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Isaac0427

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Isaac0427

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blue_leaf77

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Moreover, this doesn't seem like a quantum mechanics problem. You don't even need ##\hbar## as it can be canceled out and the ##\nabla^2## term is also missing, it's just a math eigenvalue problem.

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Isaac0427

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I know from previous threads that ##\hat{E} \psi =E\psi## is not a correct QM equation, yet I just "derived" it. What did I do wrong?

Moreover, this doesn't seem like a quantum mechanics problem. You don't even need ##\hbar## as it can be canceled out and the ##\nabla^2## term is also missing, it's just a math eigenvalue problem.

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blue_leaf77

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$$

i\hbar \frac{\partial}{\partial t} \psi(\mathbf{r},t) = H(\mathbf{r},t) \psi(\mathbf{r},t)

$$

It may reduce to the time independent Schroedinger equation when the Hamiltonian is time-independent and ##\psi## is an eigenfunction of the Hamiltonian.

However, the equation you have there is not a Schroedinger equation since it misses the kinetic energy term.

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Isaac0427

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Would my equation work?

$$

i\hbar \frac{\partial}{\partial t} \psi(\mathbf{r},t) = H(\mathbf{r},t) \psi(\mathbf{r},t)

$$

It may reduce to the time independent Schroedinger equation when the Hamiltonian is time-independent and ##\psi## is an eigenfunction of the Hamiltonian.

However, the equation you have there is not a Schroedinger equation since it misses the kinetic energy term.

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blue_leaf77

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Isaac0427

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So this equation is correct for time dependent quantum systems, right? And because ##\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{x})##, for stationary states we get ##\hat{H}\psi=E\psi## and for time dependent systems we get ##\hat{H}\psi=\hat{E}\psi##. Is that correct?

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blue_leaf77

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There is nothing which can be associated with a quantum system since your equation is not a Schroedinger equation.So this equation is correct for time dependent quantum systems, right?

Time-dependent system is known to be difficult to solve analytically. Most time-dependent cases are dealt with the help of some perturbation theory. I can't be certain myself that a concept of eigenstates also applies in TDSE. There is probably not even a stationary state, take for example the time evolution of states in a time-dependent Hamiltonian in which the Hamiltonian in any given time commutes with the Hamiltonian at other times. The time evolution readsfor time dependent systems we get ^Hψ=^EψH^ψ=E^ψ\hat{H}\psi=\hat{E}\psi. Is that correct?

$$

\psi(x,t) = \exp\left( -\frac{i}{\hbar} \int_0^t H(t') dt' \right) \psi(x,0)

$$

I don't see how states that evolve that way can yield time-independent expectation value for any observable for this state to deserve the calling "stationary".

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Well, the Schrödinger equation indeed reads

$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}).$$

The equation can be separated in the form

$$\psi(t,\vec{x})=T(t) \Psi(\vec{x}),$$

if the Hamiltonian is not time dependent, because then

$$\mathrm{i} \hbar \partial_t \psi = \mathrm{i} \hbar \dot{T}(t) \Psi(\vec{x})= T(t) \hat{H} \Psi(\vec{x}).$$

Now devide this equation by ##T(t) \Psi(\vec{x})##. Then you get

$$\frac{1}{T}(t) \mathrm{i} \hbar \dot{T}(t) =\frac{1}{\Psi(\vec{x})} \hat{H} \Psi(\vec{x}).$$

