2014-10-29 08:19:03 8 Comments

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:$$V(r)=-V_0\quad(r<a)$$$$ V(r)=0\quad(r>a)$$ and found that there is no bound state when $V_0a^2<\pi^2\hbar^2/8m$.

So what's the condition that we have at least one bound state for 3D and 2D?

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## 3 comments

## @Qmechanic 2014-10-29 14:13:51

The precise theorem is the following, cf. e.g. Ref. 1.

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.$$ V~\longrightarrow~ \tilde{V}~:=~\frac{V}{mc^2}, \qquad {\bf r}~\longrightarrow~\tilde{\bf r}~:=~ \frac{mc}{\hbar}{\bf r},\tag{3} $$

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

$$ \psi_{\varepsilon}(r)~=~e^{-f_{\varepsilon}(r)}~\nearrow ~e^{-1}\quad\text{for}\quad \varepsilon ~\searrow ~0^{+} , \tag{4}$$

where

$$ f_{\varepsilon}(r)~:=~ (r+1)^{\varepsilon} ~\searrow ~1\quad\text{for}\quad \varepsilon ~\searrow ~0^{+} \tag{5} $$

$r$-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:

$$0~\leq~\qquad\langle\psi_{\varepsilon}|\psi_{\varepsilon} \rangle ~=~ \int_{\mathbb{R}^n} d^nr~|\psi_{\varepsilon}(r)|^2 ~\propto~ \int_{0}^{\infty} \! dr ~r^{n-1} |\psi_{\varepsilon}(r)|^2$$ $$~\leq~ \int_{0}^{\infty} \! dr ~(r+1)^{n-1} e^{-2f_{\varepsilon}(r)} ~\stackrel{f=(1+r)^{\varepsilon}}{=}~ \frac{1}{\varepsilon} \int_{1}^{\infty}\!df~f^{\frac{n}{\varepsilon}-1} e^{-2f} ~<~\infty,\qquad \varepsilon~> ~0.\tag{6} $$

The kinetic energy vanishes

$$ 0~\leq~\qquad\langle\psi_{\varepsilon}|K|\psi_{\varepsilon} \rangle ~=~ \frac{1}{2}\int_{\mathbb{R}^n}\! d^nr~ |{\bf \nabla}\psi_{\varepsilon}(r) |^2 ~=~ \frac{1}{2}\int_{\mathbb{R}^n}\! d^nr~ \left|\psi_{\varepsilon}(r)\frac{df_{\varepsilon}(r)}{dr} \right|^2 $$ $$~\propto~ \varepsilon^2\int_{0}^{\infty}\! dr~ r^{n-1} (r+1)^{2\varepsilon-2}|\psi_{\varepsilon}(r)|^2~ \leq~\varepsilon^2 \int_{0}^{\infty} \!dr ~ (r+1)^{2\varepsilon+n-3}e^{-2f_{\varepsilon}(r)}$$ $$~\stackrel{f=(1+r)^{\varepsilon}}{=}~ \varepsilon \int_{1}^{\infty}\! df ~ f^{1+\frac{\color{Red}{n-2}}{\varepsilon}} e^{-2f} ~\searrow ~0\quad\text{for}\quad \varepsilon ~\searrow ~0^{+}, \tag{7}$$ when $\color{Red}{n\leq 2}$, while the potential energy

$$0~\geq~\qquad\langle\psi_{\varepsilon}|V|\psi_{\varepsilon} \rangle ~=~ \int_{\mathbb{R}^n} \!d^nr~|\psi_{\varepsilon}(r)|^2~V({\bf r}) $$ $$ ~\searrow ~e^{-2}\int_{\mathbb{R}^n} \!d^nr~V({\bf r})~<~0 \quad\text{for}\quad \varepsilon ~\searrow ~0^{+} ,\tag{8} $$

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

$$\exists L_0>0 \forall L\geq L_0:~~0~\leq~ \langle\psi|K|\psi\rangle ~ \stackrel{(12)}{\leq} ~-\frac{v}{3}~>~0\tag{13}$$

if $\color{Red}{n<2}$.

The un-normalized potential energy tends to a negative constant

$$\langle\psi| V|\psi \rangle ~\searrow~\int_{\mathbb{R}^n} \! \mathrm{d}^nx ~V(x)~=:~v~<~0\quad\text{for}\quad L~\to~ \infty.\tag{14}$$

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

$$\frac{\langle\psi|H|\psi\rangle}{||\psi||^2}~=~\frac{\langle\psi|K|\psi\rangle+\langle\psi|V|\psi\rangle}{||\psi||^2}~\stackrel{(13)+(15)}{\leq}~ \frac{v}{3||\psi||^2}~<~0\tag{16}$$

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.--

$^1$ 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.

## @an offer can't refuse 2014-10-31 14:36:49

Very detailed and accurate answer!

## @Qmechanic 2014-11-05 20:45:41

Related: physics.stackexchange.com/q/134719/2451

## @arivero 2015-08-26 01:14:08

Is the reciprocal true? Bound state implies negative integral of V(x)?

## @Qmechanic 2015-08-31 00:25:55

$ \uparrow$ No. Think e.g. of an 1D infinite potential well.

## @Emilio Pisanty 2017-06-21 15:32:02

@Qmechanic I'm confused by this answer. If you take a very shallow well in 1D, say, $V(x) = -V_0 \mathrm{sech}(x)$, or a finite square well of depth $V_0$, and then make $V_0$ be very small, then at some point you expect the well to stop supporting bound states. Can you clarify how this fits in with the theorem?

## @Emilio Pisanty 2017-06-21 17:07:25

@Qmechanic Hmmm. It seems I was wrong about that intuition, then - the ground-state wavefunction becomes arbitrarily flat but it never gets pushed off the well, it seems. Oh well, live & learn. Thanks!

## @Qmechanic 2017-06-21 17:13:33

@Emilio Pisanty: If it is any consolation: we have all been there. Look at v1 of my answer :)

## @Johannes 2014-10-29 16:52:12

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.

## @Adam 2014-10-29 17:50:28

This argument is kind of strange, since $R_0\gg r_0$, and the potential has range $r_0$. The the potential energy should be of order $1/r_0^D$, shouldn't it ?

## @Johannes 2014-10-29 17:58:30

@Adam - the portion of the wave function 'covering' the potential scales as $(r_0/R_0)^D$, hence the $1/R_0^D$ scaling of the potential energy.

## @Adam 2014-10-29 21:30:10

ok, that makes sense. Maybe that could be added in the answer ?

## @Johannes 2014-10-30 02:30:18

@Adam - good suggestion. Made the scaling of both energy components more explicit.

## @Mateus Sampaio 2014-10-29 18:19:02

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 it's 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.

## @an offer can't refuse 2014-10-30 05:57:18

The last expression is the condition for finite potential well in $n$ dimension? What is $a$ in this expression, the region that potential is $-V_0$ when $r<a$?

## @Mateus Sampaio 2014-10-30 10:38:47

No, in the last sentences, $V$ is a general potential (the exact condition is that itmust be relatively compact with respect to the kinetic energy $H_0=-\nabla^2$), and if it satisfies the condition for a given $a>0$, than there exists a bound state.

## @an offer can't refuse 2014-10-30 12:47:25

I see, this is a sufficient condition; for I can not see this is always true for any negative potential in 2D

## @Mateus Sampaio 2014-10-30 13:26:14

You are right, this condition is always true, only for the one-dimensional case, not the two-dimensional. For this case, you can do an analogous derivation as for the three dimensional case and check that there's no restrictions or use other functions different from the gaussian to check the existence of the bound state, like in the answer and references by @Qmechanic.