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The Schrodinger equation in Hilbert space is expressed as : $$\frac{\partial}{\partial t} \psi(t) = \frac{-i}{\hbar}H\psi(t). $$

Here $\frac{\partial}{\partial t} \psi(t) \equiv \psi'(t) \equiv\lim \limits_{h \to 0} [\frac{\psi(t+h)-\psi(t)}{h}]$, and because $\psi(t)$ is a Hilbert space vector, the limit is defined using convergence via the norm. (In other words, for any $\epsilon>0$ there exists an $h_\epsilon>0$, such that $\big|\big|\psi'(t)-(\frac{\psi(t+h)-\psi(t)}{h})\big|\big| < \epsilon $ for all $|h|<h_\epsilon$). So the convergence depends on the vector as a whole.

But in the wavefunction realization (for a single particle), the Schrodinger equation is expressed as

$$\frac{\partial}{\partial t} \psi(x;t) = \frac{-i}{\hbar}\left [ - \frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V \right ] \psi(x;t). $$

Here, however, $\frac{\partial}{\partial t} \psi(x;t)$ is a pointwise partial derivative with respect to $t$, and so the convergence depends only on each individual point $x$ of $\psi(x)$ separately.

If we take the abstract Hilbert space expression as definitive (axiomatically), then how can it be shown that the wavefunction realization actually expresses the same thing, given the different meanings of $\frac{\partial}{\partial t}$?

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    $\begingroup$ In general these will not hold, particularly since the Hilbert spaces, in their full generality, are quite permissive with (discontinuous) value changes as long as they are on sets of measure zero. So, a more useful question is "what extra hypotheses are required for the two derivatives to coincide?", and related questions. $\endgroup$ – Emilio Pisanty May 18 '14 at 21:24
  • $\begingroup$ I don't understand why there should be any difference, because in both cases we speak of mappings from a subset of R to basically the same Hilbert space (the abstract space and its realization as L^2 are isomorphic). The x (or p) variable is 'frozen' when discussing the partial derivative in the case of the wavefunction and is 'hidden' and also 'frozen' when the abstract space is considered... $\endgroup$ – DanielC May 18 '14 at 23:34
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    $\begingroup$ Building on Emilio's remarks, I just tried for an hour to prove that the two notions are the same provided each $\psi(t)$ is smooth and $L^2(\mathbb R)$, and I couldn't do it, although I suspect that it's true. I guess all that proves is that my analysis is super rusty. Nice question. $\endgroup$ – joshphysics May 19 '14 at 2:51
  • $\begingroup$ @DanielC : L^2 is isomorphic to the abstract space under the inner product of each respective space. (I.e. InnerL2(f,g) iff InnerAbstract(m(f),m(g)), where m is the isomorphism function). Then the notions of limit in each space are also defined using each space's inner product, which is not the case with the d/dt of the wavefunction. $\endgroup$ – Tim May 19 '14 at 3:02
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    $\begingroup$ @joshphysics Hi Josh! I added the proofs, maybe you are interested in them. $\endgroup$ – Valter Moretti Jun 6 '14 at 8:38
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I henceforth assume $\hbar=1$. The correct way to think of Schroedinger equation in a Hilbert space $\cal H$ (referring to a self adjoint Hamiltonian $H : D(H) \to \cal H$, with $D(H)\subset \cal H$ a dense subspace) is the one where the time derivative refers to the topology of the Hilbert space (as correctly noticed at the beginning of the posted question):

$$\frac{d}{dt} \psi_t = -i H \psi_t\tag{1}\:.$$

Above $\psi_t := e^{-itH} \psi$ and $\psi \in D(H)$. This last requirement guarantees that $\psi_t \in D(H)$ for every $t \in \mathbb R$ and that the $t$-derivative $\frac{d}{dt} \psi_t$ exists in the sense of the topology of the Hilbert space, and, finally, that (1) holds true.

However, a naive interpretation of Schroedinger equation assumes that the $t$-derivative is in the standard sense for a wavefunction $\psi=\psi(t,x)$ sufficiently smooth in both variables, and the equation itself is interpreted in the sense of standard PDE, supposing that $H$ is the (hopefully unique) self-adjoint extension of a differential operator $H_x = -\frac{1}{2m} \Delta_x + V(x)$ with $V$ at least continuous:

$$\frac{\partial}{\partial t} \psi(t,x) = -i H_x \psi(t,x)\tag{2}\:.$$

This second interpretation is untenable in the general case for various reasons. In particular it is false that all solutions of (1) solve (2), because the wavefunctions solving (1) are elements of $D(H) \subset {\cal H} = L^2(\mathbb R^3, dx)$ and thus (a) they are defined up to zero measure set and (b), in general, it is not possible to obtain a continuous function changing the initial one on a zero measure set. (The reason is that the self-adjoint extension $H$ of $H_x$ ceases to be a differential operator.)

