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Show that the operator \(\hat{p}^{\prime}\) obtained by applying a translation to the operator \(\hat{p}\) is \(\hat{p}^{\prime}=\hat{T}^{\dagger} \hat{p} \hat{T}=\hat{p}\)

Short Answer

Expert verified
The translated operator \( \hat{p}^{\prime} = \hat{p} \).

Step by step solution

01

Define the Problem

We need to find the expression for the operator \( \hat{p}^{\prime} \) obtained by translating the operator \( \hat{p} \) using the translation operator \( \hat{T} \), and verify that this expression is equivalent to \( \hat{p} \).
02

Understand Translation Operator

The translation operator \( \hat{T}(a) = e^{-i \hat{p} a / \hbar} \) translates a state by a displacement \( a \). Its adjoint is \( \hat{T}^{\dagger}(a) = e^{i \hat{p} a / \hbar} \).
03

Apply Translation to Operator

The translated momentum operator is given by \( \hat{p}^{\prime} = \hat{T}^{\dagger} \hat{p} \hat{T} \). We substitute the forms of \( \hat{T} \) and its adjoint to get \( \hat{p}^{\prime} = e^{i \hat{p} a / \hbar} \hat{p} e^{-i \hat{p} a / \hbar} \).
04

Simplify the Expression

To simplify \( \hat{p}^{\prime} = e^{i \hat{p} a / \hbar} \hat{p} e^{-i \hat{p} a / \hbar} \), we note that \( \hat{T} \) contains \( \hat{p} \), which commutes with itself. This makes the exponential operators cancel inside the expression, thereby leaving \( \hat{p}^{\prime} = \hat{p} \).
05

Conclusion

Since the momentum operator commutes with the translation operator's components, it remains invariant under translation, resulting in \( \hat{p}^{\prime} = \hat{p} \). This completes the demonstration.

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Key Concepts

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

Translation Operator
In quantum mechanics, the translation operator is a significant mathematical tool used to describe how states change when a system is spatially shifted. Imagine you have a quantum state described on a number line. Now, if you want to shift this state by some distance, you use the translation operator, often represented as \( \hat{T}(a) \). This operator is defined as \( \hat{T}(a) = e^{-i \hat{p} a / \hbar} \), where \( \hat{p} \) is the momentum operator, \( a \) is the distance or displacement you want to translate the state by, and \( \hbar \) is the reduced Planck's constant.

The beauty of the translation operator lies in its ability to move the entire wave function of a quantum state in space. When applied to a function \( \psi(x) \), it results in a new function \( \psi(x - a) \). This can be visualized as picking up the wave function and moving it along the \( x \)-axis by a distance \( a \).
Some crucial points to remember about the translation operator:
  • It preserves the probability distribution of the quantum state, merely displacing it.
  • It is unitary, meaning \( \hat{T}^{\dagger}(a) \hat{T}(a) = \hat{I} \), where \( \hat{I} \) is the identity operator.
  • The adjoint of \( \hat{T}(a) \) is \( \hat{T}^{\dagger}(a) = e^{i \hat{p} a / \hbar} \), a valuable property for reversing translations.
Momentum Operator
The momentum operator in quantum mechanics \( \hat{p} \) is one of the fundamental operators that corresponds to the classical concept of momentum. In the position basis, this operator is expressed as \( \hat{p} = -i \hbar \frac{d}{dx} \), where \( \hbar \) signifies the reduced Planck's constant, and \( \frac{d}{dx} \) represents a derivative with respect to position \( x \).

What makes the momentum operator special is its role in determining a particle's momentum when it acts on a wave function \( \psi(x) \). The resulting operation gives the momentum representation, helping us learn about a particle's motion and dynamic properties. It's analogous to calculating speed by looking at the change in position with respect to time in classical physics.
Important characteristics of the momentum operator:
  • It is Hermitian, ensuring that observable properties (like momentum) are real.
  • As a generator of translations, it connects directly with the translation operator.
  • Makes up part of the essential algebra governing quantum systems, particularly through its commutation relations with other operators.
Commutation Relations
Commutation relations form the foundation for much of the algebra in quantum mechanics. They define how two operators interact, particularly whether their order of application matters. If for two operators \( \hat{A} \) and \( \hat{B} \), the commutation relation \( [\hat{A}, \hat{B}] = \hat{A} \hat{B} - \hat{B} \hat{A} = 0 \) holds, then \( \hat{A} \) and \( \hat{B} \) commute, meaning their order of application is interchangeable.

