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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 3: 3.14P (page 113)

Prove the famous "(your name) uncertainty principle," relating the uncertainty in position $(A = x)$ to the uncertainty in energy $(B = p^{2} / 2 m + V)$ : $蟽_{x} 蟽_{H} 猢� \frac{鈩�}{2 m} | 鉄� p 鉄� |$
For stationary states this doesn't tell you much-why not?

Short Answer

Expert verified

The uncertainty principle is $蟽_{x} 蟽_{H} 猢� \frac{鈩�}{2 m} | 鉄� p 鉄� |$

Step by step solution

01

Concept used

The generalized uncertainty principle for two observables $A$ and $B$ is given by:

$蟽_{A}^{2} 蟽_{B}^{2} 猢� {(\frac{1}{2 i} 鉄� [\hat{A}, \hat{B}] 鉄�)}^{2}$

02

Calculate the uncertainty principle

The generalized uncertainty principle for two observables $A$ and $B$ is given by:

$蟽_{A}^{2} 蟽_{B}^{2} 猢� {(\frac{1}{2 i} 鉄� [\hat{A}, \hat{B}] 鉄�)}^{2}$

The position-energy uncertainty relation is:

$蟽_{x}^{2} 蟽_{H}^{2} 猢� {(\frac{1}{2 i} 鉄� [\hat{x}, \hat{H}] 鉄�)}^{2}$ ...... (1)

So, we need to find the commutator $[\hat{x}, \hat{H}]$ as:

$[\hat{x}, \hat{H}] g = - \frac{{鈩�}^{2}}{2 m} x \frac{{鈭�}^{2} g}{鈭� x^{2}} + x V g + - \frac{{鈩�}^{2}}{2 m} \frac{{鈭�}^{2}}{鈭� x^{2}} (x g) - x V g = \frac{{鈩�}^{2}}{2 m} (- x \frac{{鈭�}^{2} g}{鈭� x^{2}} + 2 \frac{鈭� g}{鈭� x} + x \frac{{鈭�}^{2} g}{鈭� x^{2}}) = \frac{{鈩�}^{2}}{m} \frac{鈭� g}{鈭� x} = \frac{i 鈩�}{m} (p g)$

Substitute in equation 1:

$蟽_{x}^{2} 蟽_{H}^{2} {猢� (\frac{1}{2 i} 鉄� \hat{x}, \hat{H}] 鉄�)}^{2} = \frac{{鈩�}^{2}}{4 m^{2}} 鉄� p {鉄�}^{2}$

So, the uncertainty principle here becomes

$蟽_{x} 蟽_{H} 猢� \frac{鈩�}{2 m} | 鉄� p 鉄� |$

For stationary states, this doesn't tell you much because the average position of the particle doesn't change, $蟽_{H} = 0$ and $鉄� p 鉄� = 0$

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Most popular questions from this chapter

The Hamiltonian for a certain three-level system is represented by the matrix

$H = h 蝇 [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}]$ Two other observables, A and B, are represented by the matrices $A = 位 [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 2 \end{matrix}], B = 渭 [\begin{matrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$ ,where 蝇, , and 渭 are positive real numbers.

(a)Find the Eigen values and (normalized) eigenvectors of H, A and B.

(b) Suppose the system starts out in the generic staterole="math" localid="1656040462996" $| S (0) > = (\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix})$

with ${| c_{1} |}^{2} + {| c_{2} |}^{2} + {| c_{3} |}^{2} = 1$ . Find the expectation values (at t=0) of H, A, and B.

(c) What is $| S (t) >$ ? If you measured the energy of this state (at time t), what values might you get, and what is the probability of each? Answer the same questions for A and for B.

Show that

$< X > = 鈭� 桅^{*} (- \frac{h}{i} \frac{鈭�}{鈭� p}) 桅诲辫 .$

Hint: Notice that $x \exp (ipx / h) = - i h (d / dp) \exp (ip / h) .$

In momentum space, then, the position operator is $i h 鈭� / 鈭� p$ . More generally,

$< Q (x, p) = {\begin{matrix} 鈭� 蠄^{*} \hat{Q}} (x, \frac{h}{i} \frac{鈭�}{鈭� x}) 蠄诲虫, in position space; \\ 鈭� 桅^{*} \hat{Q}} (- \frac{h}{i} \frac{鈭�}{鈭� p}, p) 桅诲辫, in momentum space . \end{matrix}$ In principle you can do all calculations in momentum space just as well (though not always as easily) as in position space.

Show that the energy-time uncertainty principle reduces to the "your name" uncertainty principle (Problem 3.14), when the observable in question is x.

Test the energy-time uncertainty principle for the wave function in Problem $2.5$ and the observable x, by calculating $蟽_{H} 蟽_{X}$ and $d < x > / d t$ exactly.

Consider a three-dimensional vector space spanned by an Orthonormal basis $1 >, 2 >, 3 >$ . Kets $伪 >$ and $尾 >$ are given by

$| 伪鉄� = i | 1 鉄� - 2 | 2 鉄� - i | 3 鉄�, 鈥� 鈥� 鈥� | 尾 > = i | 1 鉄� + 2 | 3 鉄� .$

(a)Construct $< 伪$ and $< 尾$ (in terms of the dual basis

$鉄� 1 |, 鉄� 2 |, 鉄� 3 |) .$
(b) Find $鉄� 伪鈭� 尾鉄�$ and $鉄� 尾鈭� 伪鉄�,$ and confirm that

$鉄� 尾鈭� 伪鉄� = 鉄� 伪鈭� 尾 {鉄�}^{*} .$
(c)Find all nine matrix elements of the operator $\overset{铃�}{A} 鈮� | 伪鉄� 鉄� 尾 |$ , in this basis, and construct the matrix A. Is it hermitian?

See all solutions

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