/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Calculate \((z \hat{H} z),\) in ... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate \((z \hat{H} z),\) in the ground state of hydrogen. Hint: This takes two pages and six integrals, or four lines and no integrals, depending on how you set it up. To do it the quick way, start by noting that \([z,[H, z]]=2 z H z-H z^{2}-z^{2} H^{25}\).

Short Answer

Expert verified
In the ground state, the expectation value of \((zHz)\) is zero.

Step by step solution

01

Understand the Commutator Property

The exercise gives a commutator identity: \([z, [H, z]] = 2zHz - H(z^2) - (z^2)H\). This suggests that applying the Hamiltonian \(H\) to terms including \(z\) and understanding the commutators will simplify the calculations. Note that \(z\) represents the position operator, and \(H\) is the Hamiltonian of the hydrogen atom.
02

Apply the Identity to the Given Expression

You need to calculate \(zHz\) in the hydrogen ground state. Using the given commutator identity helps:\(2zHz = [z,[H,z]] + H(z^2) + (z^2)H\)Therefore, rearrange to find \(zHz\) by substituting this equation directly.
03

Evaluate in the Ground State

The ground state of hydrogen has a simple wave function, \(\psi_1(r) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a}\), where \(a\) is the Bohr radius. In the ground state, observables like momenta are zeroes. Specifically, within the context of this commutator, the expectation values can result in zeroes, simplifying the computation.
04

Solve for the Expectation Value

Recognize that in the computation of \(zHz\), the commutator suggests symmetries and zero-valued contributions typical in ground states with radial symmetry and spherical harmonics:\(\langle zHz \rangle = \frac{1}{2}(\langle Hz^2 \rangle + \langle z^2H \rangle)\).Using the symmetry of the situation and simplifying through integration and algebra, we see that the expectation value becomes \(\langle \psi_1 | zhz | \psi_1 \rangle = 0\) due to symmetric cancellations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Commutator Identity
In quantum mechanics, commutators play a crucial role in understanding the behavior of quantum systems. The commutator identity given in the exercise,
  • \[ [z, [H, z]] = 2zHz - H(z^2) - (z^2)H \]
helps in simplifying calculations tremendously. Here, \(z\) is the position operator, and \(H\) is the Hamiltonian, an operator that represents the total energy of the system. Commutators describe how two operators don't commute; that is, applying them in one order differs from applying them in the reverse order.
Such identities are not just mathematical sleight of hand--they reveal the underlying physical principles governing the quantum world. In many cases, leveraging these relationships, especially when dealing with operators like the position operator and the Hamiltonian, can lead directly to simplifications otherwise not immediately apparent.
Thus, understanding the role of commutators helps in reducing complex integrals to more manageable terms, capitalizing on inherent system symmetries.
Hydrogen Atom Ground State
The hydrogen atom ground state presents a fundamental study in quantum mechanics. In the ground state, the atom is at its lowest energy level. Its wave function,
  • \[ \psi_1(r) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a} \]
where \(a\) is the Bohr radius, describes the electron's spatial distribution around the nucleus.
Understanding the hydrogen atom's ground state sheds light on basic quantum principles due to its simplicity and symmetry. For instance, the radial symmetry of the wave function results in simplification of calculations involving expectation values, as examined in this exercise.
The radial symmetry implies equal probability distribution in all directions from the nucleus, causing symmetric cancellations in calculations of certain expectation values. This makes certain terms in quantum mechanical equations negligible, streamlining problem-solving.
Expectation Values
Expectation values are an essential concept in quantum mechanics, providing the "average" outcome of measuring a given quantity. Calculating the expectation value for an observable, like energy or position, involves integrating over the probability distribution of the state in question.
In the exercise, the expectation value,
  • \[ \langle zHz \rangle = \frac{1}{2}(\langle Hz^2 \rangle + \langle z^2H \rangle) \]
is derived by recognizing symmetries inherent in the hydrogen atom's ground state. Because of these symmetrical properties, complicated expressions may simplify significantly.
This simplification often leads to zero expectation values for certain operators in symmetric systems. The example shows that understanding and applying expectation values correctly in symmetric states, such as the ground state of hydrogen, can yield straightforward and insightful results regarding the system's behavior.
Position Operator
The position operator, denoted often as \(z\) in one-dimensional cases, represents a fundamental observable in quantum mechanics. It helps in determining where a particle might be located spatially in a quantum system.
The position operator has special commutation relations with other operators like momentum or Hamiltonian. Here, the commutator relation
  • \[ [z, [H, z]] = 2zHz - H(z^2) - (z^2)H \]
is particularly useful. It allows physicists to understand how action sequences, involving energy and position, interplay at a quantum level.
Using operators such as \(z\) and understanding their commutation dynamics with the Hamiltonian is critical in solving quantum mechanics problems. Through such operators, physicists translate abstract equations into meaningful descriptions of physical phenomena.

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Most popular questions from this chapter

Construct the matrix \(S_{r}\) representing the component of spin angular momentum along an arbitrary direction \(\hat{r} .\) Use spherical coordinates, for which \(\hat{r}=\sin \theta \cos \phi \hat{\imath}+\sin \theta \sin \phi \hat{\jmath}+\cos \theta \hat{k}\). Find the eigenvalues and (normalized) eigenspinors of \(\mathrm{S}_{r} .\) Answer: \(\chi_{+}^{(r)}=\left(\begin{array}{c}\cos (\theta / 2) \\ e^{i \phi} \sin (\theta / 2)\end{array}\right) ; \quad \chi_{-}^{(r)}=\left(\begin{array}{c}e^{-i \phi} \sin (\theta / 2) \\ -\cos (\theta / 2)\end{array}\right)\). Note: You're always free to multiply by an arbitrary phase factor-say, \(e^{i \phi-\text { so }}\) your answer may not look exactly the same as mine.

Find the matrix representing \(S_{x}\) for a particle of spin \(3 / 2\) (using as your basis the eigenstates of \(S_{z}\) ). Solve the characteristic equation to determine the eigenvalues of \(S_{x}\).

(a) Construct the wave function for hydrogen in the state \(n=4, \ell=3\), \(m=3 .\) Express your answer as a function of the spherical coordinates \(r\), \(\theta,\) and \(\phi\). (b) Find the expectation value of \(r\) in this state. (As always, look up any nontrivial integrals.) (c) If you could somehow measure the observable \(L_{x}^{2}+L_{y}^{2}\) on an atom in this state, what value (or values) could you get, and what is the probability of each?

Quarks carry spin \(1 / 2 .\) Three quarks bind together to make a baryon (such as the proton or neutron); two quarks (or more precisely a quark and an antiquark) bind together to make a meson (such as the pion or the kaon). Assume the quarks are in the ground state (so the orbital angular momentum is zero). (a) What spins are possible for baryons? (b) What spins are possible for mesons?

(a) Prove that for a particle in a potential \(V(\mathbf{r})\) the rate of change of the expectation value of the orbital angular momentum \(\mathbf{L}\) is equal to the expectation value of the torque: \(\frac{d}{d t}\langle\mathbf{L}\rangle=\langle\mathbf{N}\rangle\), where \(\mathbf{N}=\mathbf{r} \times(-\nabla V)\). (Thisis the rotational analog to Ehrenfest's theorem.) (b) Show that \(d\langle\mathbf{L}\rangle / d t=0\) for any spherically symmetric potential. (This is one form of the quantum statement of conservation of angular momentum.)

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