A spherical potential well is defined by V(r)= 0 for r 0 .

A spherical potential well is defined by V(r)= 0 for ...

A spherical potential well is defined by $$ V(r)= \begin{cases}0 & \text { for } r<a \\ V_{0} & \text { otherwise }\end{cases} $$ where $V_{0}>0 .$ Consider a stationary state with angular-momentum quantum number $l .$ By writing the wavefunction $\psi(\mathbf{x})=R(r) \mathrm{Y}_{l}^{m}(\theta, \phi)$ and using $p^{2}=p_{r}^{2}+\hbar^{2} L^{2} / r^{2}$, show that the state's radial wavefunction $R(r)$ must satisfy $$ -\frac{\hbar^{2}}{2 m}\left(\frac{\mathrm{d}}{\mathrm{d} r}+\frac{1}{r}\right)^{2} R+\frac{l(l+1) \hbar^{2}}{2 m r^{2}} R+V(r) R=E R $$ Show that in terms of $S(r) \equiv r R(r)$, this can be reduced to $$ \frac{\mathrm{d}^{2} S}{\mathrm{~d} r^{2}}-l(l+1) \frac{S}{r^{2}}+\frac{2 m}{\hbar^{2}}(E-V) S=0 $$ Assume that $V_{0}>E>0 .$ For the case $l=0$ write down solutions to this equation valid at (a) $r<a$ and (b) $r>a$. Ensure that $R$ does not diverge at the origin. What conditions must $S$ satisfy at $r=a ?$ Show that these conditions can be simultaneously satisfied if and only if a solution can be found to $k \cot k a=-K$, where $\hbar^{2} k^{2}=2 m E$ and $\hbar^{2} K^{2}=2 m\left(V_{0}-E\right) .$ Show graphically that the equation can only be solved when $\sqrt{2 m V_{0}} a / \hbar>\pi / 2$. Compare this result with that obtained for the corresponding one-dimensional potential well. The deuteron is a bound state of a proton and a neutron with zero angular momentum. Assume that the strong force that binds them produces a sharp potential step of height $V_{0}$ at interparticle distance $a=$ $2 \times 10^{-15} \mathrm{~m}$. Determine in $\mathrm{MeV}$ the minimum value of $V_{0}$ for the deuteron to exist. Hint: remember to consider the dynamics of the reduced particle.

A spherical potential well is defined by
$$
V(r)= \begin{cases}0 & \text { for } r<a \\ V_{0} & \text { otherwise }\end{cases}
$$
where $V_{0}>0 .$ Consider a stationary state with angular-momentum quantum number $l .$ By writing the wavefunction $\psi(\mathbf{x})=R(r) \mathrm{Y}_{l}^{m}(\theta, \phi)$ and using $p^{2}=p_{r}^{2}+\hbar^{2} L^{2} / r^{2}$, show that the state's radial wavefunction $R(r)$ must satisfy
$$
-\frac{\hbar^{2}}{2 m}\left(\frac{\mathrm{d}}{\mathrm{d} r}+\frac{1}{r}\right)^{2} R+\frac{l(l+1) \hbar^{2}}{2 m r^{2}} R+V(r) R=E R
$$
Show that in terms of $S(r) \equiv r R(r)$, this can be reduced to
$$
\frac{\mathrm{d}^{2} S}{\mathrm{~d} r^{2}}-l(l+1) \frac{S}{r^{2}}+\frac{2 m}{\hbar^{2}}(E-V) S=0
$$
Assume that $V_{0}>E>0 .$ For the case $l=0$ write down solutions to this equation valid at (a) $r<a$ and (b) $r>a$. Ensure that $R$ does not diverge at the origin. What conditions must $S$ satisfy at $r=a ?$ Show that these conditions can be simultaneously satisfied if and only if a solution can be found to $k \cot k a=-K$, where $\hbar^{2} k^{2}=2 m E$ and $\hbar^{2} K^{2}=2 m\left(V_{0}-E\right) .$ Show graphically that the equation can only be solved when $\sqrt{2 m V_{0}} a / \hbar>\pi / 2$. Compare this result with that obtained for the corresponding one-dimensional potential well.

The deuteron is a bound state of a proton and a neutron with zero angular momentum. Assume that the strong force that binds them produces a sharp potential step of height $V_{0}$ at interparticle distance $a=$ $2 \times 10^{-15} \mathrm{~m}$. Determine in $\mathrm{MeV}$ the minimum value of $V_{0}$ for the deuteron to exist. Hint: remember to consider the dynamics of the reduced particle.

Key Concepts

Graphical Solution and Critical Potential Depth

The transcendental quantization condition is often solved graphically to illustrate how the discrete energy levels emerge. By plotting the functions on both sides of the equation, one can see that a solution exists only when the potential well is sufficiently deep or wide, i.e., when a parameter like ?(2mV?) a/? exceeds a critical value (in this case, ?/2). This concept demonstrates that a minimum potential depth is required to support a bound state, analogous to the one-dimensional case.

Transcendental Quantization Condition

The matching conditions at the boundary typically lead to a transcendental equation, such as k cot(ka) = -K in this problem, where k and K are related to the energy of the state in the well and outside it respectively. This condition arises from equating the logarithmic derivatives of the solutions in the two regions, and it quantizes the allowed energy levels. Only certain discrete energies satisfy this relation, providing the spectrum of bound states.

