Hey there! Let's sit down and really break down these core topics from your syllabus. These "simple systems" are the absolute bedrock of quantum chemistry. Why? Because they are the first problems we can solve exactly with the Schrödinger equation. Understanding them isn't just about memorizing formulas; it's about seeing how the principles of quantum mechanics (like wavefunctions and boundary conditions) naturally lead to the weird and wonderful rules of the quantum world, like quantized energy.
We'll go through each one, step-by-step, just like we would in a tutoring session. Our main tool will be the Time-Independent Schrödinger Equation (TISE):
$$\hat{H}\psi = E\psi$$...which, for a single particle in one dimension, is:
$$- \frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)$$Let's get started.
This is the "hello, world!" of quantum mechanics. We're modeling a particle (like an electron) that is completely free to move, but only within a strictly defined region (a "box"). Think of an electron trapped on a tiny wire.
First, we have to define the potential energy, $V(x)$:
Because the particle can't exist where $V=\infty$, its wavefunction $\psi(x)$ must be zero outside the box. This gives us our all-important boundary conditions:
Inside the box, $V(x)=0$, so the Schrödinger equation becomes much simpler:
$$- \frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} = E\psi(x)$$Let's rearrange this into a standard form you'll see a lot in math:
$$\frac{d^2\psi(x)}{dx^2} = -\frac{2mE}{\hbar^2}\psi(x)$$To make this look nicer, we'll define a constant $k^2 = \frac{2mE}{\hbar^2}$. (We use $k^2$ because we know $E$ must be positive, so $k^2$ is positive).
Our equation is now: $\frac{d^2\psi}{dx^2} = -k^2\psi$.
This is a classic second-order differential equation. The general solution (the family of all functions that satisfy this equation) is:
$\psi(x) = A\sin(kx) + B\cos(kx)$
Now, we apply those boundary conditions we just talked about. This is where the magic happens!
The sine function is zero at integer multiples of $\pi$. Therefore, the quantization condition is:
$kL = n\pi$, where $n = 1, 2, 3, ...$
Tutor's Note: Why not $n=0$? Because if $n=0$, then $k=0$, and $\psi(x) = A\sin(0) = 0$. Again, no particle. We also don't need negative $n$ values (like -1, -2, ...) because $\sin(-n\pi x/L) = -\sin(n\pi x/L)$. This just flips the wavefunction, but the probability (the square of the function) is the same, so it's not a unique physical state. We stick with $n > 0$.
Now we finally see where quantized energy comes from. We just found $k = n\pi/L$. Let's plug this back into our definition of $k^2$:
$k^2 = \frac{n^2\pi^2}{L^2} = \frac{2mE_n}{\hbar^2}$
Now, we just solve for $E_n$ (we add the subscript 'n' because the energy now depends on $n$):
$$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$$Since $\hbar = h / (2\pi)$, we can substitute that in to get the most common form:
$$E_n = \frac{n^2\pi^2(h^2 / 4\pi^2)}{2mL^2} = \frac{n^2h^2}{8mL^2}$$This is our first huge result! The energy of the particle is quantized. It can only exist in discrete levels $E_1$, $E_2$, $E_3$, etc., which are proportional to $n^2$. It cannot have any energy between these values.
Our final wavefunctions are $\psi_n(x) = A\sin(n\pi x / L)$. We find the constant $A$ by normalizing the wavefunction. This means we set the total probability of finding the particle in the box to 1:
$\int_0^L \psi_n^*(x) \psi_n(x) dx = \int_0^L A^2 \sin^2\left(\frac{n\pi x}{L}\right) dx = 1$
Solving this standard integral gives $A^2(L/2) = 1$, so the normalization constant is $A = \sqrt{2/L}$.
The normalized wavefunctions are:$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$$
Let's interpret these solutions:
This is a natural extension of the 1D case. Now, the particle is trapped in a 3D box with side lengths $L_x, L_y, L_z$. The potential $V$ is 0 inside and $\infty$ outside.
The TISE becomes 3-dimensional, using partial derivatives:
$$- \frac{\hbar^2}{2m} \left( \frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2} \right) = E\psi$$(where $\psi = \psi(x,y,z)$)
This looks terrifying, but we can solve it with a powerful technique called separation of variables. We propose a solution that is a product of three independent 1D functions:
$\psi(x,y,z) = X(x) \cdot Y(y) \cdot Z(z)$
If we substitute this into the TISE and then divide the whole equation by $\psi = XYZ$, we get a beautiful simplification:
$$\left[ -\frac{\hbar^2}{2m} \frac{1}{X(x)} \frac{d^2X(x)}{dx^2} \right] + \left[ -\frac{\hbar^2}{2m} \frac{1}{Y(y)} \frac{d^2Y(y)}{dy^2} \right] + \left[ -\frac{\hbar^2}{2m} \frac{1}{Z(z)} \frac{d^2Z(z)}{dz^2} \right] = E$$Think about this: the first term only depends on $x$, the second only on $y$, and the third only on $z$. For their sum to equal a constant ($E$) for all $x,y,z$, each term itself must be a constant.
So, the total energy is $E = E_x + E_y + E_z$.
And look at the equation for $X(x)$: $-\frac{\hbar^2}{2m} \frac{d^2X(x)}{dx^2} = E_x X(x)$. This is exactly the 1D particle-in-a-box equation! We've broken the 3D problem into three separate 1D problems.
We can just write down the solutions by analogy with the 1D case. The particle now needs three quantum numbers ($n_x, n_y, n_z$) to define its state.
The total energy is:
$$E_{n_x, n_y, n_z} = E_x + E_y + E_z = \frac{h^2}{8m} \left( \frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2} \right)$$The total wavefunction is:
$$\psi_{n_x, n_y, n_z}(x,y,z) = \sqrt{\frac{8}{L_x L_y L_z}} \sin\left(\frac{n_x \pi x}{L_x}\right) \sin\left(\frac{n_y \pi y}{L_y}\right) \sin\left(\frac{n_z \pi z}{L_z}\right)$$Now for a new, crucial concept: degeneracy. This occurs when different states (i.e., different wavefunctions, different sets of quantum numbers) have the exact same energy.
This is best seen in a cubic box, where $L_x = L_y = L_z = L$. The energy equation simplifies to:
$$E_{n_x, n_y, n_z} = \frac{h^2}{8mL^2} (n_x^2 + n_y^2 + n_z^2)$$Let's look at the first few energy levels:
This "symmetry degeneracy" is a direct result of the box's dimensions being identical. If $L_x \ne L_y \ne L_z$, the degeneracy would be "lifted" or broken.
This is arguably the most important "simple system" in chemistry. Why? Because it's the fundamental model for bond vibrations. Near the equilibrium bond length, the potential energy of a chemical bond looks very much like a parabola.
Classically, a harmonic oscillator (like a mass on a spring) follows Hooke's Law: the restoring force is $F = -kx$, where $k$ is the force constant (a stiff bond has a large $k$).
