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Chapter 4 The Postulates——a General Discussion
2024-08-06
Solutions to Principles of Quantum Mechanics
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目录

Chapter 4 The Postulates——a General Discussion
4.1 The Postulates
4.2 Discussion of Postulates I-III
4.3 The Schrödinger Equation (Dotting Your $i$'s and Crossing your $\hbar$'s)

Chapter 4 The Postulates——a General Discussion

4.1 The Postulates

4.2 Discussion of Postulates I-III

Exercise 4.2.1 Consider the following operators on a Hilbert space V3(C)\mathbb{V}^{3}(C):

Lx=121/2(010101010)Ly=121/2(0−i0i0−i0i0)Lz=(10000000−1)L_{x}=\frac{1}{2^{1/2}}\begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}\qquad L_{y}=\frac{1}{2^{1/2}}\begin{pmatrix} 0&-i&0\\ i&0&-i\\ 0&i&0 \end{pmatrix}\qquad L_{z}=\begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}

(1) What are the possible values one can obtain if LzL_{z} is measured?

(2) Take the state in which Lz=1L_{z}=1. In this state what are ⟨Lx⟩\langle L_{x}\rangle, ⟨Lx2⟩\langle L_{x}^{2}\rangle and ΔLx\Delta L_{x}?

(3) Find the normalized eigenstates and the eigenvalues of LxL_{x} in the LzL_{z} basis.

(4) If the particle is in the state with Lz=−1L_{z}=-1, and LxL_{x} is measured, what are the possible outcomes and their probabilities? (5) Consider the state

∣ψ⟩=(1/21/21/21/2)|\psi\rangle=\begin{pmatrix} 1/2\\ 1/2\\ 1/2^{1/2} \end{pmatrix}

in the LzL_{z} basis. If Lz2L_{z}^{2} is measured in this state and a result +1+1 is obtained, what is the state after the measurement? How probable was this result? If LzL_{z} is measured, what are the outcomes and respective probabilities?

(6) A particle is in a state for which the probabilities are P(Lz=1)=1/4P(L_{z}=1)=1/4, P(Lz=0)=1/2P(L_{z}=0)=1/2, and P(Lz=−1)=1/4P(L_{z}=-1)=1/4. Convince yourself that the most general, normalized state with this property is

∣ψ⟩=eiδ12∣Lz=1⟩+eiδ221/2∣Lz=0⟩+eiδ32∣Lz=−1⟩|\psi\rangle=\frac{e^{i \delta_1}}{2}\left|L_z=1\right\rangle+\frac{e^{i \delta_2}}{2^{1 / 2}}\left|L_z=0\right\rangle+\frac{e^{i \delta_3}}{2}\left|L_z=-1\right\rangle

It was stated earlier on that if ∣ψ⟩|\psi\rangle is a normalized state then the state eiθ∣ψ⟩e^{i \theta}|\psi\rangle is a physically equivalent normalized state. Does this mean that the factors eiδie^{i \delta_i} multiplying the LzL_z eigenstates are irrelevant? [Calculate for example P(Lx=0)P\left(L_x=0\right).]

Solution. (1) The possible values one can obtain if LzL_{z} is measured are its eigenvalues

Lz=(10000000−1)L_{z}=\begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}

Eigenvalues are 1,0,−1{1,0,-1}.

(2) The state in which Lz∣ψ⟩=1⋅∣ψ⟩L_{z}|\psi\rangle=1\cdot|\psi\rangle is the corresponding eigenvector

