Previously I thought this is a universal theorem, for one can prove it in the one dimensional case using variational principal.
However, today I'm doing a homework considering a potential like this:$$V(r)=-V_0\quad(ra)$$ and found that there is no bound state when V0a2<π2ℏ2/8m.
So what's the condition that we have at least one bound state for 3D and 2D?
The precise theorem is the following, cf. e.g. Ref. 1.
Theorem 1: Given a non-positive (=attractive) potential V≤0 with negative spatial integral v := ∫Rndnr V(r) < 0,
then there exists a bound state1 with energy E<0 for the Hamiltonian H = K+V,K = −ℏ22m∇2
if the spatial dimension n≤2 is smaller than or equal to two.
The theorem 1 does not hold for dimensions n≥3. E.g. it can be shown that already a spherically symmetric finite well potential does not2 always have a bound state for n≥3.
Proof of theorem 1: Here we essentially use the same proof as in Ref. 2, which relies on the variational method. We can for convenience use the constants c, ℏ and m to render all physical variables dimensionless, e.g.
V ⟶ ˜V := Vmc2,r ⟶ ˜r := mcℏr,
and so forth. The tildes are dropped from the notation from now on. (This effectively corresponds to setting the constants c, ℏ and m to 1.)
Consider a 1-parameter family of trial wavefunctions
ψε(r) = e−fε(r) ↗ e−1forε ↘ 0+,
where
fε(r) := (r+1)ε ↘ 1forε ↘ 0+
r-pointwise. Here the ↗ and ↘ symbols denote increasing and decreasing limit processes, respectively. E.g. eq. (4) says in words that for each radius r≥0, the function ψε(r) approaches monotonically the limit e−1 from below when ε approaches monotonically 0 from above.
It is easy to check that the wavefunction (4) is normalizable:
0 ≤ ⟨ψε|ψε⟩ = ∫Rndnr |ψε(r)|2 ∝ ∫∞0dr rn−1|ψε(r)|2
≤ ∫∞0dr (r+1)n−1e−2fε(r) f=(1+r)ε= 1ε∫∞1df fnε−1e−2f < ∞,ε > 0.
The kinetic energy vanishes
0 ≤ ⟨ψε|K|ψε⟩ = 12∫Rndnr |∇ψε(r)|2 = 12∫Rndnr |ψε(r)dfε(r)dr|2
∝ ε2∫∞0dr rn−1(r+1)2ε−2|ψε(r)|2 ≤ ε2∫∞0dr (r+1)2ε+n−3e−2fε(r)
f=(1+r)ε= ε∫∞1df f1+n−2εe−2f ↘ 0forε ↘ 0+,
when
n≤2, while the potential energy
0 ≥ ⟨ψε|V|ψε⟩ = ∫Rndnr |ψε(r)|2 V(r)
↘ e−2∫Rndnr V(r) < 0forε ↘ 0+,
remains non-zero due to assumption (1) and Lebesgue's monotone convergence theorem.
Thus by choosing ε↘0+ smaller and smaller, the negative potential energy (8) beats the positive kinetic energy (7), so that the average energy ⟨ψε|H|ψε⟩⟨ψε|ψε⟩<0 eventually becomes negative for the trial function ψε. A bound state1 can then be deduced from the variational method.
Note in particular that it is absolutely crucial for the argument in the last line of eq. (7) that the dimension n≤2. ◻
Simpler proof for n<2: Consider an un-normalized (but normalizable) Gaussian test/trial wavefunction
ψ(x) := e−x22L2,L > 0.
Normalization must scale as
||ψ|| (9)∝ Ln2.
The normalized kinetic energy scale as
0 ≤ ⟨ψ|K|ψ⟩||ψ||2 ∝ L−2
for dimensional reasons. Hence the un-normalized kinetic scale as
0 ≤ ⟨ψ|K|ψ⟩ (10)+(11)∝ Ln−2.
Eq. (12) means that
∃L0>0∀L≥L0: 0 ≤ ⟨ψ|K|ψ⟩ (12)≤ −v3 > 0
if n<2.
The un-normalized potential energy tends to a negative constant
⟨ψ|V|ψ⟩ ↘ ∫Rndnx V(x) =: v < 0forL → ∞.
Eq. (14) means that
∃L0>0∀L≥L0: ⟨ψ|V|ψ⟩ (14)≤ 2v3 < 0.
It follows that the average energy
⟨ψ|H|ψ⟩||ψ||2 = ⟨ψ|K|ψ⟩+⟨ψ|V|ψ⟩||ψ||2 (13)+(15)≤ v3||ψ||2 < 0
of trial function must be negative for a sufficiently big finite L≥L0 if n<2. Hence the ground state energy must be negative (possibly −∞). ◻
References:
K. Chadan, N.N. Khuri, A. Martin and T.T. Wu, Bound States in one and two Spatial Dimensions, J.Math.Phys. 44 (2003) 406, arXiv:math-ph/0208011.
K. Yang and M. de Llano, Simple variational proof that any two‐dimensional potential well supports at least one bound state, Am. J. Phys. 57 (1989) 85.
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1 The spectrum could be unbounded from below.
2 Readers familiar with the correspondence ψ1D(r)=rψ3D(r) between 1D problems and 3D spherically symmetric s-wave problems in QM may wonder why the even bound state ψ1D(r) that always exists in the 1D finite well potential does not yield a corresponding bound state ψ3D(r) in the 3D case? Well, it turns out that the corresponding solution ψ3D(r)=ψ1D(r)r is singular at r=0 (where the potential is constant), and hence must be discarded.