The potential energy for this system is $V(x) = \frac{1}{2}kx^2$. This is our parabola.
Plugging this into the 1D TISE gives the Schrödinger equation for the SHO:
$$- \frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} + \frac{1}{2}kx^2\psi(x) = E\psi(x)$$Solving this equation is significantly more difficult than the PIB (it's often solved with "ladder operators" or a power series method, which are beyond this review). For an MSc course, you need to know the results and their interpretations intimately.
Just like the PIB, applying the boundary conditions (this time, $\psi(x)$ must go to 0 as $x \to \pm\infty$) restricts the energies to discrete values.
The allowed energy levels for the 1D simple harmonic oscillator are:
$$E_v = \left(v + \frac{1}{2}\right) \hbar\omega \quad \text{where} \quad v = 0, 1, 2, ...$$Let's unpack this:
Tutor's Note: This is a huge result. Let's compare it to the PIB:
- Zero-Point Energy: The lowest energy state is $v=0$. Its energy is $E_0 = (0 + 1/2)h\nu = \frac{1}{2}h\nu$. Like the PIB, the SHO has a non-zero ground state energy. The molecule never stops vibrating.
- Energy Spacing: What's the gap between adjacent levels?$\Delta E = E_{v+1} - E_v = \left[(v+1) + \frac{1}{2}\right]h\nu - \left[v + \frac{1}{2}\right]h\nu = h\nu$.The energy levels are equally spaced! This is completely different from the PIB, where $E_n \propto n^2$ and the gaps get wider. This "equal-spacing" rule is the basis for understanding fundamental IR and Raman spectroscopy (where the selection rule is $\Delta v = \pm 1$).
The SHO wavefunctions are a product of three parts: a normalization constant ($N_v$), a special set of polynomials called Hermite polynomials ($H_v(y)$), and a Gaussian function ($e^{-y^2/2}$), where $y$ is a simplified coordinate $y = (\sqrt{mk}/\hbar)^{1/2} x$.
$\psi_v(y) = N_v H_v(y) e^{-y^2/2}$
You should be familiar with the first few:
Just like the PIB, the state $v$ has $v$ nodes.
If you plot the probability $\psi_v^2(x)$ and overlay the parabolic potential $V(x)$, you see something amazing. The "tails" of the Gaussian function extend beyond the classical "turning points" (the points where $E_{total} = V(x)$).
This means there is a non-zero probability of finding the particle in the "classically forbidden region," where its potential energy is greater than its total energy (implying negative kinetic energy, a classical impossibility).
This is quantum mechanical tunnelling. It's a direct consequence of the wave-nature of matter and is crucial for explaining phenomena from alpha decay to the function of enzymes.
This is our model for molecular rotation. We model a diatomic molecule (like CO or HCl) as two masses ($m_1, m_2$) connected by a rigid rod of a fixed length $r$.
This two-body problem is simplified by using the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$. The problem becomes equivalent to a single particle of mass $\mu$ rotating on the surface of a sphere of radius $r$.
Since the radius $r$ is fixed (the "rigid" part), the potential energy $V$ is constant (we can set $V=0$). The particle only has kinetic energy of rotation. This problem is best solved in spherical coordinates ($r, \theta, \phi$).
The TISE involves the Hamiltonian for rotation. This Hamiltonian is related to the angular momentum operator, $\hat{L}$.$$\hat{H}\psi(\theta, \phi) = \frac{\hat{L}^2}{2I} \psi(\theta, \phi) = E\psi(\theta, \phi)$$
Here, $I = \mu r^2$ is the moment of inertia. This equation is an "eigenvalue equation" which says "when the $\hat{L}^2$ operator acts on the wavefunction $\psi$, it returns the square of the angular momentum $L^2$, multiplied by $\psi$."
When this equation is solved (again, a complex differential equation), the wavefunctions $\psi(\theta, \phi)$ are found to be a very famous set of functions: the Spherical Harmonics, $Y_{J, M_J}(\theta, \phi)$.
Tutor's Note: This is a major connection. These are the exact same angular functions that describe atomic orbitals!
This isn't a coincidence. The angular motion of a rigid rotor is identical to the angular motion of an electron in a hydrogen atom.
- $Y_{0,0}$ is the $s$ orbital (spherically symmetric).
- The three $Y_{1, M_J}$ functions correspond to the $p_x, p_y, p_z$ orbitals.
- The five $Y_{2, M_J}$ functions correspond to the $d$ orbitals.
The boundary conditions (the wavefunction must be continuous and single-valued) force the solutions to be described by two quantum numbers:
Now, let's find the energy. From our Hamiltonian, $E = L^2 / (2I)$. Since $L^2$ is quantized as $\hbar^2 J(J+1)$, the energy $E$ must be:
$$E_J = \frac{\hbar^2}{2I} J(J+1) \quad \text{where} \quad J = 0, 1, 2, ...$$This is the key result for the rigid rotor.
But notice something: the energy only depends on $J$, not on $M_J$!
This means that for any given $J$, all $(2J+1)$ states (with their different $M_J$ values) have the exact same energy. This is another example of degeneracy.
This is the $(2J+1)$-fold degeneracy of rotational energy levels. This is a "symmetry degeneracy" that exists because, in the absence of an external field, all orientations in space are equivalent.
Let's do a final comparison of the energy level spacing, as this is what determines their spectra.
Mastering these three models gives you the entire conceptual toolkit for understanding translational, vibrational, and rotational quantum mechanics, which are the three pillars of molecular spectroscopy.
Hello again! Let's get our hands dirty with the differential equations. We're going to see how applying physical, common-sense boundary conditions (like "the wavefunction can't explode to infinity") forces the solutions to adopt specific, quantized energies.
This is the "brute force" way to solve the SHO's Schrödinger equation. It's a classic example of how to tackle a complex differential equation.
We start with the TISE for the SHO:
$$-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} + \frac{1}{2}kx^2\psi(x) = E\psi(x)$$This equation is messy. We can clean it up by introducing dimensionless variables.Let $\omega = \sqrt{k/m}$. The equation becomes:$$-\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2x^2\psi = E\psi$$
Now, let's define a dimensionless position variable $y$ and a dimensionless energy $\epsilon$:
We need to change our derivative from $\frac{d}{dx}$ to $\frac{d}{dy}$ using the chain rule:$\frac{d\psi}{dx} = \frac{d\psi}{dy} \frac{dy}{dx} = \frac{d\psi}{dy} \sqrt{\frac{m\omega}{\hbar}}$$\frac{d^2\psi}{dx^2} = \frac{d}{dx}\left(\frac{d\psi}{dy} \sqrt{\frac{m\omega}{\hbar}}\right) = \frac{d}{dy}\left(\frac{d\psi}{dy} \sqrt{\frac{m\omega}{\hbar}}\right) \frac{dy}{dx} = \frac{d^2\psi}{dy^2} \left(\frac{m\omega}{\hbar}\right)$
Now, substitute this back into the TISE:
$$-\frac{\hbar^2}{2m} \left( \frac{m\omega}{\hbar} \frac{d^2\psi}{dy^2} \right) + \frac{1}{2}m\omega^2\left( \frac{\hbar}{m\omega} y^2 \right)\psi = E\psi$$Simplify the constants:
$$-\frac{\hbar\omega}{2} \frac{d^2\psi}{dy^2} + \frac{\hbar\omega}{2} y^2\psi = E\psi$$Divide the entire equation by $-\frac{\hbar\omega}{2}$:
$$\frac{d^2\psi}{dy^2} - y^2\psi = -\frac{2E}{\hbar\omega}\psi$$Substitute our dimensionless energy $\epsilon$:
$$\frac{d^2\psi}{dy^2} - y^2\psi = -\epsilon\psi \quad \implies \quad \frac{d^2\psi}{dy^2} + (\epsilon - y^2)\psi = 0$$This is our clean, canonical equation. Now we just have to solve it.
