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1

Basically any measurement is on wave function |ψ⟩ is done by operator X such that X|ψ⟩ results observable x with some probability. is not entirely correct. In quantum mechanics a system is described by a state $|\psi\rangle\in\mathcal{H}$ and a set of self-adjoint operators $A_1,\ldots,A_n$ representing the observables that we want to measure. Moreover, ...


-2

When a quantum mechanical object is subjected to a measurement its wavefunction has to be coupled to a measurement device. That coupling will change both the wavefunction of the object and the wavefunction of the measurement device (as well as the wavefunction of the environment!). It is important to understand that a measured quantum object is not the same ...


2

What is a wave function? It is the solution of a quantum mechanical equation ( with the appropriate potentials),on which boundary conditions are imposed to make it specific to a system . $|\psi\rangle$ by itself is not independent of the environment the way that the operators X are. Thus the answer depends on the system under consideration. I like ...


0

All evolution in Quantum Mechanics is done by the Hamiltonian. However, measurements of an operator result in a single eigenvalue, and the wavefunction collapses (As to how, why, etc, it is unknown as of now, Copenhagen is the most common interpretation) to an eigenstate of the operator measured with the same eigenvalue. Specifically, it collapses to the ...


0

There is no need of a potential for the Schrödinger equation to have a solution, namely $$ i\hbar \frac{\partial}{\partial t} |\psi\rangle = H |\psi\rangle $$ does possess solutions even when the Hamiltonian contains no potential. Questions of normalisability may arise, but that is another point (and can anyway be solved by expanding in Fourier terms). ...


0

Yes, a free particle has a wavefunction. If it has sharp momentum $p$, it is given by the plane wave $$ \psi_p(x) = \mathrm{e}^{\mathrm{i}px}$$ which might look a bit strange, because it is non-normalizable/not square-integrable, but that should not be all too troubling - no momentum uncertainty at all is rather unphysical, after all. The general ...


0

With this transformation $\rho = kr$ you can't possibly change the form of the equation, because it's a scale transformation that does nothing special to the derivatives, so I think that you have computed the derivatives wrong.


2

The standard procedure is the following: starting from $ \langle nlm|\,\partial^2_z\,| n'l'm'\rangle $ insert the identity operator with respect to the position basis $$ 1 = \int d\textbf{r} |\textbf{r}\rangle\otimes\langle \textbf{r}| $$ to have $$ \int d\textbf{r}\, \langle nlm\, |\,\partial^2_z\,|\textbf{r}\rangle\cdot\langle \textbf{r}|n'l'm'\rangle. ...


0

To keep things simple, let's talk about plane acoustic waves in one dimension. If we solve the wave equation in one dimension , we find that the acoustic pressure as a function of space and time is of the form $$P(x,t) = Ae^{i(kx -\omega t)}$$ where $A$ is the maximum amplitude, $x$ and $t$ are the displacement and time respectively, $\omega$ is the ...


1

Solving time-dependent SE as danielsmw mentioned above good starting point. $$ i\hbar\frac{\partial\psi}{\partial t}=H\psi $$ $$ \frac{d\psi}{\psi}=\frac{H}{i\hbar}dt $$ $$ log(\psi)\mid^{\psi}_{\psi_{0}}=-\frac{iHt}{\hbar}\mid^{t}_{t_0} $$ suppose $\psi=|\alpha,t>$ and $\psi_{0}=|\alpha, t_{0}>$ $$ \psi=e^{-\frac{iH(t-t_{0})}{\hbar}}\psi_0 $$ Under ...


0

You may consider reading about Aharonov-Bohm effect. This is one of those cases, where the phase of the wave function, in sum with the electromagnetic 4-potential, is extremely important, as it gives different physical results. This effect was also checked experimentally, so it is not a pure theoretical abstraction.


1

You are making confusion between two different things: the first is why the wave function of two identical particles is what it is, the second is the probability of an event happening in several ways (which has nothing to do with physics but is just the definition of unions of probabilities). As per the first question: given $\mathcal{H}_1$ and ...


