic oscillators, it appears there. In particilar, $$\frac{d^2f}{dx^2} = -k^2f$$ $$\frac{d}{dx}\left(\frac{df}{dx} \right) = -k^2f$$ ...


1

How come a free particle (take a free electron) has a representation as a plane wave and not as something with spherical symmetry? Because we specified a single outward momentum vector $\vec{k}$, with a direction to it. This breaks spherical symmetry, but it's not required. We can do spherical waves too, and we often do in scattering theory: $$ ...


1

Solving $$e^{ikL}-e^{-ikL}=0$$ we write: $$e^{ikL}=e^{-ikL}$$ then, dividing both sides of the equation by $e^{-ikL}$ we find that $$\frac{e^{ikL}}{e^{-ikL}}=e^{2ikL}=1$$ From Euler's Formula, $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ the solution is that k is quantized: $k =\frac{n\pi}{L}$ for positive integer n. Thus $$\psi (x)= A(e^{ikx}-e^{-ikx}) = ...


1

So as you say, a planar wave has a constant amplitude. This then implies that they do extend all the way to infinity in both directions--there is no edge to it. Of course, this is never physically realized, but it's a good first (or zeroth) approximation in a lot of cases--just like it is in electromagnetism, if you've dealt with that before. The reason ...


1

A spread of momentum is sometimes used literally (as in, the momentum is definitely bigger than $a$ and definitely less than $b$ so the spread of momentum of is $b-a$), and sometimes it is used colloquially just to say that most of the time the momentum is between an $a$ and a $b$ such that $b-a$ equals your spread of momentum. The reason having a spread of ...


1

This is actually not surprising. Even an electron orbit in the Bohr atom has a zero expectation value for the momentum if you average over whole orbits. Think of what it would mean for there to be a finite expectation value for momentum: you'd see an electron that's escaping from the nucleus! Perhaps what you want to calculate is the root mean square speed ...


1

Paraphrasing WillO in the comments above, $x$ is an element of a vector space, and $\psi$ is an element of the dual space, so it follows immediately that: $\psi(x) = \langle x\ |\ \psi\rangle$. More precisely, "$x$" is a vector in the infinite vector space of positions, whose basis are Dirac deltas, while $\psi$ is a complex function of these vectors, itself ...


1

What you're asking for is impossible. The energy eigenvalues are given by the zeros of the determinant (as you surmised) and cannot be related to the zeros of the airy function. $$Ai(\zeta_a)Bi(\zeta_0)-Ai(\zeta_0)Bi(\zeta_a)=0$$ For the odd solution (and with derivatives on $Ai$ for the even solution). Also be careful, your $z$ is not dimensionless as it ...


1

Indeed, there is no getting around having to solve this problem numerically. Departing from the determinant method originally posted, perhaps an alternative approach here would be to solve the system using only the energy. We start with the full form of our points of interest taking the odd wavefunctions as an example: $$z(0) = ...


1

Your solution is right. What you get verifying it is that $\psi$ is also an eigenfunction of the momentum operator, which means $$\hat p\psi=p\psi,$$ where $\hat p=-i\hbar\nabla$ is momentum operator, and $p$ is its eigenvalue. Now, applying $\hat p$ twice and dividing by $2m$, you can get $$\frac1{2m}\hat p^2\psi=\frac1{2m}p^2\psi,$$ which is just ...



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