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Context : In quantum mechanics, time evolution is described by a one-parameter unitary group, acting on the Hilbert space of states. Under Stone's theorem (with the right hypotheses), this group has a self-adjoint generator, which is called the Hamiltonian.

For a particle living in a 1D box, the Hilbert space is $L^2([0,\ell])$ (with $\ell>0$ the length of the box) and the Hamiltonian operator is : $$ H = -\frac{\hbar^2}{2m}\frac{d}{d x^2} + V(x) \tag{1}$$

We also need to specify the domain on which $H$ is defined and self-adjoint. Now, it is pointed out here and in this P.SE post, that this is equivalent to fixing boundary conditions.

Then, the (unitary) time evolution operator is $U(t) = e^{-iHt/\hbar}$ and is well defined on the whole Hilbert space.

Questions:

  • Concretely, how does changing the domain for $H$ change the time evolution operator $U(t)$ ?
  • Presumably, $\psi = 1/\sqrt{\ell} \in L^2([0,\ell])$ is a valid wave-function, whichever boundary condition we choose, since it lives in the Hilbert space. How do boundary conditions determine its time evolution?
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Here is the deal: canonical quantization forces your $(1)$ to be a quantum Hamiltonian. This operator drives (generates) time evolution iff it is self-adjoint. An operator $A$, if unbounded and defined on a (complex) separable Hilbert space $\mathcal{H}$, can be correctly defined only on a proper subset of $\mathcal{H}$ (called the maximal domain of $A$), if we require the operator to be closed. Self-adjoint operators are always closed, therefore their maximal domain is always proper: $D(A)\subsetneq \mathcal{H}$. This is basic stuff.

Come back to $(1)$. We want $H$ to be self-adjoint when seen as operator $H:D(H)\rightarrow L^2 ([0,l])$. For any well behaved real function $V(x)$, $D(H) = D\left(\frac{d^2}{d x^2}\right)\subsetneq L^2 ([0,l])$.

We can choose only one boundary condition for non-constant elements in $L^2 ([0,l])$, which renders $H$ self-adjoint on its maximal domain. So your first question is almost trivial: $U(t)$ exists only when $H$ is self-adjoint, which happens only in the presence of a particular boundary condition: $$\psi\in L^2 ([0,l]),~ \psi (0) = \psi (l)=0 \tag{2}$$ or the trivial (normalized) wavefunction: $\psi$ is constant, i.e. $$\psi (x) = \frac{\exp{i\alpha}}{\sqrt{l}},\alpha\in\mathbb{R} \tag{3}$$ As soon as you consider functions from the Hilbert space outside these two possible boundary conditions, then the Hamiltonian is no longer self-adjoint (not even symmetric, because you cannot make the boundary term vanish), so that $U(t)$ does not exist. Existence of constant wavefunctions is an artifact of using an unphysical restraining of motion in a box with "hard" walls. Usually PIB involve no dynamics at all ($V(x) \equiv 0$), case in which a constant wavefunction is trivially the $0$ vector, so that only option $(2)$ remains.

The second question gets an answer immediately: Since the constant function $(3)$ is a trivial solution to the spectral equation (energy spectrum = numeric value of $V(x), x\in [0,1]$), then there is no impact of choosing particular boundary conditions on the exact form of time evolution: $U(t) = \exp (-i H(x)t /\hbar)$ which acts trivially on a constant wavefunction, as a phase-factor containing the numeric value of the potential.

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  • $\begingroup$ I think a better example is to consider the operator densely defined on $L^2((0,l))$ (the open interval). You can impose periodic or infinite-well boundary conditions (by extending the domain) and get different time evolutions. You seem to be assuming that it is already defined on functions on the closed interval, so you only get one extension. $\endgroup$ – Keith McClary Apr 22 at 19:23
  • $\begingroup$ Aren't $L^2([0,\ell])$ and $L^2((0,\ell))$ isometric ? (Since the value of $f\in L^2$ is not well defined) $\endgroup$ – SolubleFish Apr 23 at 7:54

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