3

The visualization method you choose is directly and completely determined by the information you need to see regarding your state. For the states of a single bosonic mode, there are multiple different visualization methods, and they all have their pros and cons. In particular, there is a direct trade-off between the amount of information you can display on a ...


0

The question is begging a simple answer: Because nature can be represented as resonators that respond in a delayed way we adopt to study it using complex numbers. When an electronic circuit is excited by a regular periodic source of tension (measured as a real function - $V_0\cos(\omega\cdot\ t)$ for instance) the most common answer is a variation in the ...


2

I don't want to be too much precise, but the Schrödinger equation ($i \dot{\psi}= H\psi$ to avoid confusion) has at most an unique solution under very general assumptions on the Hamiltonian operator $H$, even if you see it as a liner equation in the more general setting of Banach spaces. In particular, it is not a priori necessary that $H$ is self-adjoint ...


3

To understand what these orbitals are, you first have to understand the notion of superposition in quantum mechanics. In regular classical physics, a particle or a system must be in a definite state. A car is at a particular mile marker on a highway, moving at a particular speed. The Moon orbits around the Earth with a particular velocity at a particular ...


0

Yes,both the set of equalities are true but only in the position representation where operator(x)=x.In the momentum representation,where operator(x) takes a different form they are not true


2

The probability of just one of your atoms getting through the wall is very low, multiply that probability (which is way lower than 1) by the probabilty that all of your atoms go through, you get a number so small that the universe would almost certainly have ended long before you got through the wall. A 70 Kg human body has approximately 7 X $10^{27}$ ...


0

According to the diagram you provided, $|a\rangle$ is injected into a beam-splitter, resulting in the out-put state, \begin{equation} |\psi_{o}\rangle =B|a\rangle= e^{\frac{i\pi}{2}}|b\rangle +|c\rangle. \end{equation} Here the beam-splitter unitary is denoted by $B$. I have expressed $i$ as $e^{\frac{i\pi}{2}}$, to highlight the fact that $|b\rangle$ picks ...


0

Generally speaking, bound states in a system give rise to resonant behaviour. The number of states defines the number of different resonances that can be observed. For example, consider the absorption of photons by an electron trapped in a quantum well: If the well only contains one bound state then it could absorb any photon with any energy greater than ...


1

Integration over pure quantum states usually refers to the Haar measure, i.e., the unitarily invariant measure. Vaguely speaking, you assign the same volume (=weight in the integral) to any two set of states which are related by an arbitrary unitary rotation $U$. In the case of one qubit, this is equivalent to integrating over the Bloch sphere; i.e., we ...


1

If I understand the question. Assume you have a quantum oscillator, then if the system is an eigenvector of the hamiltonian, then a series of measurements will give a series of results reflecting eigenvalues of the system. The appearance of these eigenvalues must obey at a number of many measurements the probabilities of each eigenvalue to appear. It's ...


3

Given the wavefunction $\psi(x)$, the probability to find any particle within an interval $[x_0,x_0+\Delta x]$ is $$ P([x_0,x+\Delta x_0]) = \int_{x_0}^{x_0+\Delta x} \lvert\psi(x)\rvert^2\mathrm{d}x$$ i.e. $\lvert\psi(x)\rvert^2$ is not a probability, but a probability density that has to be integrated over a set of non-zero measure to yield a notion of ...


5

The square of the wavefunction, $|\Psi(x,t)|^2$, is the probability density function for finding the particle. This means that the probability of finding the particle in an interval of (infinitesimal) width $\mathrm dx$ at position $x$ equals $|\Psi(x,t)|^2\mathrm dx$. On occasion, however, authors will drop the $\mathrm dx$ if it is convenient and does not ...