In many quantum systems, especially where momentum and position operators are involved, commutation relations reveal the underlying principles that differentiate classical from quantum mechanics. For example, the fundamental commutation relation \( [\hat{x}, \hat{p}] = i \hbar \) highlights the Heisenberg uncertainty principle, indicating why certain properties can't simultaneously have precise values.
Key considerations for commutation relations:
  • They help identify symmetries and conservation laws within a system.
  • Operators that commute generally indicate simultaneous observability; those that don't suggest an inherent quantum uncertainty.
  • Invariance properties, such as how translators leave momentum unchanged, are often elucidated through checking specific commutations.

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Most popular questions from this chapter

Consider the parity operator in three dimensions. (a) Show that \(\hat{\Pi} \psi(\mathbf{r})=\psi^{\prime}(\mathbf{r})=\psi(-\mathbf{r})\) is equivalent to a mirror reflection followed by a rotation. (b) Show that, for \(\psi\) expressed in polar coordinates, the action of the parity operator is \(\hat{\Pi} \psi(r, \theta, \phi)=\psi(r, \pi-\theta, \phi+\pi).\) (c) Show that for the hydrogenic orbitals, \(\hat{\Pi} \psi_{n \ell m}(r, \theta, \phi)=(-1)^{\ell} \psi_{n \ell m}(r, \theta, \phi).\) That is, \(\psi_{n \ell m}\) is an eigenstate of the parity operator, with eigenvalue \((-1)^{\ell} .\) Note: This result actually applies to the stationary states of any central potential \(V(\mathbf{r})=V(r) .\) For a central potential, the eigenstates may be written in the separable form \(R_{n \ell}(r) Y_{\ell}^{m}(\theta, \phi)\) where only the radial function \(R_{n \ell}-\) which plays no role in determining the parity of the state-depends on the specific functional form of \(V(r).\)

In this problem you will establish the correspondence between Equations 6.30 and 6.31. (a) Diagonalize the matrix \(^{16}\) \(M=\left(\begin{array}{cc}1 & -\varphi / N \\ \varphi / N & 1\end{array}\right)\) to obtain the matrix \(\mathrm{M}^{\prime}=\mathrm{SMS}^{-1}\) where \(S^{-1}\) is the unitary matrix whose columns are the (normalized) eigenvectors of M. (b) Use the binomial expansion to show that \(\lim _{N \rightarrow \infty}\left(\mathrm{M}^{\prime}\right)^{N}\) is a diagonal matrix with entries \(e^{-i \varphi \text { and } e^{i \varphi} \text { on the diagonal. }}\) (c) Transform back to the original basis to show that \(\lim _{N \rightarrow \infty} \mathrm{M}^{N}=\mathrm{S}^{-1}\left[\lim _{N \rightarrow \infty}\left(\mathrm{M}^{\prime}\right)^{N}\right] \mathrm{S}\) agrees with the matrix in Equation 6.31.

Show that, for a Hermitian operator \(\hat{Q},\) the operator \(\hat{U}=\exp [i \hat{Q}]\) is unitary. Hint: First you need to prove that the adjoint is given by \(\hat{U}^{\dagger}=\exp [-i \hat{Q}] ;\) then prove that \(\hat{U}^{\dagger} \hat{U}=1 .\) Problem \(\underline{3.5}\) may help.

Consider a free particle of mass \(m .\) Show that the position and momentum operators in the Heisenberg picture are given by \(\hat{x}_{H}(t)=\hat{x}_{H}(0)+\frac{1}{m} \hat{p}_{H}(0) t\) \(\hat{p}_{H}(t)=\hat{p}_{H}(0).\) Comment on the relationship between these equations and the classical equations of motion. Hint: you will first need to evaluate the commutator \(\left[\hat{x}, \hat{H}^{n}\right] ;\) this will allow you to evaluate the commutator \([\hat{x}, \hat{U}].\)

Consider two particles of mass \(m_{1}\) and \(m_{2}\) (in one dimension) that interact via a potential that depends only on the distance between the particles \(V\left(\left|x_{1}-x_{2}\right|\right),\) so that the Hamiltonian is \(\hat{H}=-\frac{\hbar^{2}}{2 m_{1}} \frac{\partial^{2}}{\partial x_{1}^{2}}-\frac{\hbar^{2}}{2 m_{2}} \frac{\partial^{2}}{\partial x_{2}^{2}}+V\left(\left|x_{1}-x_{2}\right|\right)\) Acting on a two-particle wave function the translation operator would be \(\hat{T}(a) \psi\left(x_{1}, x_{2}\right)=\psi\left(x_{1}-a, x_{2}-a\right)\) (a) Show that the translation operator can be written \(\hat{T}(a)=e^{-\frac{i a}{\hbar} \hat{P}}\) where \(\hat{P}=\hat{p}_{1}+\hat{p}_{2}\) is the total momentum. (b) Show that the total momentum is conserved for this system.

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