Why the statement "there exist at least one bound state for negative/attractive potential" doesn't hold for 3D case?
The precise theorem is the following, cf. e.g. Ref. 1.
Theorem 1: Given a non-positive (=attractive) potential $V\leq 0$ with negative spatial integral $$ v~:=~\int_{\mathbb{R}^n}\! d^n r~V({\bf r}) ~<~0 ,\tag{1} $$ then there exists a bound state$^1$ with energy $E<0$ for the Hamiltonian $$\begin{align} H~=~&K+V, \cr K~=~& -\frac{\hbar^2}{2m}{\bf \nabla}^2\end{align}\tag{2} $$ if the spatial dimension $\color{Red}{n\leq 2}$ is smaller than or equal to two.
The theorem 1 does not hold for dimensions $n\geq3$. E.g. it can be shown that already a spherically symmetric finite well potential does not$^2$ always have a bound state for $n\geq3$.
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 $c$, $\hbar$ and $m$ to render all physical variables dimensionless, e.g.
$$\begin{align} V~\longrightarrow~& \tilde{V}~:=~\frac{V}{mc^2}, \cr {\bf r}~\longrightarrow~&\tilde{\bf r}~:=~ \frac{mc}{\hbar}{\bf r},\end{align}\tag{3} $$
and so forth. The tildes are dropped from the notation from now on. (This effectively corresponds to setting the constants $c$, $\hbar$ and $m$ to 1.)
Consider a 1-parameter family of trial wavefunctions
$$\begin{align} \psi_{\varepsilon}(r)~=~&e^{-f_{\varepsilon}(r)}~\nearrow ~e^{-1}\cr &\text{for}\quad \varepsilon ~\searrow ~0^{+} , \end{align}\tag{4}$$
where
$$\begin{align} f_{\varepsilon}(r)~:=~& (r+1)^{\varepsilon} ~\searrow ~1\cr &\text{for}\quad \varepsilon ~\searrow ~0^{+}\end{align} \tag{5} $$
$r$-pointwise. Here the $\nearrow$ and $\searrow$ symbols denote increasing and decreasing limit processes, respectively. E.g. eq. (4) says in words that for each radius $r \geq 0$, the function $\psi_{\varepsilon}(r)$ approaches monotonically the limit $e^{-1}$ from below when $\varepsilon$ approaches monotonically $0$ from above.
It is easy to check that the wavefunction (4) is normalizable:
$$\begin{align}0~\leq~~&\langle\psi_{\varepsilon}|\psi_{\varepsilon} \rangle\cr ~=~~& \int_{\mathbb{R}^n} d^nr~|\psi_{\varepsilon}(r)|^2 \cr ~\propto~~& \int_{0}^{\infty} \! dr ~r^{n-1} |\psi_{\varepsilon}(r)|^2\cr ~\leq~~& \int_{0}^{\infty} \! dr ~(r+1)^{n-1} e^{-2f_{\varepsilon}(r)} \cr ~\stackrel{f=(1+r)^{\varepsilon}}{=}&~ \frac{1}{\varepsilon} \int_{1}^{\infty}\!df~f^{\frac{n}{\varepsilon}-1} e^{-2f}\cr ~<~~&\infty,\qquad \varepsilon~> ~0.\end{align}\tag{6} $$
The kinetic energy vanishes
$$\begin{align} 0~\leq~~&\langle\psi_{\varepsilon}|K|\psi_{\varepsilon} \rangle \cr ~=~~& \frac{1}{2}\int_{\mathbb{R}^n}\! d^nr~ |{\bf \nabla}\psi_{\varepsilon}(r) |^2\cr ~=~~& \frac{1}{2}\int_{\mathbb{R}^n}\! d^nr~ \left|\psi_{\varepsilon}(r)\frac{df_{\varepsilon}(r)}{dr} \right|^2 \cr ~\propto~~& \varepsilon^2\int_{0}^{\infty}\! dr~ r^{n-1} (r+1)^{2\varepsilon-2}|\psi_{\varepsilon}(r)|^2\cr ~\leq~~&\varepsilon^2 \int_{0}^{\infty} \!dr ~ (r+1)^{2\varepsilon+n-3}e^{-2f_{\varepsilon}(r)}\cr ~\stackrel{f=(1+r)^{\varepsilon}}{=}&~ \varepsilon \int_{1}^{\infty}\! df ~ f^{1+\frac{\color{Red}{n-2}}{\varepsilon}} e^{-2f}\cr ~\searrow ~~&0\quad\text{for}\quad \varepsilon ~\searrow ~0^{+},\end{align} \tag{7}$$ when $\color{Red}{n\leq 2}$, while the potential energy
$$\begin{align}0~\geq~&\langle\psi_{\varepsilon}|V|\psi_{\varepsilon} \rangle\cr ~=~& \int_{\mathbb{R}^n} \!d^nr~|\psi_{\varepsilon}(r)|^2~V({\bf r}) \cr ~\searrow ~& e^{-2}\int_{\mathbb{R}^n} \!d^nr~V({\bf r})~<~0 \cr &\text{for}\quad \varepsilon ~\searrow ~0^{+} ,\end{align}\tag{8} $$
remains non-zero due to assumption (1) and Lebesgue's monotone convergence theorem.
