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Following von Neumann, the measurement process is just a special type of interaction between two systems, that follows special rules when it comes to averaging a specific observable $X$. Let $H$ be a Hilbert space, $(\Omega,\mathscr{B})$ a Borel space, with $\Omega\subseteq \mathbb{R}$ and $\mathscr{B}$ a Borel $\sigma$-algebra on $\Omega$. By means of the ...

4

It is your Eq 1 and 2 that are mathematically inconsistent -- just take complex conjugate on both sides of Eq 1 and you will see. I think you confused a state being an element of the Hilbert space and a state satisfying the schrodinger equation. I can write down plenty of elements of the Hilbert space that does not satisfy the Schrodinger equation, for ...

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Is it true that if $u_n(x)$ and $u_m(x)$ are orthogonal (which is true), then $u_n(x)$ and $u_m(x)f(x)$ will be also orthogonal? No. The simplest example of this is the case $f(x) = u_n(x)/u_m(x)$ for whatever $n$ and $m$ you're considering. More broadly, the result you're trying to prove is false. Consider the infinite square well between $\pm ... 3 Consider first the particle in a 1D infinite potential well: The probability of finding the particle must be zero where the potential is infinite, so the wavefunction$\Psi$must be zero at the edges of the box.$\Psi$is non-zero somewhere inside the box, so it must have a form something like the red line. I've just drawn a random squiggle for the red ... 2 The wavefunction: $$\Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)}$$ is an infinite plane wave. So it describes a particle that has an infinite extent in both time and space. That is, it exists for$-\infty \le x \le \infty$and for$-\infty \le t \le \infty$. Unsurprisingly, if the particle has an infinite extent then it's amplitude is everywhere zero and ... 2 This is a straightforward substitution. In momentum space the position operator is: $$\hat{x} = i \hbar \frac{d}{dp}$$ 2 First note the following: If$\psi(x,0)$is any normalizable wave-function in your Hilbert space, then$\psi(x,t) = e^{-iHt} \psi(x,0)$(I've set$\hbar = 1$) satisfies the time-dependent Schrodinger equation (we call this "time evolution"). This can be seen by using the fact that the stationary states form a basis for your Hilbert space and so$\psi(x,0)$... 2 We prove this by a reductio ad absurdum. We start by assuming that the wavefunction of a non-degenerate ground state is complex, then show this means the wavefunction must be degenerate. Suppose we have a complex ground state. Then we can write it as a sum of real and imagniary parts: $$\psi = \psi_r + i\psi_i \tag{1}$$ The ground state obeys ... 2 The "independent" in "time-independent Schrödinger equation" doesn't mean that the wavefunction$\psi(x,t)$is independent of time, but that the quantum state it defines doesn't change with time. Since$\psi(x)$and$\mathrm{e}^{\mathrm{i}\phi}\psi(x)$for any$\phi\in\mathbb{R}$define the same quantum state, this does not imply$\partial_t\psi(x,t) = 0$. ... 2 There's a meaningful difference between the boundary conditions. For the infinite square well, the probability at the boundaries is required to be 0. This limits us to sines as opposed to cosines (assuming the standard convention of one boundary is at position 0, and there is no nonzero phase in our solutions). Since$\sin 0 = 0$,$\sin \left ( \frac{n \pi ...

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The situation here is a bit complicated. The problem is that with a measurement process, you are not allowed to consider only the evolution of the measured system, but you have to take into account the evolution of the global system formed by the measured system and measurement apparatus. For that system, the (global) wavefunction always evolves by means of ...

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In the Heisenberg picture, the correct description of a dissipative process (of which the collapse is just the the simplest model) is through a quantum stochastic process. There is an extended literature on this. Properly designed, these processes preserve the commutation relations between key observables during the time evolution, which is an essential ...

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In the case of periodic boundary conditions it is necessary to ensure the continuity of the solution not only in value on the boundary, $\Psi(0) = \Psi(L)$, but also in derivative, $\partial_x \Psi(0) = \partial_x \Psi(L)$. This is simply to rule out kinks and ensure that the solution is translationally invariant. The derivative condition as applied to  ...

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First, you are not talking about a phase space, but configuration space. Now, the space of wavefunctions of a single particle in 3D space is the space of Lebesgue square-integrable functions $L^2(\mathbb{R}^3)$ on the configuration space $\mathbb{R}^3$. So already for a single particle, there is no generic requirement for the wavefunction to be smooth, ...

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You should also match the derivatives at $x=0$ so that they took into account the $\delta$-function. If you take smooth well $V_\epsilon(x)$ and consider small region near zero $(-\epsilon,\epsilon)$ where the "meat" of the well is concentrated you may then integrate the Schrodinger equation at that region, ...

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