However it could happen that one finds a solution of (2) $\psi=\psi(t,x)$ which is differentiable in $t$ for every $x$ and sufficiently regular in $x$ depending on the regularity of $V$ in order to belong to $D(H_x)\subset D(H)$. Does $\psi$ solve (1)?

The only thing to check is if, almost everywhere in $x$ and for a given $t \in \mathbb R$,

$$\left(\frac{d}{dt} \psi_t\right)(x) = \frac{\partial}{\partial t} \psi(t,x)\:.\tag{3}$$

We have a pair of elementary facts:

(A) If both sides of (3) exist, and the right hand side belongs to $L^2(\mathbb R^3, dx)$, (3) holds.

(B) If there are $\epsilon>0$ and $g_{t} \in L^2(\mathbb R^3, dx)$, with $$\left|\frac{\partial}{\partial \tau} \psi(\tau,x)\right| \leq |g_{t}(x)|\quad \mbox{almost everywhere in $x$, } \forall \tau \in (t-\epsilon, t+\epsilon)$$ then the left-hand side of (3) exists (and (3) holds true for (A)).

ADDENDUM.

SKETCH OF PROOF

Regarding (A), since the $t$ derivative exists in the sense of the Hilbert space topology, we know that $$\lim_{h\to 0} \int \left| \frac{1}{h}(\psi_{t+h}(x)-\psi_t(x)) - \frac{d \psi_t(x)}{dt}\right|^2 dx=0$$ A know result of $L^p$ spaces theory says that if $f_n \to f$ as $n\to +\infty$ in $L^p$, there is a subsequence with $f_{n_k} \to f$ almost everywhere as $k \to +\infty$. Therfore there is a sequence $h_k \to 0$ as $k\to +\infty$, such that, almost everywhere in $x$, $$\frac{1}{h_k}(\psi_{t+h_k}(x)-\psi_t(x)) \to \frac{d \psi_t(x)}{dt}\:.\tag{4}$$ On the other hand we know that, just because $\frac{\partial \psi(t,x)}{\partial t}$ exists, for every $x$ we also have $$\frac{1}{h}(\psi_{t+h}(x)-\psi_t(x)) \to \frac{\partial \psi(t,x)}{\partial t}\quad \mbox{if $h\to 0$}.$$ Therefore, in particular, again for every $x$, $$\frac{1}{h_k}(\psi_{t+h_k}(x)-\psi_t(x)) \to \frac{\partial \psi(t,x)}{\partial t}\quad \mbox{if $k\to \infty$}\:.\tag{5}$$ Comparing (4) and (5), we conclude that (A) holds: $$\left(\frac{d}{dt} \psi_t\right)(x) = \frac{\partial}{\partial t} \psi(t,x)$$ almost everywhere in $x$. So that (A) is true.

Regarding (B), what has to be proved is that: $$\lim_{h\to 0} \int \left| \frac{1}{h}(\psi(t+h,x)-\psi(t,x)) - \frac{\partial \psi(t,x)}{\partial t}\right|^2dx =0\:.\tag{6}$$ Lagrange's theorem allows us to re-write the integral as: $$\int \left| \frac{\partial \psi(\tau,x)}{\partial \tau}|_{\tau= t_{x,h}} - \frac{\partial \psi(\tau,x)}{\partial \tau}|_{\tau=t}\right|^2 dx$$ where $t_{x,h} \in [t-h,t+h]$. In our hypotheses we also have that: $$\left| \frac{\partial \psi(\tau,x)}{\partial \tau}|_{\tau= t_{x,h}} - \frac{\partial \psi(\tau,x)}{\partial \tau}|_{\tau=t}\right|^2 \leq 2|g_t(x)|^2$$ almost everywhere in $x$ and for all sufficiently small $h$. Lebesgue's dominated convergence theorem implies that the symbol of integral and that of limit can be swapped in the RHS of (6), obtaining $$\lim_{h\to 0} \int \left| \frac{1}{h}(\psi(t+h,x)-\psi(t,x)) - \frac{\partial \psi(t,x)}{\partial t}\right|^2 dx$$ $$ = \int \lim_{h\to 0}\left| \frac{1}{h}(\psi(t+h,x)-\psi(t,x)) - \frac{\partial \psi(t,x)}{\partial t}\right|^2 dx =0\:,$$ as wanted.

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