Thus by choosing $ \varepsilon \searrow 0^{+}$ smaller and smaller, the negative potential energy (8) beats the positive kinetic energy (7), so that the average energy $\frac{\langle\psi_{\varepsilon}|H|\psi_{\varepsilon}\rangle}{\langle\psi_{\varepsilon}|\psi_{\varepsilon}\rangle}<0$ eventually becomes negative for the trial function $\psi_{\varepsilon}$. A bound state$^1$ 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 $\color{Red}{n\leq 2}$. $\Box$
Simpler proof for $\color{Red}{n<2}$: Consider an un-normalized (but normalizable) Gaussian test/trial wavefunction
$$\psi(x)~:=~e^{-\frac{x^2}{2L^2}}, \qquad L~>~0.\tag{9}$$
Normalization must scale as
$$||\psi|| ~\stackrel{(9)}{\propto}~ L^{\frac{n}{2}}.\tag{10}$$
The normalized kinetic energy scale as
$$0~\leq~\frac{\langle\psi| K|\psi \rangle}{||\psi||^2} ~\propto ~ L^{-2}\tag{11}$$
for dimensional reasons. Hence the un-normalized kinetic scale as
$$0~\leq~\langle\psi| K|\psi \rangle ~\stackrel{(10)+(11)}{\propto} ~ L^{\color{Red}{n-2}}.\tag{12}$$
Eq. (12) means that
$$\begin{align}\exists L_0>0 \forall L\geq L_0:~~0~\leq~& \langle\psi|K|\psi\rangle\cr ~ \stackrel{(12)}{\leq} ~&-\frac{v}{3}~>~0\end{align}\tag{13}$$
if $\color{Red}{n<2}$.
The un-normalized potential energy tends to a negative constant
$$\begin{align}\langle\psi| V|\psi \rangle ~\searrow~&\int_{\mathbb{R}^n} \! \mathrm{d}^nx ~V(x)~=:~v~<~0\cr &\quad\text{for}\quad L~\to~ \infty.\end{align}\tag{14}$$
Eq. (14) means that
$$\exists L_0>0 \forall L\geq L_0:~~ \langle\psi| V|\psi\rangle ~\stackrel{(14)}{\leq}~ \frac{2v}{3} ~<~ 0.\tag{15}$$
It follows that the average energy
$$\begin{align}\frac{\langle\psi|H|\psi\rangle}{||\psi||^2} ~=~~&\frac{\langle\psi|K|\psi\rangle+\langle\psi|V|\psi\rangle}{||\psi||^2}\cr ~\stackrel{(13)+(15)}{\leq}&~ \frac{v}{3||\psi||^2}~<~0\end{align}\tag{16}$$
of trial function must be negative for a sufficiently big finite $L\geq L_0$ if $\color{Red}{n<2}$. Hence the ground state energy must be negative (possibly $-\infty$). $\Box$
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 $\psi_{1D}(r)=r\psi_{3D}(r)$ between 1D problems and 3D spherically symmetric $s$-wave problems in QM may wonder why the even bound state $\psi_{1D}(r)$ that always exists in the 1D finite well potential does not yield a corresponding bound state $\psi_{3D}(r)$ in the 3D case? Well, it turns out that the corresponding solution $\psi_{3D}(r)=\frac{\psi_{1D}(r)}{r}$ is singular at $r=0$ (where the potential is constant), and hence must be discarded.
