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Suppose we lived in a universe in which the Schrödinger equation contains second order time derivatives, $$i\hbar \partial_t^2|\varphi(t)\rangle = \mathbb{H} | \varphi(t)\rangle.$$ Would it be true that the norm of $|\varphi(t)\rangle$ is time independent?

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    $\begingroup$ Your Hamiltonian, then, is antihermitean? $\endgroup$ – Cosmas Zachos Sep 12 '18 at 21:54
  • $\begingroup$ Consider the kernel of $(i\hbar \partial_t -\mathbb{K})$ for hermitean time-independent $\mathbb{K}$. It will also be in the kernel of $(i\hbar \partial_t^2 -\mathbb{K}^2/i\hbar)$. So $\mathbb{H}=-i\mathbb{K}^2/\hbar$ should do the trick. $\endgroup$ – Cosmas Zachos Sep 12 '18 at 22:05
  • $\begingroup$ ...that is, e.g., for $\mathbb{K}=\hbar \omega$, you have $\mathbb{H}=-i\hbar \omega^2$, so you get oscillatory solutions $\exp(\pm i\omega t)$ preserving the norm; while their trigonometric combinations, naturally, do not. $\endgroup$ – Cosmas Zachos Sep 13 '18 at 0:45
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No, the norm would not be preserved. For simplicity, let's assume the Hamiltonian is time-independent with discrete (but possibly degenerate) eigenvalues $E_n$. Then we have $$\begin{align*} |\psi(t)\rangle &= \sum_n c_n(t)\, |E_n\rangle \\ i \hbar \partial_t^2 |\psi(t)\rangle &= H |\psi(t)\rangle \\ i \hbar \sum_n \ddot{c}_n(t)\, |E_n\rangle &= \sum_n c_n(t)\, E_n |E_n\rangle \\ \sum_n (i \hbar \ddot{c}_n(t)- c_n(t)\, E_n) |E_n\rangle &= 0 \\ \ddot{c}_n(t) + \frac{i}{\hbar} E_n c_n(t) &= 0 \\ c_n(t) &= \exp \left( \sqrt{-\frac{i E_n}{\hbar}} t \right) = \exp \left( \sqrt{\frac{E_n}{\hbar}} e^{-\frac{i \pi}{4}} t \right) \\ |c_n(t)|^2 &= c_n(t)\, c_n^*(t) = \exp \left( \sqrt{\frac{2 E_n}{\hbar}} t \right) \end{align*}$$

and $|\psi(t)|^2$ grows exponentially over time. The problem is that $e^{-i E_n / \hbar}$ evolves in a norm-preserving way over time, but the second derivative brings in a square root that rotates the pure imaginary exponent into having a real part, which messes things up.

But your proposed modification isn't really the physically natural one anyway. For one thing, note that your "$\hbar$" has different units than the physical one. Rather than leaving the $i\hbar$ as-is but changing the $\partial_t$ to a $\partial_t^2$, the more natural modification to the Schrodinger equation that makes it second-order is to "square" the operators on both sides, i.e. to change $i \hbar \partial_t$ to $(i \hbar \partial_t)^2 = -\hbar^2 \partial_t^2$ and $H$ to $H^2$, to get $$-\hbar^2 \partial_t^2 |\psi(t)\rangle = H^2 |\psi(t)\rangle.$$ In the case of the Hamiltonian $H^2 = (cp)^2 + (mc^2)^2$ for a relativistic particle, this is known as the Klein-Gordon equation, and while it has the same problem of non-constant norm, it turns out to accurately describe the time evolution of a scalar relativistic quantum field (not particle, as one might expect).

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  • $\begingroup$ "tparker" could you detail the step from row 4 to row 5? I did not understand: \begin{align*} | \\ \sum_n (i \hbar \ddot{c}_n(t)- c_n(t)\, E_n) |E_n\rangle &= 0 \\ \ddot{c}_n(t) + \frac{i}{\hbar} E_n c_n(t) &= 0 \\ . \end{align*} $\endgroup$ – mike Sep 13 '18 at 2:08
  • $\begingroup$ @mike All tparker did was equate the $|E_n\rangle$ coefficients. $\endgroup$ – J.G. Sep 13 '18 at 6:29
  • $\begingroup$ @mike Right - I used the fact that the eigenstates are linearly independent, so if a linear combination of them is zero then each coefficient must equal zero. $\endgroup$ – tparker Sep 13 '18 at 10:18
  • $\begingroup$ I'm curious: in the "properly" squared equation you give, in what kind of space does [math]|\psi(t)\rangle[/math] live? If it's a quantum field it can't be the usual kind of wave function, so what is it when constructed explicitly? $\endgroup$ – The_Sympathizer Sep 14 '18 at 1:48
  • $\begingroup$ @The_Sympathizer That's a great question, which I deliberately glossed over. In classical field theory the answer's easy: it's just scalar functions $\varphi(t, {\bf x}):\mathbb{R}^4 \to \mathbb{R}$ defined on spacetime. In quantum field theory you're asking about the mathematical structure of the Hilbert space, and frankly no one really knows at any mathematically rigorous level. In the Lagrangian formalism the fields $\varphi(x)$ are just plain old commuting scalar fields as in the classical case, but the mathematical pathology gets pushed into this horribly ill-defined path integral. ... $\endgroup$ – tparker Sep 14 '18 at 3:53
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The answer to your question is that for generic initial conditions the answer is no. I will give an explicit counter example below.