As $y$ gets very large, $y^2$ will be much, much larger than the constant $\epsilon$. So, the equation becomes:
$$\frac{d^2\psi}{dy^2} \approx y^2\psi \quad (\text{for large } y)$$A function that, when differentiated twice, gives itself back times $y^2$ is a Gaussian. The solutions are $\psi \approx e^{+y^2/2}$ and $\psi \approx e^{-y^2/2}$.
The $e^{+y^2/2}$ solution explodes at infinity, which is physically impossible (it's not normalizable). Therefore, our wavefunction must have the $e^{-y^2/2}$ behavior at large $y$.
Based on this, let's "peel off" the asymptotic part. We propose a general solution of the form:
$$\psi(y) = H(y) e^{-y^2/2}$$Here, $H(y)$ is (we hope) a simpler function, a polynomial, that "corrects" the wavefunction for small values of $y$. We need to substitute this back into our canonical equation $\frac{d^2\psi}{dy^2} + (\epsilon - y^2)\psi = 0$. This requires some calculus:
Now, plug $\psi''$ and $\psi$ into the canonical equation:
$$(H'' - 2yH' + (y^2-1)H)e^{-y^2/2} + (\epsilon - y^2)H e^{-y^2/2} = 0$$Divide by $e^{-y^2/2}$ (which is never zero) and watch the $y^2$ terms cancel out:
$$H'' - 2yH' + (y^2-1)H + (\epsilon - y^2)H = 0$$$$H'' - 2yH' + (\epsilon - 1)H = 0$$This is Hermite's Differential Equation. Our problem is now reduced to solving this.
We solve this by assuming $H(y)$ is a power series:
$$H(y) = \sum_{j=0}^{\infty} a_j y^j = a_0 + a_1 y + a_2 y^2 + \dots$$We need its derivatives:
Substitute these three series into Hermite's equation:
$$\sum_{j=2}^{\infty} j(j-1) a_j y^{j-2} - 2y\sum_{j=1}^{\infty} j a_j y^{j-1} + (\epsilon - 1)\sum_{j=0}^{\infty} a_j y^j = 0$$$$\sum_{j=2}^{\infty} j(j-1) a_j y^{j-2} - \sum_{j=1}^{\infty} 2j a_j y^j + \sum_{j=0}^{\infty} (\epsilon - 1) a_j y^j = 0$$We need all terms to have the same power, $y^k$. Let's re-index the first sum.Let $k = j-2$, so $j = k+2$. When $j=2, k=0$.$$\sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2} y^k - \sum_{k=1}^{\infty} 2k a_k y^k + \sum_{k=0}^{\infty} (\epsilon - 1) a_k y^k = 0$$
For this entire polynomial to equal zero, the coefficient of each power of $y$ must be zero.Let's look at the $k \ge 1$ terms first, which are in all three sums:
$$(k+2)(k+1) a_{k+2} - 2k a_k + (\epsilon - 1) a_k = 0$$$$(k+2)(k+1) a_{k+2} - (2k - \epsilon + 1) a_k = 0$$This gives us the famous recursion relation for the coefficients:
$$a_{k+2} = \frac{2k + 1 - \epsilon}{(k+2)(k+1)} a_k$$This relation links $a_0 \to a_2 \to a_4 \dots$ (the even series) and $a_1 \to a_3 \to a_5 \dots$ (the odd series).
Tutor's Note: This is the most important step! What happens if this series goes on forever? For large $k$, the relation becomes $a_{k+2} \approx \frac{2k}{k^2} a_k \approx \frac{2}{k} a_k$. This is the power series for $e^{y^2}$.If $H(y)$ behaves like $e^{y^2}$, our full solution $\psi(y) = H(y) e^{-y^2/2}$ would behave like $\psi \approx e^{y^2} e^{-y^2/2} = e^{+y^2/2}$.This explodes at infinity! Our solution is not normalizable. It is physically unacceptable.
The only way to have a valid solution is if the power series terminates.
How do we stop it? We force the numerator of the recursion relation to be zero for some integer $k=v$.
$$2v + 1 - \epsilon = 0$$If we set this condition, then $a_{v+2} = 0$, which means $a_{v+4} = 0$, $a_{v+6} = 0$, and so on. The series stops at $a_v$, becoming a finite polynomial (which is what we wanted!)
This termination condition is the quantization condition.
$$\epsilon = 2v + 1 \quad \text{where } v = 0, 1, 2, \dots$$Now, just substitute back our original definition for $\epsilon$:
$$\epsilon = \frac{2E_v}{\hbar\omega}$$$$\frac{2E_v}{\hbar\omega} = 2v + 1$$Solving for $E_v$ gives the quantized energy levels:
$$E_v = \left(v + \frac{1}{2}\right)\hbar\omega$$And that's it! The energy quantization isn't an assumption; it's a consequence of the boundary condition that the wavefunction can't explode at infinity. The resulting polynomials $H(y)$ are the Hermite Polynomials.
This derivation shows how the two quantum numbers, $J$ and $M_J$, arise naturally from the geometry of the system.