Why does the statement "any negative potential supports a bound state" hold in 1D, but not in 3D?
In short, this is because for a bound state to occur, any positive kinetic energy needs to be fully offset by a negative potential energy. Achieving a large negative potential energy requires the particle to be localized in the volume where the potential is negative, while a low kinetic energy requires the particle to 'spread out'. This battle between potential and kinetic energy can always be won in 1D, but not in high spatial dimensions (3D or higher dimensions).
The above can be made explicit in a simple scaling argument:
Suppose in $D$ dimensions a negative potential with range $r_0$ has a bound state described by a wave function that extends over a range $R_0$. In the limit $R_0 \gt \gt r_0$ the kinetic energy $K$ scales as $\hbar^2/m R_0^2$ and can therefore be made arbitrarily small.
This would suggest that indeed any slightly negative potential would suffice to create a total energy $K+V$ that is negative, and hence a bound state. However, because the wave function extends far beyond the range of the potential, the potential energy $V$ scales with $(r_0/R_0)^D$. For $D<2$ and $R_0$ sufficiently large, the negative potential energy (~ $R_0^{-D}$) can always beat the positive kinetic energy (~ $R_0^{-2}$). For $D>2$, such is not the case.
To study bound states, we have to find solutions to the Schrödinger time-independent equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi$$ Using separation of variables, in spherical coordinates $$\psi(r,\theta,\phi)=Y^m_l(\theta,\phi)\frac{u(r)}{r},$$ where $Y^m_l(\theta,\phi)$ are the spherical harmonics, the radial part can be shown after substitution and some algebra to yield $$-\frac{\hbar^2}{2m}\frac{d^2u(r)}{dr^2}+\left[\frac{\hbar^2 l(l+1)}{2mr^2}+V(r)\right]u(r)=Eu(r)$$ that is, it reduces to an equation of the same form of a one-dimensional system but with effective potential $$V_{\text{eff}}(r)=\frac{\hbar^2 l(l+1)}{2mr^2}+V(r)$$ and boundary condition $u(0)=0$, otherwise $\psi$ would have a singularity of the form $r^{-1}$, which is not accepted (see for example, Ballentine's section 4.5). In practice, we have a repulsive potential combined with the negative potential of the well. So it's not true that in general, we'll always have a bound state, since the potential needs to be "strong enough to make" a bound state against the repulsive one. The better case is when $l=0$. Even though, the condition $u(0)=0$ gives the solution $$u(r)=\left\{\begin{array}{ll}C\frac{\sin kr}{\sin ka},& r\leq a\\ Ce^{-\alpha(r-a)},&r>a\end{array}\right.$$ where $C$ is a normalization constant, $k^2=-\dfrac{2(E-V_0)m}{\hbar^2}$ and $\alpha^2=-\dfrac{2Em}{\hbar^2}$. Imposing continuity of $u$ and $u'$, gives a condition for $\alpha$ for the existance of a bound state, which translates in the one you put in the question.
For the one-dimensional (and two-dimensional) case, the conditions of continuity in the wave function and its derivative for a square-well can always be satisfied for at least one value of $E$.
In fact, it can be shown that for any potential $V(x)\leq0$ that is negative in an interval, has a negative eigenvalue, i.e. a bound state. A better condition even gives that if $$\int_\Bbb{R^n}V(x)e^{-2a^2x^2}dx<-na^{2-n}\left(\frac{\pi}{2}\right)^{n/2},$$ holds for a well-behaved $V$, there is at least one bound state. For the proof check section 11.4.4 of this book.