∣ψ⟩=(100)|\psi\rangle=\begin{pmatrix} 1\\0\\0 \end{pmatrix}

Then in ∣ψ⟩|\psi\rangle

⟨Lx⟩=⟨ψ∣Lx∣ψ⟩=(1 0 0)12(010101010)(100)=12(1 0 0)(010)=0⟨Lx2⟩=⟨ψ∣Lx2∣ψ⟩=(1 0 0)12(010101010)(010101010)(100)=12(0 1 0)(010)=12ΔLx=⟨Lx2⟩−(⟨Lx⟩)2=(12)−02=12\begin{aligned} \langle L_{x}\rangle&=\langle\psi|L_{x}|\psi\rangle=(1~0~0)\frac{1}{\sqrt{2}}\begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}\begin{pmatrix} 1\\0\\0 \end{pmatrix}=\frac{1}{\sqrt{2}}(1~0~0)\begin{pmatrix} 0\\1\\0 \end{pmatrix}=0\\ \langle L_{x}^{2}\rangle&=\langle\psi|L_{x}^{2}|\psi\rangle=(1~0~0)\frac{1}{2}\begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}\begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}\begin{pmatrix} 1\\0\\0 \end{pmatrix}=\frac{1}{2}(0~1~0)\begin{pmatrix} 0\\1\\0 \end{pmatrix}=\frac{1}{2}\\ \Delta L_{x}&=\sqrt{\langle L_{x}^{2}\rangle-(\langle L_{x}\rangle)^{2}}=\sqrt{\left(\frac{1}{2}\right)-0^{2}}=\frac{1}{\sqrt{2}} \end{aligned}

(3) The characteristic equation for LxL_{x} is

0=det⁡(Lx−λ)=det⁡(−λ12012−λ12012−λ)=λ−λ3⇒λ∈{1,0,−1}.0=\det(L_{x}-\lambda)=\det\begin{pmatrix} -\lambda & \frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}&-\lambda&\frac{1}{\sqrt{2}}\\ 0&\frac{1}{\sqrt{2}}&-\lambda \end{pmatrix}=\lambda-\lambda^{3}\qquad \Rightarrow \qquad \lambda\in\{1,0,-1\}.

The corresponding eigenvectors ∣λ⟩|\lambda\rangle, then satisfy

0=(Lx−λI)∣λ⟩=(−λ12012−λ12012−λ)(abc)=(−λa+b2a2−λb+c2b2−λa)0=(L_{x}-\lambda I)|\lambda\rangle=\begin{pmatrix} -\lambda & \frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}&-\lambda&\frac{1}{\sqrt{2}}\\ 0&\frac{1}{\sqrt{2}}&-\lambda \end{pmatrix}\begin{pmatrix} a\\b\\c \end{pmatrix}= \begin{pmatrix} -\lambda a+\frac{b}{\sqrt{2}}\\ \frac{a}{\sqrt{2}}-\lambda b+\frac{c}{\sqrt{2}}\\ \frac{b}{\sqrt{2}}-\lambda a \end{pmatrix}

where we have parameterized the components of ∣λ⟩|\lambda\rangle by (a b c)(a~b~c). For λ=1\lambda=1, we can solve for bb and cc in terms of aa by solving the following equations:

{−a+b2=0a2−b+c2=0b2−a=0\left\{\begin{aligned} -a+\frac{b}{\sqrt{2}}=0\\ \frac{a}{\sqrt{2}}-b+\frac{c}{\sqrt{2}}=0\\ \frac{b}{\sqrt{2}}-a=0 \end{aligned}\right.

We get

{b=2ac=a\left\{ \begin{aligned} b&=\sqrt{2}a\\ c&=a \end{aligned} \right.

We then determine aa by normalizing ∣λ=1⟩|\lambda=1\rangle:

∣λ=1⟩=(a2aa)⇒1=⟨λ=1∣λ=1⟩=(a∗ 2a∗ a∗)(a2aa)=4∣a∣2⇒a=12\begin{aligned} &|\lambda=1\rangle=\begin{pmatrix} a\\\sqrt{2}a\\a \end{pmatrix}\\ \Rightarrow\quad &1=\langle\lambda=1|\lambda=1\rangle=(a^{*}~\sqrt{2}a^{*}~a^{*})\begin{pmatrix} a\\\sqrt{2}a\\a \end{pmatrix}=4|a|^{2}\\ \Rightarrow\quad &a=\frac{1}{2} \end{aligned}

(where I have chosen the arbitrary phase to be 11).

We could do the same thing for λ=0\lambda=0:

{b2=0a2+c2=0b2=0\left\{ \begin{aligned} \frac{b}{\sqrt{2}}=0\\ \frac{a}{\sqrt{2}}+\frac{c}{\sqrt{2}}=0\\ \frac{b}{\sqrt{2}}=0 \end{aligned} \right.

has a solution:

{b=0c=−a\left\{ \begin{aligned} b&=0\\ c&=-a \end{aligned} \right.