The TISE is $\hat{H}\psi = E\psi$. For the rigid rotor, $\hat{H} = \frac{\hat{L}^2}{2I}$ (potential $V=0$).The $\hat{L}^2$ operator in spherical coordinates is a beast:
$$\hat{L}^2 = -\hbar^2 \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right]$$So, our TISE is:$$-\frac{\hbar^2}{2I} \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial\psi}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\phi^2} \right] = E\psi$$
Let's rearrange it. Multiply by $-\frac{2I}{\hbar^2}$ and define a new constant $\beta = \frac{2IE}{\hbar^2}$:
$$\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial\psi}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\phi^2} = -\beta\psi$$The wavefunction $\psi$ depends on two variables, $\theta$ and $\phi$. Let's propose a solution that is a product of two functions, one for each variable:
$$\psi(\theta, \phi) = \Theta(\theta) \Phi(\phi)$$Now, substitute this into our TISE. (Note: $\frac{\partial\psi}{\partial\theta} = \Phi \frac{d\Theta}{d\theta}$ and $\frac{\partial^2\psi}{\partial\phi^2} = \Theta \frac{d^2\Phi}{d\phi^2}$)
$$\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \Phi \frac{d\Theta}{d\theta}\right) + \frac{1}{\sin^2\theta}\Theta \frac{d^2\Phi}{d\phi^2} = -\beta \Theta \Phi$$$$\frac{\Phi}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta \frac{d\Theta}{d\theta}\right) + \frac{\Theta}{\sin^2\theta}\frac{d^2\Phi}{d\phi^2} = -\beta \Theta \Phi$$Now, divide the whole equation by $\Theta\Phi$ to separate the variables:
$$\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\left(\sin\theta \frac{d\Theta}{d\theta}\right) + \frac{1}{\Phi\sin^2\theta}\frac{d^2\Phi}{d\phi^2} = -\beta$$Multiply by $\sin^2\theta$ to finish the separation:
$$\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\left(\sin\theta \frac{d\Theta}{d\theta}\right) + \beta\sin^2\theta = -\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2}$$This is the crucial step.
The only way a function of $\theta$ can equal a function of $\phi$ for all values is if both sides are equal to the same constant. Let's call this separation constant $m^2$.
Let's take the "easy" RHS first:
$$-\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2} = m^2 \quad \implies \quad \frac{d^2\Phi}{d\phi^2} = -m^2\Phi$$This is our old friend, the same equation as the PIB. The general solution is:
$$\Phi(\phi) = A e^{im\phi}$$Now we apply a boundary condition. The wavefunction must be "single-valued," meaning if we rotate by $2\pi$ radians (a full circle), we must end up with the same value we started with.
$$\Phi(\phi + 2\pi) = \Phi(\phi)$$$$A e^{im(\phi + 2\pi)} = A e^{im\phi}$$$$e^{im\phi} e^{im2\pi} = e^{im\phi} \quad \implies \quad e^{im2\pi} = 1$$Using Euler's identity ($e^{ix} = \cos(x) + i\sin(x)$), this means $\cos(m2\pi) + i\sin(m2\pi) = 1$. This is only true if $m$ is an integer.
$$m = 0, \pm 1, \pm 2, \pm 3, \dots$$This is our first quantum number! We call it $M_J$ (the magnetic quantum number) in this context.
Now we take the "hard" LHS and set it equal to the same constant $m^2$:
$$\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\left(\sin\theta \frac{d\Theta}{d\theta}\right) + \beta\sin^2\theta = m^2$$Rearrange this to get the standard form. Divide by $\sin^2\theta$ and move $m^2$ to the left:
$$\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\left(\sin\theta \frac{d\Theta}{d\theta}\right) + \beta - \frac{m^2}{\sin^2\theta} = 0$$$$\frac{1}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta \frac{d\Theta}{d\theta}\right) + \left(\beta - \frac{m^2}{\sin^2\theta}\right)\Theta = 0$$This is the famous Associated Legendre Equation.
Tutor's Note: This is the other point where, like the SHO, we must use a physical boundary condition. Solving this equation with a power series is extremely* complex (even more so than the SHO). The key is that the solution $\Theta(\theta)$ must be finite and well-behaved at the "poles" ($\theta=0$ and $\theta=\pi$).
It is a standard result from mathematical physics that the solutions to the Associated Legendre Equation are only well-behaved if the constant $\beta$ is quantized as:
$$\beta = J(J+1)$$...where $J$ is an integer ($J=0, 1, 2, \dots$) and, for a non-trivial solution, it must also be true that $J \ge |m|$. This is why $M_J$ (our $m$) is restricted to values from $-J$ to $+J$.
The solutions $\Theta(\theta)$ that satisfy this condition are the Associated Legendre Polynomials, $P_J^{|m|}(\cos\theta)$.
This is the final, easy step. The quantization of $\beta$ is the quantization of energy. We just substitute back our original definition of $\beta$:
$$\beta = \frac{2IE}{\hbar^2}$$Now set the two equal:
$$\frac{2IE_J}{\hbar^2} = J(J+1)$$Solving for $E_J$ gives the quantized rotational energy levels:
$$E_J = \frac{\hbar^2}{2I} J(J+1)$$And we're done!
The total wavefunction $\psi(\theta, \phi) = \Theta(\theta)\Phi(\phi)$ is the (normalized) product of these two solutions, which are called the Spherical Harmonics, $Y_{J, M_J}(\theta, \phi)$.
We start with the Time-Independent Schrödinger Equation (TISE) for a particle of mass $m$ in one dimension $x$:
$$-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)$$1. Define the System (Potential and Boundary Conditions):
Inside the box (from $x=0$ to $x=L$), the particle is free, so $V(x) = 0$. Outside the box, the potential $V(x) = \infty$.Because the particle cannot exist in a region of infinite potential, the wavefunction $\psi(x)$ must be zero outside the box. For the wavefunction to be continuous, it must also be zero at the walls. This gives our boundary conditions:
2. Solve the TISE Inside the Box:
Inside the box, $V(x)=0$, so the TISE simplifies to:
$$-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} = E\psi(x)$$Let's rearrange this into a standard form. Let $k^2 = \frac{2mE}{\hbar^2}$ (we use $k^2$ because $E$ must be positive, so $k$ is real):
$$\frac{d^2\psi(x)}{dx^2} = -k^2\psi(x) \quad \implies \quad \frac{d^2\psi}{dx^2} + k^2\psi = 0$$This is a standard second-order differential equation. The general solution is:
$$\psi(x) = A\sin(kx) + B\cos(kx)$$3. Apply the Boundary Conditions:
4. Find Quantized Energy ($E_n$):
The condition $\sin(kL)=0$ is only true when $kL$ is an integer multiple of $\pi$.
$$kL = n\pi \quad \text{where } n = 1, 2, 3, ...$$Note: $n=0$ is not allowed because it would make $k=0$ and $\psi(x)=0$ (the "trivial solution"). Negative $n$ values are redundant as they only flip the sign of the wavefunction, which doesn't change the probability $\psi^2$.
