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54

The atomic orbitals are eigenstates of the Hamiltonian $$H_0(\boldsymbol P,\boldsymbol R)=\frac{\boldsymbol P^2}{2m}+\frac{e}{R}$$ On the other hand, the Hamiltonian of Nature is not $H_0$: there is a contribution from the electromagnetic field as well $$H(\boldsymbol P,\boldsymbol R,\boldsymbol A)=H_0(\boldsymbol P+e\boldsymbol A,\boldsymbol R)+\frac12\... 50 I think this question makes hidden, inarticulated assumptions about reality. In physics, we make observations and then try to find models that match them. The models, though, belong only to us and exist in our heads and textbooks. We perform the calculations required to make our predictions in our models. We cannot say whether nature makes similar ... 27 The hydrogen atom in an excited state is not really in an energy eigenstate. There are two ways of looking at it. One way is to recognize that the atom is not isolated. It is always coupled to the electromagnetic field. Even if field itself is in the ground state, there are "zero-point" fluctuations in the field amplitude. Thus, the atom is always ... 22 I) The solution to the time-dependent Schrödinger equation (TDSE) is$$ \Psi(t_2) ~=~ U(t_2,t_1) \Psi(t_1),\tag{A}$$where the (anti)time-ordered exponentiated Hamiltonian$$\begin{align} U(t_2,t_1)~&=~\left\{\begin{array}{rcl} T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right] &\text{for}& t_1 ~<~t_2 \cr\cr AT\exp\left[-\frac{i}{\...

22

Nobody is "doing the normalization". Normalization is not even necessary. We often normalize for convenience, since that means that the Born rule for $\lvert \psi \rangle$ being the state $\lvert \phi \rangle$ reads $$P(\psi,\phi) = \lvert \langle\psi\vert\phi\rangle \rvert ^2$$ which is certainly easier to recall/write than $$P(\psi,\phi) = \frac{\... 21 Yes, you are on the right track. The series you have there is called Dyson's series. First note that the n'th term looks like$$ U_n = (-\frac{i}{\hbar})^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n) $$The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing ... 21 Is it true that two different states cannot evolve into the same final state? That depends on exactly what you mean. If we consider the total state of a closed system, then two different states will never simultaneously evolve into the same state at any later time. You may have learned that quantum states evolve with a unitary transformation, i.e.$$ \lvert ...

18

Well, I think you said the answer yourself when you used the words "projection operator." In the Heisenberg picture the operators get projected down to a subspace at the time of the collapse. In other words, the operator 'collapses' by picking up a projection piece that kills the unphysical part of the state. Forget about pictures for a second, the physical ...

18

One can indeed motivate the Schrödinger equation along the lines you suggest. The crucial point you are missing is time shift invariance of your quantum system. It is this that lets you write down $U = \exp(i\alpha\,H\,t)$. To explain further: The fact of $\psi(t) = U(t) \psi(0)$ is simply a linearity assumption. The evolution wrought by your state ...

17

The general situation is the following one. There is a self-adjoint operator $H :D(H) \to \cal H$, with $D(H) \subset \cal H$ a dense linear subspace of the Hilbert space $\cal H$. (An elementary case is ${\cal H} = L^2(\mathbb R, dx)$, but what follows is valid in general for every complex Hilbert space $\cal H$ associated to a quantum physical system.) It ...

14

Because you do equilibrium statistical mechanics. In the usual ensemble theory we associate to a system (a macrostate) a big number of corresponding microstates, each microstate is a point in phase space, that point is called representative point. Now you want to study how these points move in phase space. First of all the situation of equilibrium for the ...

13

Where did you get that formula? The correct normalization does not involve the time integral. Denoting with $| \psi(t) \rangle$ the state at time $t$, the normalization condition reads $$\tag{A} \langle \psi(t) | \psi(t) \rangle = \int d^3 x | \psi(x,t) |^2 = 1, \forall t.$$ What this is telling you is that at any time $t$ the particle must be somewhere. ...

13

$\rho_\psi$, the density matrix, is not an observable/operator evolving in the sense of the Heisenberg equation of motion $$\mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t} A = - [H,A]$$ since it is defined, as you correctly write, as a projector on states, hence it is time-dependent in the Schrödinger picture (since there the states it projects on are time-...

13

One way to think about it is that a particle "sniffs out" its immediate surroundings and reacts to gradient: a trend like a declining potential in one direction. Single-celled organisms do this. Plant orient toward the sun. A rock on an incline "senses" that it's center-of-mass is slight off from the point of contact with the ground. This is all loosely ...

12

I will try to make it as simple and intuitive as possible. In the Schrödinger picture, the expectation value of a given operator $\hat{\xi}$ (which itself is frozen in time) is defined as follows (with $\psi(t)$ the wavefunction of our system at time $t$): $$\langle \hat{\xi} (t) \rangle = \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle$$ Which ...