Normalizing:

∣λ=0⟩=(a0−a)⇒1=⟨λ=0∣λ=0⟩=(a∗ 0 −a∗)(a0−a)=2∣a∣2⇒a=12\begin{aligned} &|\lambda=0\rangle=\begin{pmatrix} a\\ 0\\ -a \end{pmatrix}\\ \Rightarrow\quad & 1=\langle \lambda=0|\lambda=0\rangle=(a^{*}~0~-a^{*})\begin{pmatrix} a\\0\\-a \end{pmatrix}=2|a|^{2}\\ \Rightarrow\quad & a=\frac{1}{\sqrt{2}} \end{aligned}

And for λ=−1\lambda=-1:

{a+b2=0a2+b+c2=0b2+a=0\left\{\begin{aligned} a+\frac{b}{\sqrt{2}}=0\\ \frac{a}{\sqrt{2}}+b+\frac{c}{\sqrt{2}}=0\\ \frac{b}{\sqrt{2}}+a=0 \end{aligned}\right.

We get

{b=−2ac=a\left\{ \begin{aligned} b&=-\sqrt{2}a\\ c&=a \end{aligned} \right.

Normalizing:

∣λ=−1⟩=(a−2aa)⇒1=⟨λ=−1∣λ=−1⟩=(a∗ −2a∗ a∗)(a−2aa)=4∣a∣2⇒a=12\begin{aligned} &|\lambda=-1\rangle=\begin{pmatrix} a\\ -\sqrt{2}a\\ a \end{pmatrix}\\ \Rightarrow\quad & 1=\langle \lambda=-1|\lambda=-1\rangle=(a^{*}~-\sqrt{2}a^{*}~a^{*})\begin{pmatrix} a\\-\sqrt{2}a\\a \end{pmatrix}=4|a|^{2}\\ \Rightarrow\quad & a=\frac{1}{2} \end{aligned}

Therefore,

∣λ=1⟩=12(121)∣λ=0⟩=12(10−1)∣λ=−1⟩=12(1−21)|\lambda=1\rangle=\frac{1}{2}\begin{pmatrix} 1\\\sqrt{2}\\1 \end{pmatrix}\quad|\lambda=0\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 1\\0\\-1 \end{pmatrix}\quad|\lambda=-1\rangle=\frac{1}{2}\begin{pmatrix} 1\\-\sqrt{2}\\1 \end{pmatrix}

Next, we should compute the components of these 33 LxL_{x}-eigenstate in the {∣1⟩,∣0⟩,∣−1⟩}\{|1\rangle,|0\rangle,|-1\rangle\}-basis of LzL_{z}-eigenstates. But since LzL_{z} is diagonal in the basis in which LxL_{x}, LyL_{y} and LzL_{z} are given, the basis that Shankar used to write down the matrix elements of LxL_{x}, LyL_{y}, LzL_{z} is the LzL_{z}-eigenbasis. So the components of ∣Lx=1,0,−1⟩|L_{x}=1,0,-1\rangle in the given basis that we just calculated are their components in the LzL_{z}-eigenbasis.

(4) The eigenvectors of LzL_{z} corresponding to Lz=−1L_{z}=-1 is (001)\begin{pmatrix} 0\\0\\1 \end{pmatrix}. If we measure LxL_{x} in any state, the possible outcomes are any one of the eigenvalues Lx=±1,0L_{x}=\pm 1, 0.