This is the quantization condition. Now we solve for $E$ by substituting our quantized $k = n\pi/L$ back into its definition ($k^2 = 2mE/\hbar^2$):
$$\left(\frac{n\pi}{L}\right)^2 = \frac{2mE_n}{\hbar^2} \quad \implies \quad \frac{n^2\pi^2}{L^2} = \frac{2mE_n}{\hbar^2}$$Solving for energy $E_n$:
$$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$$Using $\hbar = h / (2\pi)$, this becomes the more common form:
$$E_n = \frac{n^2h^2}{8mL^2}$$5. Find Normalized Wavefunctions ($\psi_n$):
Our wavefunction is $\psi_n(x) = A\sin\left(\frac{n\pi x}{L}\right)$. We find $A$ by normalizing, which means the total probability of finding the particle in the box is 1.
$$\int_0^L \psi_n^*(x) \psi_n(x) dx = 1 \quad \implies \quad A^2 \int_0^L \sin^2\left(\frac{n\pi x}{L}\right) dx = 1$$This is a standard integral: $\int_0^L \sin^2(ax) dx = L/2$. Thus:
$$A^2 \left( \frac{L}{2} \right) = 1 \quad \implies \quad A = \sqrt{\frac{2}{L}}$$Therefore, the final normalized wavefunctions are:
$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$$The solutions $\psi_n$ and $E_n$ are not just mathematical results; they have profound physical interpretations:
a) Normalized wavefunction for $n=2$:
$$\psi_2(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{2\pi x}{L}\right)$$b) Probability $P$ in the region $L/2 \le x \le 3L/4$:
Probability is found by integrating the probability density, $\psi^*(x)\psi(x) = \psi^2(x)$, over the specified region.
$$P = \int_{L/2}^{3L/4} \psi_2^2(x) dx = \int_{L/2}^{3L/4} \frac{2}{L} \sin^2\left(\frac{2\pi x}{L}\right) dx$$To solve this, we use the trigonometric identity: $\sin^2(\theta) = \frac{1}{2} (1 - \cos(2\theta))$.Here, $\theta = \frac{2\pi x}{L}$, so $2\theta = \frac{4\pi x}{L}$.
$$P = \frac{2}{L} \int_{L/2}^{3L/4} \frac{1}{2} \left[ 1 - \cos\left(\frac{4\pi x}{L}\right) \right] dx$$$$P = \frac{1}{L} \left[ \int_{L/2}^{3L/4} 1 dx - \int_{L/2}^{3L/4} \cos\left(\frac{4\pi x}{L}\right) dx \right]$$Now, we integrate:
$$P = \frac{1}{L} \left[ x - \frac{L}{4\pi} \sin\left(\frac{4\pi x}{L}\right) \right]_{L/2}^{3L/4}$$Evaluate at the upper limit ($x=3L/4$):
$$\frac{1}{L} \left[ \frac{3L}{4} - \frac{L}{4\pi} \sin\left(\frac{4\pi (3L/4)}{L}\right) \right] = \frac{1}{L} \left[ \frac{3L}{4} - \frac{L}{4\pi} \sin(3\pi) \right] = \frac{1}{L} \left[ \frac{3L}{4} - 0 \right] = \frac{3}{4}$$Evaluate at the lower limit ($x=L/2$):
$$\frac{1}{L} \left[ \frac{L}{2} - \frac{L}{4\pi} \sin\left(\frac{4\pi (L/2)}{L}\right) \right] = \frac{1}{L} \left[ \frac{L}{2} - \frac{L}{4\pi} \sin(2\pi) \right] = \frac{1}{L} \left[ \frac{L}{2} - 0 \right] = \frac{1}{2}$$Subtract the lower limit from the upper limit:
$$P = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$$The probability is 1/4 or 25%. This makes intuitive sense, as the $\psi_2^2$ function has two identical "humps" of probability, one from $0$ to $L/2$ and one from $L/2$ to $L$. This question asks for the probability in the first half of the second hump (from $L/2$ to $3L/4$), which by symmetry is exactly one-quarter of the total probability.
c) Positions of maximum probability density:
Probability density is $P(x) = \psi_2^2(x) = \frac{2}{L} \sin^2\left(\frac{2\pi x}{L}\right)$.This is a maximum when $\sin^2(\dots)$ is maximum, which is 1. This occurs when $\sin(\frac{2\pi x}{L}) = \pm 1$.This condition is met when the argument is $\pi/2, 3\pi/2, 5\pi/2, \dots$
The probability density is a maximum at $x = L/4$ and $x = 3L/4$.
The expectation value (or average value) of position $\langle x \rangle$ is calculated by "sandwiching" the position operator $\hat{x} = x$ between $\psi^*$ and $\psi$ and integrating over all space.
$$\langle x \rangle_n = \int_0^L \psi_n^*(x) \hat{x} \psi_n(x) dx = \int_0^L \left( \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right) \right) x \left( \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right) \right) dx$$$$\langle x \rangle_n = \frac{2}{L} \int_0^L x \sin^2\left(\frac{n\pi x}{L}\right) dx$$Let's analyze the integrand: $f(x) = x \cdot \sin^2\left(\frac{n\pi x}{L}\right)$.
Let's make a substitution $u = x - L/2$, so $x = u + L/2$. The integral becomes:
$$\langle x \rangle_n = \frac{2}{L} \int_{-L/2}^{+L/2} (u + L/2) \cdot \underbrace{\sin^2\left(\frac{n\pi (u+L/2)}{L}\right)}_{\text{an even function of } u} du$$$$\langle x \rangle_n = \frac{2}{L} \left[ \int_{-L/2}^{+L/2} \underbrace{u \cdot (\text{even fn.})}_{\text{odd fn.}} du + \int_{-L/2}^{+L/2} \underbrace{(L/2) \cdot (\text{even fn.})}_{\text{even fn.}} du \right]$$So, the expression becomes: $\langle x \rangle_n = \frac{2}{L} \left[ 0 + (L/2) \cdot 1 \right] = \frac{2}{L} \cdot \frac{L}{2} = 1$.
Wait, re-check that. The integral is $\int (L/2) \cdot \psi_n^2 dx$.$\langle x \rangle_n = \frac{2}{L} \left[ 0 + \frac{L}{2} \int_0^L \psi_n^2(x) dx \right] = \frac{2}{L} \left[ \frac{L}{2} \cdot \left(\frac{L}{2} \cdot \frac{1}{A^2}\right) \right]$ ... No, that's too complex.Let's restart the second part.The second integral is $\frac{L}{2} \int_{-L/2}^{+L/2} (\text{even fn.}) du$. This is $\frac{L}{2} \int_0^L \psi_n^2(x) dx$.We know $\int_0^L \psi_n^2(x) dx = \int_0^L A^2 \sin^2(...) dx = A^2 (L/2) = 1$.Ah, $\int \psi_n^2 dx = 1$. The $A^2 = (2/L)$ is part of $\psi_n^2$.So the second integral is $\int_0^L (L/2) \cdot \psi_n^2(x) dx = (L/2) \int_0^L \psi_n^2(x) dx = (L/2) \cdot 1 = L/2$.Now, substitute back:$\langle x \rangle_n = \frac{2}{L} \left[ 0 + \frac{L}{2} \right] = 1$.Still wrong. Let's re-examine $\langle x \rangle_n = \int x \psi_n^2 dx$.$\langle x \rangle_n = \int_{-L/2}^{+L/2} (u + L/2) \cdot \psi_n^2(u) du = \int \underbrace{u \cdot \psi_n^2(u)}_{\text{odd}} du + \int (L/2) \cdot \psi_n^2(u) du$$\langle x \rangle_n = 0 + (L/2) \int \psi_n^2(u) du = (L/2) \cdot 1 = L/2$.
