Q78E When applying quantum mechanics,... [FREE SOLUTION]

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Chapter 7: Q78E (page 284)

When applying quantum mechanics, we often concentrate on states that qualify as 鈥渙rthonormal鈥�, The main point is this. If we evaluate a probability integral over all space of ${蠒_{1}}^{} 蠒_{1}$ or of ${蠒_{2}}^{} 蠒_{2}$ , we get 1 (unsurprisingly), but if we evaluate such an integral for ${蠒_{1}}^{} 蠒_{2} o r {蠒_{2}}^{} 蠒_{1}$ we get 0. This happens to be true for all systems where we have tabulated or actually derived sets of wave functions (e.g., the particle in a box, the harmonic oscillator, and the hydrogen atom). By integrating overall space, show that expression (7-44) is not normalized unless a factor of $1 / 2$ is included with the probability.

Short Answer

Expert verified

By integrating overall space, you get that, a factor of 陆 will be required to be included with the probability.

Step by step solution

Used formula:

The wave function transitional state can be represented by the following equation:

$唯_{t r a n s i t i o n} (r, t) = 唯_{i} (r, t) + 唯_{f} (r, t) 唯_{transition} (r, t) = 唯_{i} (r) e^{- {iE}_{i} t / h} + 唯_{f} (r) e^{- {iE}_{f} t / h} 鈥� . . (7 - 44)$

Where,

唯

isWave function,

唯_{i}

is the Wave function of initial state,

唯_{f}

is the Wave function of final state, r is the radius, t is time,

E_{i}

is the initial energy,

E_{f}

is final energy, h is Planck鈥檚 constant.

Integrating overall space:

Consider the given data as below.

$鈭� 唯 . 唯鈥� d V = 1$

Now, if you introduce a factor A which can be adjusted to give probability equal to 1.

$鈭� (A . {唯_{i}}^{*} (r) e^{+ i E_{i} t / h} + A . {唯_{f}}^{*} (r) e^{+ e^{+ i E_{i} t / h}}) (A . 唯_{i} (r) e^{- i E_{i} t / h} + A . 唯_{f} (r) e^{+ e^{- i E_{i} t / h}}) 鈥� d V = 1 {|A|}^{2} [{鈭� |唯_{i} (r)|}^{2} d V + e^{+ i (E_{i} + E_{f}) t / h} 鈭� {唯_{i}}^{*} (r) 唯_{f} (r) d V + e^{- i (E_{i} - E_{f}) t / h} 鈭� {唯_{f}}^{*} (r) 唯_{i} (r) d V + {鈭� |唯_{f} (r)|}^{2} d V] = 1 {|A|}^{2} (1 + 0 + 0 + 1) = 1$

$A = \frac{1}{\sqrt{2}}$

Hence, the factor $\frac{1}{\sqrt{2}}$ must be included with each wave function.

As you know that probability density is the square of wave function.

Hence, a factor of $\frac{1}{2}$ will be required to be included with the probability.

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Short Answer

Step by step solution

Used formula:

Integrating overall space:

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