The probabilities for Lx=±1,0L_{x}=\pm 1,0 in the state ∣−1⟩=∣Lz=−1⟩|-1\rangle=|L_{z}=-1\rangle are:

P(Lx=1)=∣⟨Lx=1∣Lz=−1⟩∣2=∣(12 12 12)(001)∣2=14P(Lx=0)=∣⟨Lx=0∣Lz=−1⟩∣2=∣(12 0 −12)(001)∣2=12P(Lx=−1)=∣⟨Lx=−1∣Lz=−1⟩∣2=∣(12 −12 12)(001)∣2=14\begin{aligned} &P(L_{x}=1)=|\langle L_{x}=1|L_{z}=-1\rangle|^{2}=\left|(\frac{1}{2}~\frac{1}{\sqrt{2}}~\frac{1}{2})\begin{pmatrix} 0\\0\\1 \end{pmatrix}\right|^{2}=\frac{1}{4}\\ &P(L_{x}=0)=|\langle L_{x}=0|L_{z}=-1\rangle|^{2}=\left|(\frac{1}{\sqrt{2}}~0~-\frac{1}{\sqrt{2}})\begin{pmatrix} 0\\0\\1 \end{pmatrix}\right|^{2}=\frac{1}{2}\\ &P(L_{x}=-1)=|\langle L_{x}=-1|L_{z}=-1\rangle|^{2}=\left|(\frac{1}{2}~-\frac{1}{\sqrt{2}}~\frac{1}{2})\begin{pmatrix} 0\\0\\1 \end{pmatrix}\right|^{2}=\frac{1}{4} \end{aligned}

(5) Consider the state

∣ψ⟩=(121212)|\psi\rangle=\begin{pmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{\sqrt{2}} \end{pmatrix}

in the LzL_{z} basis.

Since Lz2L_{z}^{2} is measured to be +1+1, LzL_{z} can be +1+1 or −1-1. The state after the measurement is

∣ψ⟩after=N(∣Lz=+1⟩⟨Lz=+1∣+∣Lz=−1⟩⟨Lz=−1∣)∣ψ⟩=N((100)(1 0 0)+(001)(0 0 1))(121212)=N(100000001)(121212)=N(12012)=1(12)2+(12)2(12012)=23(12012)=(13023)\begin{aligned} |\psi\rangle_{\text{after}}&=\mathcal{N}(|L_{z}=+1\rangle\langle L_{z}=+1|+|L_{z}=-1\rangle\langle L_{z}=-1|)|\psi\rangle\\ &=\mathcal{N}\left(\begin{pmatrix} 1\\0\\0 \end{pmatrix}(1~0~0)+\begin{pmatrix} 0\\0\\1 \end{pmatrix}(0~0~1)\right)\begin{pmatrix} \frac{1}{2}\\\frac{1}{2}\\\frac{1}{\sqrt{2}} \end{pmatrix}\\ &=\mathcal{N}\begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&1 \end{pmatrix}\begin{pmatrix} \frac{1}{2}\\\frac{1}{2}\\\frac{1}{\sqrt{2}} \end{pmatrix}=\mathcal{N}\begin{pmatrix} \frac{1}{2}\\0\\\frac{1}{\sqrt{2}} \end{pmatrix}\\ &=\frac{1}{\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}}}\begin{pmatrix} \frac{1}{2}\\0\\\frac{1}{\sqrt{2}} \end{pmatrix}\\ &=\frac{2}{\sqrt{3}}\begin{pmatrix} \frac{1}{2}\\0\\\frac{1}{\sqrt{2}} \end{pmatrix}\\ &=\begin{pmatrix} \frac{1}{\sqrt{3}}\\0\\ \sqrt{\frac{2}{3}} \end{pmatrix} \end{aligned}

where N\mathcal{N} normalizes the state. The probability of this result is

P(Lz2=+1)=P(Lz=+1)+P(Lz=−1)=∣⟨Lz=+1∣ψ⟩∣2+∣⟨Lz=−1∣ψ⟩∣2=∣(1 0 0)(121212)∣2+∣(0 0 1)(121212)∣2=∣12∣2+∣12∣2=14+12=34\begin{aligned} P(L_{z}^{2}=+1)&=P(L_{z}=+1)+P(L_{z}=-1)\\ &=|\langle L_{z}=+1|\psi\rangle|^{2}+|\langle L_{z}=-1|\psi\rangle|^{2}\\ &=\left|(1~0~0)\begin{pmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{\sqrt{2}} \end{pmatrix}\right|^{2}+\left|(0~0~1)\begin{pmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{\sqrt{2}} \end{pmatrix}\right|^{2}\\ &=\left|\frac{1}{2}\right|^{2}+\left|\frac{1}{\sqrt{2}}\right|^{2}\\ &=\frac{1}{4}+\frac{1}{2}\\ &=\frac{3}{4} \end{aligned}