0

Gigi Butbala's answer in terms of sines and cosines is perfectly correct. But just to prove that it can be done, here's how you'd proceed from your two equations in terms of $A$ and $B$: The first equation shows that you must have $$ B = - e^{-ikL} A $$ and plugging this in to the second equation yields $$ A e^{ikL/2} - A e^{-3 ikL/2} = 0 \quad ...


0

You will have to settle for momentum since speed is not a quantum mechanical observable (because $\dot{x}$ is not a classical observable on the phase space, which are functions of $x$ and $p$, but a function of a classical trajectory $(x(t),p(t))$, so canonical quantization does not produce a "speed" observable). The probability for a certain momentum, ...


0

I could be wrong, but I remember reading that scanning tunneling microscopy can be used to measure $|\psi|^2$. A quick Google search brought me to this paper, "http://www.ncbi.nlm.nih.gov/pubmed/19090685." I'm sure you can find more information by looking into STM online.


1

The wave function is connected to an experimentally observable value through its complex conjugate square, which gives the probability of an interaction or a decay happening; from this a crossection for the interaction can be predicted, number of events versus some variable in appropriate units. Example: the experiment can measure very many decays of the ...


1

Experimental physicists are very visual people. They are more interested in seeing the wavefunction. Calculating it is the work of a theoretical physicist. "Catching sight of the elusive wavefunction", would be a good read for you at this point. As you proceed with Griffiths, you will see that in experiments we measure certain quantities, called the ...


2

In general, the integral $$ V := \int \mathrm{d} \mu = \int 1 \mathrm{d}\mu$$ is the integration of the identity over the space the measure $\mu$ is defined on, and should be intuitively understood as the volume of the space with respect to the measure. (This is usually only finite for compact spaces.) Normalizing the measure means sending $\mu \mapsto ...


-1

you can make any state if you properly choose bases as, $|\Psi> = \alpha|0> + \beta |1>$ where $|0>$ and $|1>$ are assumed as the complete bases. In this case, $\alpha$ and $\beta$ are the probability amplitude to observe 0 or 1 respectively. Similarly if the system can be written by a continuous bases, like space $|x>$, any state can be ...


2

In the case of an infinite superposition of eigenstates it becomes more complicated but we can still write a general expression for it. If $$\psi(x) = \sum_{n=0}^\infty a_n \phi_n(x)$$ where the $\phi_n$ are the eigenstates of the Hamiltonian. The time dependent wavefunction will look like: $$\Psi(x,t) = \sum_{n=0}^\infty a_n \phi_n(x) T_n(t)$$ where $T_n = ...


-1

In that case, you have a sum of overlaps between a pair of functions having same eigenvalue index. The cross overlap will be zero because of the orthogonality of basis set which is very important.


2

So, I suppose that $Φ(k)$ is the probability density of the momentum. Is this true? Almost. $\Phi(k)$ is the probability amplitude for the momentum of the particle. The probability density is obtained as usual by squaring the amplitude, giving $|\Phi(k)|^2$. For a free particle, all values of momentum are always allowed, which enables the superposition ...


2

First, realize we are doing an approximation when we are evaluating the coefficient $A_p$ in the Fourier series $$ \psi(x) = \sum_p A_p \psi_p(x)$$ by the integral $$ A_p = \int_{-\infty}^\infty \psi_p^*(x)\psi(x)\mathrm{d}x$$ since the limits should really be the boundary of the interval on which the Fourier series is periodic. Furthermore, $\psi(x) = ...


4

The Schrodinger equation is an approximation because it ignores relativistic effects and it ignores spin. However, aside from these limits it applies to any system and not just electrons. The trouble is that the Schrodinger equation, and indeed most partial differential equations, are impossible to solve except in a few special cases. Since we have no ...


2

Your introductory claims are not correct. The usual wavefunction is not a function of spacetime, for two reasons: It is not Lorentz invariant (time is singled out, so it is not a function of a point on a spacetime manifold, and it is not even time dependent at all in the Heisenberg picture) and it can depend on more than one "set" of positions - for a system ...



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