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10

What exactly does $\Delta_{r_e}$ mean ? Your wave function isn't a field in space, it is a field on configuration space, i.e. it assigns complex numbers to a configuration. If your electron is at $(x_e,y_e,z_e)$ and your proton is at $(x_p,y_p,z_p)$ then the configuration is the point $(x_e,y_e,z_e,x_p,y_p,z_p)$ in a 6d space. A point in that 6d space ...


9

$\mathfrak{R}e$: real part. $A^*$: complex conjugate of probability amplitude $A$.


5

The index of the Laplacian tells you which of the coordinates it acts on, that is, if you write $r = (r_x,r_y,r_z)^T$ and $R = (R_x,R_y,R_z)^T$ as Cartesian coordinates, then \begin{align} \Delta_r & := \frac{\partial^2}{\partial {r_x}^2} + \frac{\partial^2}{\partial {r_y}^2} + \frac{\partial^2}{\partial {r_z}^2} \\ \Delta_R & := ...


5

Notation $\renewcommand{braket}[2]{\langle #1 | #2 \rangle}$ First let us define the term "1-form": A 1-form is a linear function which takes a single vector as it's argument. That is it. It is no more confusing or complicated than that single statement. Consider a vector $v$. Given any other vector $w$ we can form the inner product $\braket{v}{w}$. ...


5

Ok, I have a solution for $\delta'(x)$ based on a very crude limit. I'm going to neglect factors of $\hbar$, $m$, etc for the sake of eliminating clutter. Let $V_\epsilon(x)=\frac{\delta(x+\epsilon)-\delta(x)}{\epsilon}$. Then $\lim_{\epsilon\rightarrow 0}V_e(x)=\delta'(x)$. We'll solve the Schrodinger equation for finite $\epsilon$ and then take the limit ...


4

Let's simplify things, $\hbar=m=1$ and we put in the $\delta^{(n)}$ as $V(x)=V_0\delta^{(n)}/2$. Then the problem is $$-\psi'' - V_0\delta^{(n)}\psi = E \psi$$ Apart from $x=0$ the equation gives $$\psi'' = -E \psi $$ we want a bound state $E<0$ which falls off at infinity, so we get a solution $\psi_+ = A \exp(-k x)$ for $x>0$ and $\psi_- = A ...


3

The wavefunction contains all the information about the system, but it is not itself observable. Since our intuition tends to be linked to what we have observed in the past, it's not surprising that the wavefunction seems unintuitive. As you say it can be negative, but it can also be a complex number making it even less intuitive (if that's possible!). ...


2

The important thing is the relative sign between the potential and the Laplacian. Otherwise there are two square roots of $-1$ namely $\pm i.$ You could write $j=-i$ and then you have two perfectly good square roots of $-1$ you can use $i$ or you can use $j$. For square roots of positive numbers you get things like $\pm\sqrt 2$ and there is a way to pick ...


2

A wavefunction with negative sign works just like any other wave with negative sign. For example, water waves with negative height cancel out with waves of positive height. You can also make a 'negative' wave on a string by pulling the end down and back up, which will cancel with a positive wave.


1

How to get from $\left(\frac{-\hbar^2}{2m_e} \Delta_{r_e} + \frac{-\hbar^2}{2M_P} \Delta_{r_p} +V(r) \right)\Psi(\vec r_e,\vec r_p) = E \Psi(\vec r_e,\vec r_p)$ to: $\left(\frac{-\hbar^2}{2(m_e+M_p)} \Delta_{_{R}} + \frac{-\hbar^2}{2\mu} \Delta_{r} +V(r) \right)\Psi(\vec r,\vec R) = E \Psi(\vec r,\vec R)$ with: $\vec R=\frac{m_e \overrightarrow{r_e} + ...


1

Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy ...