If LzL_{z} is measured after Lz2L_{z}^{2} was measured and Lz2=+1L_{z}^{2}=+1 was found, the possible outcomes and relative probabilities are:

P(Lz=+1)after=∣⟨Lz=+1∣ψ⟩after∣2=∣(1 0 0)(13023)∣2=13P(Lz=0)after=∣⟨Lz=0∣ψ⟩after∣2=∣(0 1 0)(13023)∣2=0P(Lz=−1)after=∣⟨Lz=−1∣ψ⟩after∣2=∣(0 0 1)(13023)∣2=23\begin{aligned} P(L_{z}=+1)_{\text{after}}&=|\langle L_{z}=+1|\psi\rangle_{\text{after}}|^{2}=\left|(1~0~0)\begin{pmatrix} \frac{1}{\sqrt{3}}\\0\\ \sqrt{\frac{2}{3}} \end{pmatrix}\right|^{2}=\frac{1}{3}\\ P(L_{z}=0)_{\text{after}}&=|\langle L_{z}=0|\psi\rangle_{\text{after}}|^{2}=\left|(0~1~0)\begin{pmatrix} \frac{1}{\sqrt{3}}\\0\\ \sqrt{\frac{2}{3}} \end{pmatrix}\right|^{2}=0\\ P(L_{z}=-1)_{\text{after}}&=|\langle L_{z}=-1|\psi\rangle_{\text{after}}|^{2}=\left|(0~0~1)\begin{pmatrix} \frac{1}{\sqrt{3}}\\0\\ \sqrt{\frac{2}{3}} \end{pmatrix}\right|^{2}=\frac{2}{3} \end{aligned}

(6) A particle is in a state for which the probabilities are P(Lz=1)=1/4P(L_{z}=1)=1/4, P(Lz=0)=1/2P(L_{z}=0)=1/2, and P(Lz=−1)=1/4P(L_{z}=-1)=1/4. Suppose it has the following form

∣ψ⟩=C1∣Lz=+1⟩+C2∣Lz=0⟩+C3∣Lz=−1⟩|\psi\rangle=C_{1}|L_{z}=+1\rangle+C_{2}|L_{z}=0\rangle+C_{3}|L_{z}=-1\rangle

where C1C_{1}, C2C_{2} and C3C_{3} are complex numbers. Then we have

∣C1∣2=C1∗C1=P(Lz=+1)=14⇒C1=12eiδ1∣C2∣2=C2∗C2=P(Lz=0)=12⇒C2=12eiδ2∣C3∣2=C3∗C3=P(Lz=−1)=14⇒C3=12eiδ3\begin{aligned} |C_{1}|^{2}&=C_{1}^{*}C_{1}=P(L_{z}=+1)=\frac{1}{4} \quad &\Rightarrow & \quad C_{1}=\frac{1}{2}e^{i\delta_{1}}\\ |C_{2}|^{2}&=C_{2}^{*}C_{2}=P(L_{z}=0)=\frac{1}{2} \quad &\Rightarrow & \quad C_{2}=\frac{1}{\sqrt{2}}e^{i\delta_{2}}\\ |C_{3}|^{2}&=C_{3}^{*}C_{3}=P(L_{z}=-1)=\frac{1}{4} \quad &\Rightarrow & \quad C_{3}=\frac{1}{2}e^{i\delta_{3}} \end{aligned}

where δ1\delta_{1}, δ2\delta_{2} and δ3\delta_{3} are arbitrary real numbers. Therefore, it has the form

∣ψ⟩=eiδ12∣Lz=1⟩+eiδ221/2∣Lz=0⟩+eiδ32∣Lz=−1⟩|\psi\rangle=\frac{e^{i \delta_1}}{2}\left|L_z=1\right\rangle+\frac{e^{i \delta_2}}{2^{1 / 2}}\left|L_z=0\right\rangle+\frac{e^{i \delta_3}}{2}\left|L_z=-1\right\rangle

The values of the phases matter when measuring an observable that is incompatible with LzL_{z}, as an example:

⟨Lx=0∣Lz=1⟩=12(1 0 −1)(100)=12⟨Lx=0∣Lz=0⟩=12(1 0 −1)(010)=0⟨Lx=0∣Lz=−1⟩=12(1 0 −1)(001)=−12\begin{aligned} \langle L_{x}=0|L_{z}=1\rangle&=\frac{1}{\sqrt{2}} (1~0~-1)\begin{pmatrix} 1\\0\\0 \end{pmatrix}=\frac{1}{\sqrt{2}}\\ \langle L_{x}=0|L_{z}=0\rangle&=\frac{1}{\sqrt{2}} (1~0~-1)\begin{pmatrix} 0\\1\\0 \end{pmatrix}=0\\ \langle L_{x}=0|L_{z}=-1\rangle&=\frac{1}{\sqrt{2}} (1~0~-1)\begin{pmatrix} 0\\0\\1 \end{pmatrix}=-\frac{1}{\sqrt{2}} \end{aligned}
P(Lx=0)=∣⟨Lx=0∣ψ⟩∣2=∣(12 0 −12)(eiδ12eiδ22eiδ32)∣2=18∣eiδ1−eiδ3∣2=18(eiδ1−eiδ3)(e−iδ1−e−iδ3)=18(1−ei(δ1−δ3)−e−i(δ1−δ3)+1)=18(2−2⋅ei(δ1−δ3)+e−i(δ1−δ3)2)=14(1−cos⁡(δ1−δ3))\begin{aligned} P(L_{x}=0)&=|\langle L_{x}=0|\psi\rangle|^{2}\\ &=\left|\left(\frac{1}{\sqrt{2}}~0~-\frac{1}{\sqrt{2}}\right)\begin{pmatrix} \frac{e^{i\delta_{1}}}{2}\\ \frac{e^{i\delta_{2}}}{\sqrt{2}}\\ \frac{e^{i\delta_{3}}}{2} \end{pmatrix}\right|^{2}\\ &=\frac{1}{8}|e^{i\delta_{1}}-e^{i\delta_{3}}|^{2}\\ &=\frac{1}{8}(e^{i\delta_{1}}-e^{i\delta_{3}})(e^{-i\delta_{1}}-e^{-i\delta_{3}})\\ &=\frac{1}{8}(1-e^{i(\delta_{1}-\delta_{3})}-e^{-i(\delta_{1}-\delta_{3})}+1)\\ &=\frac{1}{8}(2-2\cdot\frac{e^{i(\delta_{1}-\delta_{3})}+e^{-i(\delta_{1}-\delta_{3})}}{2})\\ &=\frac{1}{4}(1-\cos(\delta_{1}-\delta_{3})) \end{aligned}

It depends on phases and can be measured by experiment.

 ■~\tag*{$\blacksquare$}

Exercise 4.2.2 Show that for a real wave function ψ(x)\psi(x), the expectation value of momentum ⟨P⟩=0\langle P\rangle=0. (Hint: Show that the probabilities for the momenta ±p\pm p are equal.) Generalize this result to the case ψ=cψr\psi=c \psi_r, where ψr\psi_r is real and cc an arbitrary (real or complex) constant. (Recall that ∣ψ⟩|\psi\rangle and α∣ψ⟩\alpha|\psi\rangle are physically equivalent.)

Solution. Since ψ(x)\psi(x) is real, ψ∗(x)=ψ(x)\psi^{*}(x)=\psi(x).

⟨P⟩=∫−∞+∞ψ∗(x)(−iℏ∂∂x)ψ(x) dx=−iℏ∫−∞+∞ψ(x)∂∂xψ(x) dx=−12iℏ∫−∞+∞∂∂xψ2(x) dx=−12iℏψ2(x)∣−∞+∞=0 \begin{aligned} \langle P\rangle&=\int_{-\infty}^{+\infty} \psi^{*}(x)\left(-i\hbar\frac{\partial}{\partial x}\right)\psi(x)~dx\\ &=-i\hbar\int_{-\infty}^{+\infty}\psi(x)\frac{\partial}{\partial x}\psi(x)~dx\\ &=-\frac{1}{2}i\hbar\int_{-\infty}^{+\infty}\frac{\partial}{\partial x}\psi^{2}(x)~dx\\ &=-\frac{1}{2}i\hbar \psi^{2}(x)\bigg|_{-\infty}^{+\infty}\\ &=0 \end{aligned}

Since ψ(x)→0\psi(x)\to 0, as x→±∞x\to\pm\infty.