Correct Symmetry Argument:The expectation value is $\langle x \rangle = \int_0^L x \psi_n^2(x) dx$.Let's analyze the integrand $f(x) = x \cdot \psi_n^2(x)$ around the center point $L/2$.The function $\psi_n^2(x)$ is symmetric around $L/2$.The function $x$ is not.However, the physical probability $\psi_n^2(x)$ is perfectly symmetric. For every particle found at position $x_1 < L/2$, there is an equal probability of finding it at $x_2 > L/2$ where $x_2 = L-x_1$. The average of these two positions is $(x_1 + (L-x_1))/2 = L/2$.Because the probability distribution $\psi_n^2(x)$ is symmetric about $x=L/2$, the average position must be the center of symmetry.$$\langle x \rangle_n = \frac{L}{2} \quad \text{for all } n$$
Relation to Classical Prediction:A classical particle in a box moves at constant velocity, so it spends an equal amount of time at every point in the box. Its probability distribution is flat ($P(x) = 1/L$). The average position for a classical particle is also $\int_0^L x (1/L) dx = (1/L) [x^2/2]_0^L = L^2 / (2L) = L/2$.In this case, the quantum expectation value matches the classical average position for all energy levels.
We need to prove that the integral of the product of $\psi_1$ and $\psi_2$ is zero. This is the condition for orthogonality. (We can ignore the $\psi^*$ since they are real).
$$\int_0^L \psi_1(x) \psi_2(x) dx = \int_0^L \left(\sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)\right) \left(\sqrt{\frac{2}{L}}\sin\left(\frac{2\pi x}{L}\right)\right) dx$$$$= \frac{2}{L} \int_0^L \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2\pi x}{L}\right) dx$$We use the trigonometric identity: $\sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$.
Substitute this into the integral:
$$= \frac{2}{L} \int_0^L \frac{1}{2} \left[ \cos\left(-\frac{\pi x}{L}\right) - \cos\left(\frac{3\pi x}{L}\right) \right] dx$$Since $\cos(-y) = \cos(y)$, this simplifies to:
$$= \frac{1}{L} \int_0^L \left[ \cos\left(\frac{\pi x}{L}\right) - \cos\left(\frac{3\pi x}{L}\right) \right] dx$$Now, we integrate term by term:
$$= \frac{1}{L} \left[ \frac{L}{\pi} \sin\left(\frac{\pi x}{L}\right) - \frac{L}{3\pi} \sin\left(\frac{3\pi x}{L}\right) \right]_0^L$$Evaluate at the upper limit ($x=L$):
$$\frac{1}{L} \left[ \frac{L}{\pi} \sin(\pi) - \frac{L}{3\pi} \sin(3\pi) \right] = \frac{1}{L} [ 0 - 0 ] = 0$$Evaluate at the lower limit ($x=0$):
$$\frac{1}{L} \left[ \frac{L}{\pi} \sin(0) - \frac{L}{3\pi} \sin(0) \right] = \frac{1}{L} [ 0 - 0 ] = 0$$The result is $0 - 0 = 0$. The integral is zero, proving the wavefunctions are orthogonal. (Q.E.D.)
Physical Significance:Orthogonality is a mathematical statement that the wavefunctions are fundamentally distinct and independent. It is part of the "orthonormal" nature of eigenstates. Physically, it means that if a system is in a pure stationary state $\psi_1$, the probability of measuring it and finding it to be in state $\psi_2$ is exactly zero. The states are mutually exclusive.
1. Set up the Schrödinger Equation:
For a 3D system, the TISE is:$$-\frac{\hbar^2}{2m} \left( \frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2} \right) + V(x,y,z)\psi = E\psi$$Inside a cubic box of side $L$, $V=0$. The boundary conditions are $\psi=0$ at $x,y,z = 0$ and $x,y,z = L$.
2. Method of Separation of Variables:
We propose a solution that is a product of three independent functions:$$\psi(x,y,z) = X(x)Y(y)Z(z)$$
Substitute this into the TISE:
$$-\frac{\hbar^2}{2m} \left( YZ\frac{d^2X}{dx^2} + XZ\frac{d^2Y}{dy^2} + XY\frac{d^2Z}{dz^2} \right) = E(XYZ)$$Divide the entire equation by $\psi = XYZ$:
$$-\frac{\hbar^2}{2m} \left( \frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2} + \frac{1}{Z}\frac{d^2Z}{dz^2} \right) = E$$$$\implies \left[ -\frac{\hbar^2}{2m} \frac{1}{X}\frac{d^2X}{dx^2} \right] + \left[ -\frac{\hbar^2}{2m} \frac{1}{Y}\frac{d^2Y}{dy^2} \right] + \left[ -\frac{\hbar^2}{2m} \frac{1}{Z}\frac{d^2Z}{dz^2} \right] = E$$The first term depends only on $x$. The second only on $y$. The third only on $z$. For their sum to be a constant ($E$) for all $x,y,z$, each term must itself be a constant. Let's call them $E_x, E_y, E_z$.
And the total energy is $E = E_x + E_y + E_z$.
3. Find the Solutions:
Each of these three equations is just the 1D particle-in-a-box problem! We can write down the solutions directly from Question 1.
The state of the particle is now defined by three quantum numbers ($n_x, n_y, n_z$).
Total Wavefunction $\Psi(x, y, z)$ (by multiplying $X, Y, Z$):
$$\psi_{n_x, n_y, n_z}(x,y,z) = \sqrt{\frac{8}{L^3}} \sin\left(\frac{n_x \pi x}{L}\right) \sin\left(\frac{n_y \pi y}{L}\right) \sin\left(\frac{n_z \pi z}{L}\right)$$Total Energy $E_{n_x, n_y, n_z}$ (by adding $E_x, E_y, E_z$):
$$E_{n_x, n_y, n_z} = \frac{h^2}{8mL^2} \left( n_x^2 + n_y^2 + n_z^2 \right)$$Degeneracy is the phenomenon where two or more distinct quantum states (i.e., different wavefunctions, corresponding to different sets of quantum numbers) have the exact same energy.
Using the energy expression for the 3D cubic box: $E_{n_x, n_y, n_z} = \frac{h^2}{8mL^2} \left( n_x^2 + n_y^2 + n_z^2 \right)$.Let $E_{unit} = h^2 / (8mL^2)$.
1. Lowest Energy Level (Ground State):
2. Second Lowest Energy Level (First Excited State):
3. Third Lowest Energy Level (Second Excited State):
Tutor's Note: The next level would be $(3,1,1)$, giving $E = 11 E_{unit}$ (also $g=3$). The level after that would be $(2,2,2)$, giving $E = 12 E_{unit}$ (which is non-degenerate, $g=1$).