Boundary Conditions and Matching Solutions

In quantum mechanics, the physical acceptability of a solution under a potential step requires that both the wavefunction and its derivative (appropriately normalized by any factors like r in spherical coordinates) are continuous at the boundary of the potential (r = a). These continuity conditions lead to matching equations that connect the solutions inside and outside the well, ensuring that the probability current is conserved and the solution remains finite and well-behaved.

Reduced Mass in Two-Body Systems

When dealing with bound states of two particles, such as the proton and neutron in the deuteron, it is important to account for their dynamics using the reduced mass. The use of reduced mass transforms the two-body problem into an equivalent one-body problem, reflecting the relative motion between the particles. This is essential for correctly determining the energy levels and minimum potential depth necessary for binding.

Separation of Variables in Spherical Coordinates

In problems with spherical symmetry, the wavefunction is typically separated into a product of a radial part and an angular part, using the ansatz ?(x) = R(r) Y??(?, ?). This separation leverages the orthogonality and completeness of spherical harmonics, turning the three-dimensional Schrödinger equation into a set of ordinary differential equations, one of which is the radial equation that encapsulates the dependence on the distance from the origin.

Spherical Potential Well

A spherical potential well is a quantum mechanical potential that depends only on the radial coordinate r. It is defined such that inside a sphere (r < a) the potential takes one value (typically zero) and outside (r > a) it takes another value. This symmetry simplifies the analysis, as the Schrödinger equation can be separated into radial and angular parts, and it models systems where the confining force is isotropic.

Radial Schrödinger Equation and Reduced Wavefunction

The radial part of the Schrödinger equation, when accounting for the kinetic energy operator in spherical coordinates, acquires additional terms related to angular momentum, notably l(l+1)/r². By defining the reduced radial wavefunction S(r) = r R(r), the equation is transformed into a simpler second-order differential equation without a first derivative term. This transformation is useful for eliminating the singular behavior at r = 0 and obtaining a form that resembles the one-dimensional Schrödinger equation.

Transcript

00:01 Okay, so one solution of this short -danger's time -independent equation can be written as this is z -zi and this is equal to x into x y of y and z of z.

00:23 And substituting this expression into the short injures equation and dividing by this on both side of the equation we have that one over x into x partial square x into x over partial x square plus one over y into y times partial square y into y divided by partial y square plus 1 over z into z times partial square z times z over partial z square it is equal to minus 8 pi square m over h squared times e right so this yields the partial square x over x divided by partial x square plus e sub x x into x is equal to zero and this is equal to this is equal to partial square y into y divided by partial y square plus e y y y into y this is equal to 0 and partial square z times z over partial with respect to z square plus e z in z into z is equal to zero right and we are we are x e y and e z are constants and this satisfies that e x plus e y plus ez is equal to 8 pi square m over 8 square times e right and by solving the above three equations we have that x or in terms of x is equal to cx e power a aorta square root e e x time x right and and plus d x e power minus aorta square root e x time x right and y of y is equal to c sub y power iota square root e sub y times y plus d sub y power e power a iota square root e.

03:19 Of y times y and this is z into z is equal to c z e power iota square root e z times z plus d z e power minus iota square root e z times z right and finally this finally putting the values we have that this is equal to c x e power iota square root e sub x times x plus d sub x e power minus iota x x times x and this is c y e power iota square root ey times y times y times x right and plus okay so this is at times c yota square root ey times y plus d, y, e power, minus iota square root, ey times y.

04:47 So we are just putting the values here, right? and here this is times cz, e power, iota square root, ez times z plus dze power minus iota square root, ez, e power, minus iota square root, e z times z right and here uh here uh these uh x and d x c y d y and c y and c y and c d g are constants right these are constants okay and uh by boundary conditions this i is equal to zero at x is equal to 0 and it gives c sub x is equal to minus d sub x and by the boundary condition si is equal to uh zero right and y is equal to zero it goes c y is equal to minus d sub y and by boundary conditions si is equal to zero at uh z is equal to zero so it gives us c z is equal to minus d z so we have that this is equal to this is equal to cx c y c z and this is e power iota square root x power times x minus e power minus aota square root uh e y this is e x right so this is e x times x and this is e power aota square root e sub y times y minus e y power minus iota square root e sub y times y right and this is times e power iota square root uh e power z times z minus e power minus iota square root uh e sub z time z right and finally this is equal to uh this is equal to finally this is equal to a sine square root x x x x times x so this is x right sign up square root e sub y times y and sign up square root e sub z times z right and using the above expression this expression for the boundary condition which is this is equal to 0 at x is equal to l x it gives x x is equal to l pi over l x whole square right and here here by the boundary conditions the above boundary conditions which is x is equal to l z and this gives us e z is equal to n pi over l sub z whole square right and we are l r n is equal to 0 1 2 and up to so on so we have l pi over l x whole square plus r pi over l sub y whole square plus n pi over l sub y whole square plus n pi over l sub z whole square this is equal to 8 pi square times m over it square times e so e is equal to h square over eight times m into l square or l sub x square plus r square over l sub y square plus n square over l sub z square and this is equal to so when let when l x is equal to l y and this is equal to l y and this is equal to l z and this is equal to l.

09:23 So e will be equal to each square over 8 times l square and this is l square plus r square plus n square and if e is equal to e0 par l is equal to 1, r is equal to n is equal to 0...

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