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How can you approximate the number of bound states in a potential well of depth $-V_0$ and width $-a$ to $+a$ using uncertainty principle?

Qmechanic
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robert froast
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2 Answers2

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The $n$-th state wave function has $n-1$ zeros. In a well of $2a$ width, the uncertainty of coordinate between two zeros is $$ \Delta x \sim \frac{a}{n}. $$ Hence the uncertainty of momentum is $$ \Delta p \sim \frac{\hbar}{\Delta x} \sim \frac{\hbar n}{a}. $$ Estimate of the $n$-th's state energy is $$ E_n \sim -V_0 +\frac{(\Delta p)^2}{2m} \sim -V_0 + \frac{\hbar^2n^2}{2ma^2} $$ For bound states, $E_n <0$ and we obtain an estimation of the number of bound states $$ n_b \sim \sqrt{\frac{ma^2V_0}{\hbar^2}}.\quad (1) $$

Update. Let's show the applicability of (1) in a simple example. Consider a truncated harmonic oscillator well with the potential $$ U(x) = \left\{ \begin{array}{ccc} \frac{m\omega^2 x^2}{2}-V_0 & \mbox{if} & |x| < a\\ 0 & \mbox{if} & |x| > a \end{array} \right. $$ From the $E_n = \hbar\omega(n+1/2)$ formula, which is valid for the untruncated harmonic oscillator, one expects a number of bound states $$ n_b\approx \frac{E}{\hbar\omega}.\quad (2) $$ If the vertical walls at $|x|=a$ are absent, then the parameters $a$ and $V_0$ are related to each other in the following way: $$ \frac{m\omega^2a^2}{2} = V_0\quad\longrightarrow\quad a\sim\sqrt{\frac{V_0}{m\omega^2}}. $$ Substitution of the last relation into (1) leads to $$ n_b \sim \frac{E}{\hbar\omega}. $$ This expression is equivalent to (2).

Gec
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    What if there is a simple harmonic potential of height -V and we have to give an approximate idea of the number of bound states? – robert froast May 19 '21 at 09:13
  • How does the simple harmonic potential of height $-V_0$ look like precisely? Is it $U(x) = m\omega^2 x^2/2$? Isn't for this case $a = \infty$, $V_0 = \infty$ and $n_b = \infty$? – Gec May 19 '21 at 10:31
  • May be it is just V= -mω2x2/2 and then how will you approach this problem? – robert froast May 19 '21 at 11:09
  • Thanks so much for updating it. Please include for the case of delta function also. It would be very helpful. Thankyou so much. – robert froast May 19 '21 at 15:00
  • For the delta-function potential $U(x) = -\alpha \delta(x)$ we have $V_0 = \alpha/2a$ and $a\to 0$. Hence, (1) gives $0$ in this case while it is known that there is one bound state for any $\alpha > 0$. Delta-function is a singular potential. I am not sure if ordinary arguments are valid for it. Provided the simplicity and crudeness of approximations made you may consider $0\approx 1$. – Gec May 19 '21 at 15:20
  • For the case of harmonic oscillator it is coming V0/ℏω instead of E/ℏω ? – robert froast May 20 '21 at 02:20
  • I was thinking about the free particle and there I realize that bound states can't be exist as V = 0 and E can never be less than 0. Now how can you estimate the case for the hydrogen atom V(r)=−a/r using uncertainty principle? – robert froast May 20 '21 at 02:46
  • @robertfroast in the case of the harmonic oscillator, I've cosidered the harmonic well of depth $V_0$. For the ordinary harmonic oscillator, you can take $E$ instead of $V_0$. – Gec May 20 '21 at 14:32
  • @robertfroast question about hydrogen atom is another one. Initially, you asked about the 1d potential well of finite depth. – Gec May 20 '21 at 14:35
  • But how to do for hydrogen atom any idea? – robert froast May 21 '21 at 01:24
  • @robertfroast I don't know how to generalize simple arguments like HUP and wave function's zeros for the hydrogen atom. In this case, you need to use the WKB method, just like Qmechanic wrote. – Gec May 21 '21 at 07:56
  • Can you please show it using WKB approximation? – robert froast May 21 '21 at 07:57
  • I can't. I don't understand your question about the hydrogen atom. What exactly do you want to estimate? And I don't know how to apply the WKB for the hydrogen atom. – Gec May 21 '21 at 12:15
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  1. The HUP is usually only used to give a (often crude) estimate for the ground state energy, cf. e.g. my Phys.SE answer here.

  2. To find the number of bound states, short of solving the TISE exactly, one may rely on the WKB estimate that there's 1 bound state per $h$-volume in phase space, i.e. $$ N ~\sim~\frac{2p_{\max}\cdot 2x_{\max}}{h} ~=~\frac{2\sqrt{2m|V_0|}\cdot 2a}{h},$$ cf. e.g. my Phys.SE answer here.

Qmechanic
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