11

Your second and third equations are the same equation. They just use a different notation for the time derivative. Since in this "abstract" form $|\Psi \rangle$ only depends on time perhaps it is more correct to use the last one, but it is matter of taste. In order to get your first equation (a wave equation), you must project on $\langle x|$: $$H(P=-i\hbar ... 11 "Schroedinger equation" unfortunately is a bit ambiguous word. It could refer to$$i\hbar\frac{d \psi(t)}{dt} = H_t\psi(t) \tag{1}$$but also to a more precise form like this:$$ i\hbar\frac{d \psi(t)}{dt} = \left(-\frac{1}{2m}{\bf P}^2 + V_t\right)\psi(t) \:.$$The former version does not depend on the quantum physical system you are dealing with. So, in ... 11 You are right on track. A great way to think about the Hamiltonian is that it's the thing that "causes translation in time". This just jargon for the fact, which you already observed, that the Hamiltonian tells you how the system moves forward in time, as encapsulated by the Schrodinger equation$$i\hbar d_t |\Psi\rangle = H |\Psi \rangle . $$My guess ... 11 For time-independent Hamiltonians, U(t) = \mathrm{e}^{\mathrm{i}Ht} follows from Stone's theorem on one-parameter unitary groups, since the Schrödinger equation$$ H\psi = \mathrm{i}\partial_t \psi$$is just the statement that H is the infinitesimal generator of a one-parameter group parameterized by the time t. For time-dependent Hamiltonians H(t) = ... 11 The Fourier transform \phi(k) is a function only of k and not of time because it indicates the amplitude of each plane wave that compose the wave function. The amplitudes are conserved in time, because the plane waves linear superpose among them and don't interact. The evolution in time so is not in the amplitudes \phi(k), but you can observe how ... 11 Short answer: yes, but only the phase factor has the time dependence. The spatial profile is constant in time because the eigenstates of the Hamiltonian are Stationary states. Maths: The time dependent Schroedinger equation looks like this:$$ i\hbar \frac{\partial \Psi}{\partial t} = H \Psi = \left ( -\frac{\hbar^2 }{2 m}\frac{\partial^2}{\partial x^2} + ...

10

The time ordering enters as a consequence of the definition of the Hamiltonian as the generator of time translations. In the Schödinger picture: $$|\psi(t)\rangle \approx \left(1 - \frac{i}{\hbar} H(t') [t - t'] + \mathcal{O}([t-t']^2)\right) |\psi(t')\rangle,$$ where the relationship becomes exact in the limit as $t-t' \rightarrow 0^+$. It's an exercise in ...

10

Actually in this specific case you are helped a bit because the time dependence is contained in a term proportional to the unit matrix $$\hat H = E_0e^{t/\omega_0}\hat I + E_1\left(\begin{array}{cc} 0&1 \\ 1&0 \end{array}\right) \, .$$ You can then make the time independent change of basis defined by $$U=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} ... 10 If H(t) is selfadjoint and bounded (thus everywhere defined), the theory is standard and quite easy to handle using the uniform operator topology. You can find all required proofs in the first or second volume of Reed and Simon's textbook on mathematical methods. If the operators H(t) are unbounded, the theory is much more difficult also in view of ... 9 Non-equilibrium systems are most often considered in the approximation where local equilibrium is valid, yielding a hydrodynamic or elasticity description. Local equilibrium means that equilibrium is assumed to hold on a scale large compared to the microscopic scale but small compared with the scale where observations are made. In this case, one considers a ... 9 The Stone's theorem proves the following. Consider a group of unitary operators (U(t))_{t\in\mathbb{R}} acting on a Hilbert space \mathscr{H} (i.e. satisfying U(t+s)=U(t)U(s), in more mathematical terms this is a unitary representation of the abelian group \mathbb{R} on \mathscr{H}). If in addition such group is strongly continuous, namely is such ... 9 The time dependent Schroedinger equation looks like this:$$ i\hbar \frac{\partial \Psi}{\partial t} = H \Psi = \left ( -\frac{\hbar^2 }{2 m}\frac{\partial^2}{\partial x^2} + V(x,t) \right ) \Psi(x,t) , you attempt a solution via separation of variables: $\Psi(x,t) = \psi(x) T(t)$, plug it in. If the potential $V$ is time independent such that $V(x,t) = ... 8 The result can be proved in a general way using, well, math. In particular the theory of semigroups of linear operators on Banach spaces (I know that seems advanced and maybe not physical, but it is an elegant way of proving the result you seek ;-) ). Define the Banach space$\mathscr{L}^1(\mathscr{H})$of trace class operators over a separable Hilbert ... 8 An intuitive answer to motivate the Stone theorem that ACuriousMind's answer cites. Take a quantum system. For now, let it be a finite dimensional one (the Stone theorem is needed to make the reasoning work in a separable, infinite dimensional Hilbert space). Let it evolve for time$t_1$. The evolved state is some linear operator$U(t_1)$imparted on the ... 8 Because in QM, the wave function contains all the relevant information, and$x$is the coordinate parameter. You are expecting$x$to be dependent on time because you are used to see it as the particle's coordinate, i.e. a function of time that says where the particle is for each time parameter value. But here,$\psi$has this information, and$x\$ is just ...

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