1

The wavefunction $\psi(x)$ satisfies $$ -\frac{\hbar^2}{2m}\psi'' + V_0\left(\frac{x}{L} - 1\right) \psi = E\psi, \quad 0 \leq x \leq L\\ -\frac{\hbar^2}{2m}\psi'' = E\psi, \quad x > L $$ Since the bound states have $E < 0$ let's introduce $$ k = \frac{\sqrt{-2mE}}{\hbar}\\ \varkappa = \frac{\sqrt{2mV_0}}{\hbar} $$ Then $$ \psi'' - ...


1

In response to your edit: The real part of $(A^*(S)A(T))$ is equal to the real part of $(A(S)A^*(T))$, since they are just complex conjugates of each other. So he could have written either one. Concrete example: let $(A^*(S)A(T))=a+ib$ for some $a,b$. Then $(A(S)A^*(T))=a-ib$, since it's just the complex conjugate. Then $(A^*(S)A(T))+(A(S)A^*(T))=2a$. We ...


1

Probability is the chances of occurrence of an event, say for example the event is to find an electron around the nucleus of an atom. If the atom has an electron around the nucleus then the probability of finding the electron around the nucleus is one(1). but if the electron is to be sort some place away from the vicinity of the nucleus then the probability ...


1

$\Delta_e$ is called the Laplacian operator and it is defined as $$\Delta_e \equiv \frac{\partial^2}{\partial e_x^2} + \frac{\partial^2}{\partial e_y^2}+ \frac{\partial^2}{\partial e_z^2}$$ The Laplacian operator is related to the Del operator $\nabla_e$ $$\nabla_e \equiv \frac{\partial}{\partial e_x}\hat e_x+ \frac{\partial}{\partial e_y}\hat e_y+ ...


1

Suppose you have a skewed coordinate system, with vectors that have components $v^i$ defined relative to some basis vectors $\hat e_i$ such that $\vec v = v^i \hat e_i$ in the Einstein summation convention (we sum over any index which is repeated once above, once below). The meaning of a skewed coordinate system is that $\hat e_i \cdot \hat e_j \ne ...


1

The name "1-form" is only used because there are 2-forms and 3-forms and so on. An equivalent but probably better name for 1-forms in this case is that of "dual vector." Any vector space will have associated with it the dual space consisting of all linear functions that map vectors into scalars. This is all a dual vector is: a thing that linearly maps ...


1

What you've written is only true if $\psi(x)$ is an eigenstate of $H$. For some general $\psi(x)$ that is not an eigenstate (i.e. $H \psi(x) \ne E \psi(x)$), then $\Psi(x, t)$ will be more complicated than just the time independent wavefunction multiplied by a phase factor.


1

The separable solutions are exactly the eigenstates of the Hamiltonian which are exactly the ones where the probability density does not change. However you can have a flux or flow of probability even if the probability doesn't change. This is like electromagnetism where you can have current floe through a wire even if the wire has no change of charge ...


1

Indeed the cosine function is valid for instance for other boundary conditions $$\psi'(0)=\psi'(L)=0$$ Your goal when you look for a set of solutions to the Schroedinger equation is to be able to decompose any general wavefunction as a sum over this set, and to do it consistently your boundary conditions must be the same for all solutions. So it is unlikely ...


1

It’s unrealistic to expect most students to able to derive the energy quantisation of a hydrogen atom on the spot. For hydrogen: $$E_n = \frac{-13.6 }{n^2}\rm eV$$ (ignore the minus sign for your problem). For a particle in a 1 D infinite potential well: $$E_n = \frac{n^2h^2}{8mL^2}$$ Set $n = 1$ to obtain the energies of both ground states. From the ...


1

Unfortunately, that question's answer depends on the theory of quantum gravity that you chose. But roughly here is the situation : If you take a classical spacetime and quantum matter fields ($G_{\mu\nu} = \langle T_{\mu\nu} \rangle_\omega$), also known as semiclassical gravity, then the stress energy tensor varies with the measurement. The standard ...


1

I think it is important to note that the wavefunction of quantum mechanics is not a field like the electric or magnetic field. It assigns complex numbers to configurations. So it is a function where the domain of the function is not space or even spacetime. So there is not a value of the field at a point in spacetime, so there isn't a thing to curve it. ...



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