Now, since the left-hand side is independent of ##\vec{x}## and the right-hand-side is independent of ##t##, it must be a constant, which we call ##E##. This implies the equation for ##T##

$$\mathrm{i} \hbar \dot{T}=E T \; \Rightarrow \; T(t)=T_0 \exp(-\mathrm{i} \omega t), \quad \omega=\frac{E}{\hbar},$$

and for ##\Psi##:

$$\hat{H} \Psi(\vec{x})=E \Psi(\vec{x}),$$

i.e., ##\Psi## must be an eigenvector of ##\hat{H}## which of course is NOT ##\mathrm{i} \hbar \partial_t## but some differential operator in ##\vec{x}## like the standard case for the motion of a particle in some external potential,

$$\hat{H}=-\frac{\hbar^2}{2m} \Delta + V(\vec{x}).$$

So the general solution of the time-dependent Schrödinger equation can be written in terms of the energy-eigensolutions (i.e., the solutions of the time-independent Schrödinger equation)

$$\psi(t,\vec{x})=\sum_{E} a_E \exp(-\mathrm{i} \omega t) \Psi_E(\vec{x}), \quad \hat{H} \Psi_E(\vec{x})=E \Psi_E(\vec{x}).$$

The sum goes over all energy eigenvalues. If there are continuous energy eigenvalues ("scattering solutions"), there's also an integral over the corresponding values. Of course, there are cases, where the Hamiltonian has only a continuous spectrum, as in the case of the free particle.

$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}).$$

The equation can be separated in the form

$$\psi(t,\vec{x})=T(t) \Psi(\vec{x}),$$

if the Hamiltonian is not time dependent, because then

$$\mathrm{i} \hbar \partial_t \psi = \mathrm{i} \hbar \dot{T}(t) \Psi(\vec{x})= T(t) \hat{H} \Psi(\vec{x}).$$

Now devide this equation by ##T(t) \Psi(\vec{x})##. Then you get

$$\frac{1}{T}(t) \mathrm{i} \hbar \dot{T}(t) =\frac{1}{\Psi(\vec{x})} \hat{H} \Psi(\vec{x}).$$

Now, since the left-hand side is independent of ##\vec{x}## and the right-hand-side is independent of ##t##, it must be a constant, which we call ##E##. This implies the equation for ##T##

$$\mathrm{i} \hbar \dot{T}=E T \; \Rightarrow \; T(t)=T_0 \exp(-\mathrm{i} \omega t), \quad \omega=\frac{E}{\hbar},$$

and for ##\Psi##:

$$\hat{H} \Psi(\vec{x})=E \Psi(\vec{x}),$$

i.e., ##\Psi## must be an eigenvector of ##\hat{H}## which of course is NOT ##\mathrm{i} \hbar \partial_t## but some differential operator in ##\vec{x}## like the standard case for the motion of a particle in some external potential,

$$\hat{H}=-\frac{\hbar^2}{2m} \Delta + V(\vec{x}).$$

So the general solution of the time-dependent Schrödinger equation can be written in terms of the energy-eigensolutions (i.e., the solutions of the time-independent Schrödinger equation)

$$\psi(t,\vec{x})=\sum_{E} a_E \exp(-\mathrm{i} \omega t) \Psi_E(\vec{x}), \quad \hat{H} \Psi_E(\vec{x})=E \Psi_E(\vec{x}).$$

The sum goes over all energy eigenvalues. If there are continuous energy eigenvalues ("scattering solutions"), there's also an integral over the corresponding values. Of course, there are cases, where the Hamiltonian has only a continuous spectrum, as in the case of the free particle.

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blue_leaf77

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blue_leaf77

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I thought you were deriving the general form of the wavefunction arising from a time dependent Hamiltonian. If the expansion coefficient is not a function of time, wouldn't ##\psi(t,\vec{x})## as you wrote above just the solution of a Schroedinger equation with a time-independent Hamiltonian. It's just the time evolution operator ##U(t,t_0)## applied to the initial state ##\psi(t_0,\vec{x}) = \sum_E a_E \psi_E(t_0,\vec{x})##. That ##a_E## for time dependent Hamiltonian should be a function of time, you can look at equation 5.5.4 in Modern Quantum Mechanics by Sakurai.So the general solution of the time-dependent Schrödinger equation

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blue_leaf77

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Ah sorry I misread

as "time-dependent Hamiltonian".time-dependent Schrödinger equation

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