First of all, we live in a universe where the Schrödinger equations can be written as a second-order equation: just apply $- \mathrm{i} \partial_t$ to the (first-order) Schrödinger equation \begin{align} \mathrm{i} \partial_t \psi(t) = H \psi(t) , && \psi(0) = \phi \in \mathcal{H} , \label{Schroedinger_equantion} \end{align} and obtain the second-order wave equation \begin{align} \partial_t^2 \psi(t) = - H^2 \psi(t) , \label{wave_equation} \end{align} with the initial conditions \begin{align*} \psi(0) = \phi , \; \partial_t \psi(0) = - \mathrm{i} H \psi(0) = - \mathrm{i} H \phi . \end{align*} Note that as a second-order equation, you need to specify not just $\psi(0)$ but also its time-derivative as initial condition. Moreover, to ensure that what you obtain is actually equivalent to the Schrödinger equation, you need to choose a \emph{particular} initial condition for $\partial_t \psi(0) = - \mathrm{i} H \phi$. Other choices will give you other equations, but none of them are equivalent.

As you can see, if you choose an initial condition that is compatible with the Schrödinger equation, the solutions will have the usual properties, e. g. the norm $\lVert \psi(t) \rVert = \lVert \psi(0) \rVert$. But for generic choices, this need not be true: pick $H = \sigma_3$ so that $H^2 = \mathbb{1}$. Then the wave equation is simply \begin{align*} \frac{\partial^2}{\partial t^2} \left ( \begin{matrix} \psi_1(t) \\ \psi_2(t) \\ \end{matrix} \right ) + \left ( \begin{matrix} \psi_1(t) \\ \psi_2(t) \\ \end{matrix} \right ) = 0 , \end{align*} so two copies of the usual wave equation. The solutions are \begin{align*} \psi(t) = c_- \, \mathrm{e}^{- \mathrm{i} t} + c_+ \, \mathrm{e}^{+ \mathrm{i} t} \end{align*} where the coefficients $c_{\pm} \in \mathbb{C}^2$ need to be determined from the initial conditions. Now pick $\psi(0) = c_+ + c_- = (2,2)$ and $\partial_t \psi(0) = (0,0) = \mathrm{i} (c_+ - c_-)$ as initial conditions. That leads to $c_- = c_+ = 1$, and thus \begin{align*} \psi(t) = 2 \cos t \, \left ( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right ) , \end{align*} and the norm $\lVert \psi(t) \rvert = 2 \sqrt{2} \, |\cos t| $ of this vector evidently oscillates in time and is therefore not conserved. For the special initial condition $\partial_t \psi(0) = - \ii H \psi(0)$, though, the norm is conserved.

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Let me collect my above comments so the message "sometimes, but not always" is not lost.

The usual argument for the conventional Schr eqn with hermitean, time-independent $\mathbb{K}$, $$ i\hbar\partial_t \langle \varphi|\varphi\rangle = \langle \varphi|\mathbb{K}- \mathbb{K}^\dagger|\varphi\rangle=0 , $$ is ostensibly lost, since you only get $$ i\hbar\partial_t (\langle \varphi|\dot \varphi\rangle -\langle \dot \varphi|\varphi\rangle)= \langle \varphi|\mathbb{H}+ \mathbb{H}^\dagger|\varphi\rangle $$ this way—integration by parts. So, for antihermitean $\mathbb{H}$, you get a conserved quantity, but it is not the norm.

Nevertheless, the kernel of $(iℏ∂_t−\mathbb{K})$ will also be in the kernel of $(iℏ∂^2_t−\mathbb{K}^2/iℏ)$, that is, the invariant norm solutions of the Schr eqn for $\mathbb{K}$ will also solve your equation with $\mathbb{H}=−i\mathbb{K}^2/ℏ$, which will also have "bad" solutions, as well, naturally.

You may illustrate this schematically, (leaving definition of the norm issues aside—you may trivially modulate your answer by space Gaussian filters), by the most trivial example, $ \mathbb{K}=ℏω$, hence $ \mathbb{H}=−iℏω^2$, where you get oscillatory solutions $\exp(±iωt)$ preserving the norm; while their trigonometric combinations, $\cos(ωt), ~\sin(ωt)$, naturally, do not.

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