The box has sides $L_x = L$, $L_y = 2L$, and $L_z = 2L$.
a) Energy expression for this system:
We use the general energy expression derived from separation of variables:
$$E_{n_x, n_y, n_z} = \frac{h^2}{8m} \left( \frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2} \right)$$Substitute the specific lengths:
$$E_{n_x, n_y, n_z} = \frac{h^2}{8m} \left( \frac{n_x^2}{L^2} + \frac{n_y^2}{(2L)^2} + \frac{n_z^2}{(2L)^2} \right)$$$$E_{n_x, n_y, n_z} = \frac{h^2}{8m} \left( \frac{n_x^2}{L^2} + \frac{n_y^2}{4L^2} + \frac{n_z^2}{4L^2} \right)$$Factoring out the common $1/L^2$ term gives the final expression:
$$E_{n_x, n_y, n_z} = \frac{h^2}{8mL^2} \left( n_x^2 + \frac{n_y^2}{4} + \frac{n_z^2}{4} \right)$$b) Energies of the ground state and first excited state:
Let $E_{unit} = h^2 / (8mL^2)$. So $E = E_{unit} (n_x^2 + n_y^2/4 + n_z^2/4)$.
c) Is the first excited state degenerate?
Yes. As shown above, two different states, (1,2,1) and (1,1,2), both have the same energy $E = 2.25 E_{unit}$. Because the box has a symmetry ($L_y = L_z$), swapping the quantum numbers $n_y$ and $n_z$ results in the same energy. The first excited state is two-fold degenerate ($g=2$).
We are in a cubic box, so the energy is $E = \frac{h^2}{8mL^2} \left( n_x^2 + n_y^2 + n_z^2 \right)$.We are given an energy $E = 14 \left( \frac{h^2}{8mL^2} \right)$.Therefore, we must find all sets of integers $n_x, n_y, n_z \ge 1$ that satisfy the equation:
$$n_x^2 + n_y^2 + n_z^2 = 14$$Let's test possible integers systematically. The largest $n^2$ can be is $14 - 1^2 - 1^2 = 12$, so $n$ can't be larger than 3 ($3^2=9$).
The only set of unique integers is $\{1, 2, 3\}$. The question is how many distinct permutations (which are distinct states) can be formed from these three numbers.For a set of 3 distinct numbers, the number of permutations is $3! = 3 \times 2 \times 1 = 6$.
The 6 degenerate states are:
The degeneracy of this energy level is $g=6$.
1. Schrödinger Equation:
The classical potential for an SHO (like a mass on a spring) is $V(x) = \frac{1}{2}kx^2$, where $k$ is the force constant.Substituting this into the 1D TISE gives:
$$-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} + \frac{1}{2}kx^2\psi(x) = E\psi(x)$$2. Quantized Energy Levels ($E_v$):
Solving this differential equation (which is more complex than the PIB) and applying the boundary conditions ($\psi \to 0$ as $x \to \pm\infty$) yields the quantized energy levels:
$$E_v = \left(v + \frac{1}{2}\right) \hbar\omega \quad \text{where } v = 0, 1, 2, 3, ...$$Here, $v$ is the vibrational quantum number, and $\omega$ is the classical angular frequency of the oscillator, $\omega = \sqrt{k/m}$.Since $\omega = 2\pi\nu$ and $\hbar = h / (2\pi)$, we can also write this as:$$E_v = \left(v + \frac{1}{2}\right) h\nu$$
3. Ground State Wavefunction ($\psi_0$):
The normalized wavefunction for the $v=0$ ground state is a Gaussian function:
$$\psi_0(x) = \left(\frac{\alpha}{\pi}\right)^{1/4} e^{-\alpha x^2 / 2}$$...where $\alpha = \frac{\sqrt{km}}{\hbar}$.
4. Zero-Point Energy (ZPE):
The ground state is the lowest possible energy state, $v=0$.$$E_0 = \left(0 + \frac{1}{2}\right) h\nu = \frac{1}{2}h\nu$$
Like the particle-in-a-box, the SHO has a non-zero ground state energy. The oscillator never stops vibrating, even at absolute zero. This is also a consequence of the Heisenberg Uncertainty Principle.
Below is a description of the sketches. For all plots, the $x$-axis is position, and the parabolic curve $V(x) = \frac{1}{2}kx^2$ is shown in the background. The horizontal lines $E_0, E_1, E_2$ are equally spaced.
Key Labeled Features:
Quantum Mechanical Tunnelling:
As seen in the sketches for Question 11, the wavefunction $\psi_v(x)$ does not go to zero at the classical "turning points" (where $E = V(x)$). It decays exponentially into the "classically forbidden region" where $V(x) > E$.
Since the probability density is $\psi_v^2(x)$, this means there is a finite, non-zero probability of finding the particle in a region where its total energy ($E$) is less than its potential energy ($V$). This would imply a negative kinetic energy, which is impossible in classical mechanics. This ability to "tunnel" into a barrier is a purely quantum mechanical phenomenon.
Contrast with Classical Probability:
A classical harmonic oscillator (e.g., a pendulum) has a very different probability distribution.
Therefore, the classical probability of finding the particle is minimum at the center (it spends the least time there) and maximum at the ends (it "lingers" at the turning points).
This is the exact opposite of the quantum ground state ($v=0$), which has its maximum probability at the center.This illustrates the Bohr Correspondence Principle: as the quantum number $v$ becomes very large, the quantum probability $\psi_v^2$ (which has $v+1$ peaks) begins to "blur," and the envelope of these peaks starts to resemble the classical probability (i.e., the peaks near the edges become much larger than the peaks in the center).
The Virial Theorem provides a general relationship between the average kinetic energy $\langle T \rangle$ and the average potential energy $\langle V \rangle$ for a system in a stationary state.
For a potential of the form $V(x) = cx^n$, the theorem states:$$2\langle T \rangle = n \langle V \rangle$$
For the Simple Harmonic Oscillator, the potential is $V(x) = \frac{1}{2}kx^2$. This means $n=2$.Substituting $n=2$ into the Virial Theorem gives:
$$2\langle T \rangle = 2 \langle V \rangle \quad \implies \quad \langle T \rangle = \langle V \rangle$$This is a powerful result: for any stationary state $v$ of the SHO, the average kinetic energy is exactly equal to the average potential energy.
We also know that the total energy $E_v$ is the sum of the averages:$$E_v = \langle T \rangle + \langle V \rangle$$
Since $\langle T \rangle = \langle V \rangle$, we can conclude:$$\langle T \rangle = \langle V \rangle = \frac{1}{2} E_v$$
Calculation for the $v=3$ state:
First, find the total energy for the $v=3$ state:$$E_3 = \left(v + \frac{1}{2}\right)h\nu = \left(3 + \frac{1}{2}\right)h\nu = \frac{7}{2}h\nu$$
Now, apply the Virial Theorem result:
We are given the $v=1$ wavefunction $\psi_1 = N \cdot y \cdot e^{-y^2/2}$, where $y = \alpha^{1/2}x$.The normalization condition is $\int_{-\infty}^{+\infty} \psi_1^* \psi_1 dx = 1$.
