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I am currently trying to self-study quantum mechanics. From what I have read, it is said that knowing the wave function at some instant determines its behavior at all feature instants, I came across an explanantion that the Schrödinger equation is the reason, however, in the book by Landau & Lifshitz, they made this argument before developing (or introducing) that equation (see picture attached). I would like to know whether its an assumption or not and also why they deduce that the partial time derivative must equal some linear operator acting on the wave function.

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Qmechanic
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Gauss_fan
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2 Answers2

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In the text you have pasted, they are not really making arguments, derivations or assumptions, they are making statements:

  1. The wave function determines the quantum state at all times. This can be considered a consequence of the Schrodinger equation as you point out, but here they are simply stating it as a true fact about quantum mechanics.
  2. By the principle of superposition, the relation between the time derivative and the wave function must be linear. This simply a restatement of the superposition principle, which it is stated holds for wave functions. Consider the wave function $\Psi = \psi_1 + \psi_2$. The evolution of $\Psi$ is the sum of the independent evolutions of $\psi_1$ and $ \psi_2$, so $\partial_t \Psi = \partial_t \psi_1 + \partial_t \psi_2$. The only operator for which $O(\Psi) = O(\psi_1)+O(\psi_2)$ always holds is a linear operator.
Codename 47
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If I can remember well, Landau and Lifschitz first introduce the probabilistic interpretation of the wave function, being

$$\rho(\mathbf{r},t) = \psi(\mathbf{r},t)^* \psi(\mathbf{r},t) \ ,$$

so that normalization condition holds $$1 = \int_{\mathbf{r} \in V} \psi(\mathbf{r},t)^* \psi(\mathbf{r},t) = \langle\psi(t) | \psi(t) \rangle $$

the probability density of the coordinates.

Now,

  • if we trust the superposition observed in nature the equation must be linear, and thus the equation must be of the form

$$\dfrac{d}{dt}{|\psi \rangle} = \hat{L} |\psi \rangle$$

  • if we trust (since there is an experimental evidence) the probabilistic interpretation of the wave function, and the fact that normalization holds at every time $t$, the operator of the linear equation must have the form $\hat{L} = -i c \hat{H}$, being $c = \frac{1}{\hbar}$ a constant involving Planck's constant, and $\hat{H}$ is the Hamiltonian operator, in order to match the experimental results. You can recover classical mechanics as well, as an example looking for Ehrenfest theorem. The solution of the Schrodinger equation

    $$|\dot{\psi} \rangle = -\frac{i}{\hbar}\hat{H} | \psi \rangle$$

    reads (if the Hamiltonian is independent from time)

    $$|\psi (t) \rangle = e^{-\frac{i}{\hbar} \hat{H} t} | \psi(0) \rangle \ ,$$

    being $ e^{-\frac{i}{\hbar} \hat{H} t}$ the unitary evolution operator, and the normalization condition always satisfied, if it holds at the initial time,

    $$\langle \psi(t) | \psi(t) \rangle = \langle e^{-\frac{i}{\hbar} \hat{H} t}\psi(0) | e^{-\frac{i}{\hbar} \hat{H} t} \psi(0) \rangle = \langle \psi(0) | \underbrace{e^{\frac{i}{\hbar} \hat{H} t} e^{-\frac{i}{\hbar} \hat{H} t}}_{=1} \psi(0) \rangle = \langle \psi(0) | \psi(0) \rangle= 1 \ .$$

Remarks. In the latter expression, the fact that the Hamiltonian operator is self-adjoint has been used. The unitary operator can be interpreted with a series expansion,

$$e^{-\frac{i}{\hbar} \hat{H}t} = \sum_{k=0}^{\infty} \frac{\left(-\frac{i}{\hbar} \hat{H} \right)^k}{k!} \ ,$$

and written in a "diagonal form" in the base of stationary states (here meant to be discrete), i.e. the eigenvectors of the Hamiltonian as

$$\begin{aligned} e^{-\frac{i}{\hbar} \hat{H}t} & = \underbrace{|\psi_a\rangle \langle \psi_a}_{\hat{I}} | e^{-\frac{i}{\hbar} \hat{H}t} \underbrace{|\psi_b\rangle \langle \psi_b |}_{=\hat{I}} = \\ & = \psi_a\rangle \sum_{k=0}^{\infty} \langle \psi_a | \sum_{k=0}^{\infty} \frac{\left(-\frac{i}{\hbar} \hat{H} \right)^k}{k!} |\psi_b\rangle \langle \psi_b | = \\ & = \psi_a\rangle \sum_{k=0}^{\infty} \underbrace{\frac{\left(-\frac{i}{\hbar}E_b\right)^k}{k!}}_{e^{-i \frac{E_b}{\hbar}t}} \underbrace{\langle \psi_a |\psi_b\rangle}_{=\delta_{ab}} \langle \psi_b | = \\ & = | \psi_a \rangle \langle \psi_a | e^{-\frac{i}{\hbar}E_a t} \ . \end{aligned}$$

basics
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