For general case,

⟨P⟩=∫−∞+∞[cψr(x)]∗(−iℏ∂∂x)[cψr(x)]=∣c∣2∫−∞+∞ψr(x)(−iℏ∂∂x)ψr(x) dx=∣c∣2⋅0=0 \begin{aligned} \langle P\rangle&=\int_{-\infty}^{+\infty}[c\psi_{r}(x)]^{*}\left(-i\hbar\frac{\partial}{\partial x}\right)[c\psi_{r}(x)]\\ &=|c|^{2}\int_{-\infty}^{+\infty} \psi_{r}(x)\left(-i\hbar\frac{\partial}{\partial x}\right)\psi_{r}(x)~dx\\ &=|c|^{2}\cdot 0\\ &=0 \end{aligned}
 ■~\tag*{$\blacksquare$}

Exercise 4.2.3 Show that if ψ(x)\psi(x) has mean momentum ⟨P⟩,eip0x/ℏψ(x)\langle P\rangle, e^{i p_{0} x / \hbar} \psi(x) has mean momentum ⟨P⟩+p0\langle P\rangle+p_0.

Solution.

⟨P⟩=−iℏ∫−∞+∞ψ∗(x)∂∂xψ(x) dx⟨P′⟩=−iℏ∫−∞+∞[e−ip0x/ℏψ∗(x)]∂∂x[eip0x/ℏψ(x)] dx=−iℏ∫−∞+∞[e−ip0x/ℏψ∗(x)][eip0x/ℏ⋅ip0/ℏ⋅ψ(x)+eip0x/ℏ∂∂xψ(x)] dx=−iℏ∫−∞+∞e−ip0x/ℏψ∗(x)⋅eip0x/ℏ⋅ip0/ℏ⋅ψ(x) dx    −iℏ∫−∞+∞e−ip0x/ℏψ∗(x)⋅eip0x/ℏ∂∂xψ(x) dx=p0∫−∞+∞ψ∗(x)ψ(x) dx−iℏ∫−∞+∞ψ∗(x)⋅∂∂xψ(x) dx=⟨P⟩+p0 \begin{aligned} \langle P\rangle&=-i\hbar\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\partial}{\partial x}\psi(x)~dx\\ \langle P^{\prime}\rangle&=-i\hbar\int_{-\infty}^{+\infty}[e^{-ip_{0}x/\hbar}\psi^{*}(x)]\frac{\partial}{\partial x}[e^{ip_{0}x/\hbar}\psi(x)]~dx\\ &=-i\hbar\int_{-\infty}^{+\infty}[e^{-ip_{0}x/\hbar}\psi^{*}(x)][e^{ip_{0}x/\hbar}\cdot ip_{0}/\hbar\cdot\psi(x)+e^{ip_{0}x/\hbar}\frac{\partial}{\partial x}\psi(x)]~dx\\ &=-i\hbar\int_{-\infty}^{+\infty}e^{-ip_{0}x/\hbar}\psi^{*}(x)\cdot e^{ip_{0}x/\hbar}\cdot ip_{0}/\hbar\cdot\psi(x)~dx\\ &~~~~-i\hbar\int_{-\infty}^{+\infty}e^{-ip_{0}x/\hbar}\psi^{*}(x)\cdot e^{ip_{0}x/\hbar}\frac{\partial}{\partial x}\psi(x)~dx\\ &=p_{0}\int_{-\infty}^{+\infty}\psi^{*}(x)\psi(x)~dx-i\hbar\int_{-\infty}^{+\infty}\psi^{*}(x)\cdot\frac{\partial}{\partial x}\psi(x)~dx\\ &=\langle P\rangle+p_{0} \end{aligned}
 ■~\tag*{$\blacksquare$}

4.3 The Schrödinger Equation (Dotting Your ii's and Crossing your ℏ\hbar's)