Question

In Example 4.2. neither${\text{|Ψ|}}^{\text{2}}$nor$\text{|Ψ|}$are given units—only proportionalities are used. Here we verify that the results are unaffected. The actual values given in the example are particle detection rates, in particles/second, or${\text{s}}^{\text{-1}}$. For this quantity, let us use the symbol R. It is true that the particle detection rate and the probability density will be proportional, so we may write${\text{|Ψ|}}^{\text{2}}$= bR, where b is the proportionality constant. (b) What must be the units of b? (b) What is$\left|{\text{Ψ}}_{\text{T}}\right|$at the center detector (interference maximum) in terms of the example’s given detection rate and b? (c) What would be$\left|{\text{Ψ}}_{\text{1}}\right|\text{,}{\left|{\text{Ψ}}_{\text{1}}\right|}^{\text{2}}$, and the detection rate R at the center detector with one of the slits blocked?

Short answer

a)Units of b is$\text{s}/\text{m}$

(b) The detection rate at the center detector and b is$\text{10}\sqrt{b}{s}^{-1/2}$

(c) The detection rate R is${\text{25s}}^{\text{-1}}$

Step-by-step solution

Dimensional analysis.

(a)

The dimensional analysis comes to the rescue, as we are told that the units of measurement are${\text{ψ}}^{\text{2}}$ is and$\frac{\text{1}}{\text{m}}$ that of R is $\frac{\text{1}}{\text{s}}$. Since the relation between ${\text{ψ}}^{\text{2}}$and R is${\text{ψ}}^{\text{2}}=\text{bR}$

Therefore, the units of b are$\text{s}/\text{m}$

Detection rate.

(b)

In example 4.2, we were given that the detection rate (R) is 100, hence, we can directly find${\text{ψ}}_{\text{T}}$ it from the relations given in part (a).

${\left|{\text{ψ}}_{\text{1}}\right|}^2=\left(\text{1}00b\right)s^{-1}$

$\left|{\text{ψ}}_{\text{1}}\right|=\left(\text{10}\sqrt{b}\right)s^{-1/2}$

 Step 3: Detection rate R.

(c)

The wave amplitude will be halved with just a slit open, thus we can find the new detection rate R using the relation derived in section b).

$\begin{array}{c}\left|{\text{ψ}}_{\text{1}}\right|=\frac{\text{1}}{\text{2}}\left|{\text{ψ}}_{\text{T}}\right|\\ =\frac{\text{1}}{\text{2}}\times \left(\text{10}\sqrt{b}{s}^{-1/2}\right)\end{array}$

$\begin{array}{l}{\left|{\text{ψ}}_{\text{1}}\right|}^{\text{2}}=\left(25b\right)s^{-1}\\ {\left|{\text{ψ}}_{\text{1}}\right|}^{\text{2}}={\text{R}}_{\text{1}}\text{b}\end{array}$

Comparing it we get,

${\text{R}}_{\text{1}}={\text{25\hspace{0.17em}s}}^{\text{-1}}$

Result.

(a)Units of b is$\text{s}/\text{m}$

(b) The detection rate at the center detector and b is$\text{10}\sqrt{b}{s}^{-1/2}$

(c) The detection rate R is${\text{25s}}^{\text{-1}}$