1. Substitute $\psi_1$ and Change Variables:
First, we must change the integration variable from $dx$ to $dy$.From $y = \alpha^{1/2}x$, we have $x = y / \alpha^{1/2}$, so $dx = dy / \alpha^{1/2}$.
$$\int_{-\infty}^{+\infty} (N y e^{-y^2/2}) (N y e^{-y^2/2}) dx = 1$$$$N^2 \int_{-\infty}^{+\infty} y^2 e^{-y^2} dx = 1$$Substitute $dx = dy / \alpha^{1/2}$:
$$N^2 \int_{-\infty}^{+\infty} y^2 e^{-y^2} \frac{dy}{\alpha^{1/2}} = 1$$$$\frac{N^2}{\alpha^{1/2}} \int_{-\infty}^{+\infty} y^2 e^{-y^2} dy = 1$$2. Solve the Standard Integral:
We are given the standard integral formula: $\int_{-\infty}^{+\infty} y^{2n} e^{-y^2} dy = \Gamma(n+1/2)$.
In our case, the power is $y^2$, so $2n=2$, which means $n=1$.We need to evaluate the integral $\int_{-\infty}^{+\infty} y^2 e^{-y^2} dy = \Gamma(1+1/2) = \Gamma(3/2)$.
We use the properties of the Gamma function: $\Gamma(z+1) = z\Gamma(z)$ and the standard value $\Gamma(1/2) = \sqrt{\pi}$.
$$\Gamma(3/2) = \Gamma\left(\frac{1}{2} + 1\right) = \frac{1}{2} \Gamma\left(\frac{1}{2}\right) = \frac{\sqrt{\pi}}{2}$$3. Solve for $N$:
Substitute the value of the integral back into our normalization equation:
$$\frac{N^2}{\alpha^{1/2}} \left( \frac{\sqrt{\pi}}{2} \right) = 1$$Now, solve for $N^2$:
$$N^2 = \frac{2\alpha^{1/2}}{\sqrt{\pi}}$$Finally, solve for $N$ by taking the square root of both sides:
$$N = \left( \frac{2\alpha^{1/2}}{\pi^{1/2}} \right)^{1/2} = \left( \frac{4\alpha}{\pi} \right)^{1/4}$$1. Set up the Schrödinger Equation:
The rigid rotor models a diatomic molecule with two masses $\mu_1, \mu_2$ at a fixed distance $r_0$. We simplify this to a single particle of reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$ moving on the surface of a sphere of radius $r_0$.
Since $r=r_0$ is fixed, there is no radial motion and the potential energy $V$ is constant (we set $V=0$). The TISE only involves kinetic energy of rotation and is best written in spherical coordinates ($\theta, \phi$).
The TISE is: $\hat{H}\psi(\theta, \phi) = E\psi(\theta, \phi)$.The Hamiltonian $\hat{H}$ for rotation is $\hat{H} = \hat{L}^2 / (2I)$, where $I = \mu r_0^2$ is the moment of inertia and $\hat{L}^2$ is the operator for the square of the angular momentum.
The full TISE is:$$-\frac{\hbar^2}{2I} \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial\psi}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\phi^2} \right] = E\psi$$
The term in the brackets is the $\hat{L}^2$ operator divided by $-\hbar^2$. So, the equation is compactly written as:$$\frac{\hat{L}^2}{2I} \psi(\theta, \phi) = E\psi(\theta, \phi)$$
2. Outline of Solution:
This is an eigenvalue equation. We need to find the allowed eigenvalues of the $\hat{L}^2$ operator.The equation is solved by separation of variables, letting $\psi(\theta, \phi) = \Theta(\theta) \Phi(\phi)$.This separates the TISE into two equations, one for $\theta$ (the "Legendre equation") and one for $\phi$.
3. Quantized Energy ($E_J$):
Solving these equations and applying the physical boundary conditions (that $\psi$ must be single-valued and finite) forces the introduction of two quantum numbers, $J$ and $M_J$.
The solution requires that the eigenvalues of $\hat{L}^2$ are quantized as:$$\text{Eigenvalue}(\hat{L}^2) = \hbar^2 J(J+1) \quad \text{where } J = 0, 1, 2, ...$$
To find the energy, we substitute this eigenvalue back into our Hamiltonian equation $\hat{H}\psi = E\psi$:
$$\frac{\hbar^2 J(J+1)}{2I} \psi = E_J \psi$$This directly gives the quantized rotational energy levels:
$$E_J = \frac{\hbar^2}{2I} J(J+1)$$The solutions to the rigid rotor Schrödinger equation (the Spherical Harmonics, $Y_{J, M_J}$) are defined by two quantum numbers:
1. $J$: The Total Angular Momentum Quantum Number
2. $M_J$: The $z$-Component Angular Momentum Quantum Number
Origin of Degeneracy:
The $(2J+1)$-fold degeneracy of the rigid rotor energy levels is a fundamental symmetry degeneracy.The energy expression $E_J = \frac{\hbar^2}{2I} J(J+1)$ depends only on $J$, not on $M_J$.
The $M_J$ quantum number describes the orientation of the rotation in space. In the absence of any external electric or magnetic field, space is isotropic (all directions are equivalent). Therefore, all possible orientations of the rotation for a given $J$ value must have the same energy.
For any given $J$, there are $(2J+1)$ possible values for $M_J$ (from $-J$ to $+J$). Since the energy doesn't depend on $M_J$, all $(2J+1)$ of these states are degenerate (have the same energy).
Energy Level Diagram:
(This would be a drawing, described here):
The y-axis is Energy. The x-axis is just for show.
- At the bottom, $E_0=0$, labeled $J=0$. It is a single line, labeled ($M_J=0$).
- Above $E_0$ is $E_1 = 2(\hbar^2/2I)$, labeled $J=1$. This is drawn as a single, thicker line representing 3 degenerate states, often with ($M_J = -1, 0, +1$) written next to it.
- Above $E_1$ is $E_2 = 6(\hbar^2/2I)$, labeled $J=2$. The gap between $J=1$ and $J=2$ is twice as large as the gap between $J=0$ and $J=1$. This is a single, thicker line representing 5 degenerate states ($M_J = -2, -1, 0, +1, +2$).
This comparison is key to understanding different forms of spectroscopy.
a) Particle in a 1D Box:
b) 1D Harmonic Oscillator:
c) Rigid Rotor:
Both systems exhibit degeneracy, but its origin is different.
Both the particle-in-a-box and the harmonic oscillator have a Zero-Point Energy (ZPE), a minimum ground-state energy that is greater than zero ($E_1 = h^2/8mL^2$ and $E_0 = \frac{1}{2}h\nu$, respectively).
This is a universal feature of all confined quantum systems and is a direct consequence of the Heisenberg Uncertainty Principle (HUP):$$\Delta x \cdot \Delta p_x \ge \frac{\hbar}{2}$$
Explanation:
This minimum, non-zero energy, which arises purely from the particle's confinement and the wave-like nature dictated by HUP, is the Zero-Point Energy. It is the fundamental "energy of confinement".
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