Composition Of Two Bijections Is a Bijection: Proof

A common exercise from an "Introduction To Real Analysis" course is to show that

1. if $f: A\rightarrow B$ is a bijection, and
2. if $g: B\rightarrow C$ is a bijection

then the composite $g\circ f$ is a bijective map of A onto C.

We just need to show that $g\circ f: A\rightarrow C$ is one to one and onto.

$$g\circ f = g(f(x))$$
$$f(x)=y     (y\in B)$$
$$g(y)=z    (z\in C)$$

First, show the composite is injective (one to one):
by 1, we know that $f(x_{1})=f(x_{2})$ means $x_{1}=x_{2}$ (f is one to one - injective).
by 2, we know that $g(f(x_{1}))=g(f(x_{2}))$ means $f(x_{1})=f(x_{2})$ (g is one to one - injective).

a) So, we see that $g(f(x_{1}))=g(f(x_{2}))$ means that  $x_{1}=x_{2}$, which means that $g\circ f$ is       an injective (one to one) map of A onto C.

Second, show the composite is surjective (onto):
by 1, we know that for any $b\in B$ there exists at least one $x\in A$ such that $f(x)=b$.
by 2, we know that for any $c\in C$ there exists at least one $b\in B$ such that $g(b)=c$.

b) So, we see that for any$c\in C$ there exists at least one  $x\in A$ such that $g(f(x))=c$, which means that    $g\circ f$ is a surjective (onto) map of A onto C.

With the results a) and b) we see that the composite $g\circ f$ is indeed a bijective map of A onto C.

QED

Momentum-Space Wavefunction - Expectation Values: Quantum Mechanics (Griffiths)

In D. J. Griffiths' "Introduction to Quantum Mechanics: 1st Edition," Problem 3.51 on page 117 is a 3-star problem. In the second edition, it's problem 3.12 on page 109, and is not a starred problem at all. I attribute this to being given more information about $\Psi_{x,t}$ in terms of $\Phi_{p,t}$ in the second edition. In the first edition, we are working from knowledge of the form of $\Phi$ alone. Since one is a Fourier transform of the other and vice versa, we can convert the integral used to evaluate $< x>$  from one with an integrand function containing $\Psi$ and x to one containing $\Phi$ and the x operator.

$$< x> = \int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp$$

In the first edition, we're given

$$\Phi(p,t)=\frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}e^{-ipx/\hbar}\Psi(x,t)dx        (1)$$

We know $<x>=\int_{-\infty}^{\infty}x|\Psi(x,t)|^{2}dx$, so we just need to show that

$$\int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp=\int x|\Psi(x,t)|^{2}dx$$

Easier said than done, but here we go. It's not too bad, actually...

$$\int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp$$

Using (1) gives

$$ \int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left ( -\frac{\hbar}{i} \frac{\partial }{\partial p}\right )\left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{-ipx/\hbar}\Psi dx \right ]dp$$

Simplifying a little...

$$\int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left [ \frac{1}{\sqrt{2\pi \hbar}}\int i\hbar\frac{\partial }{\partial p} e^{-ipx/\hbar}\Psi dx \right ]dp        (2)$$

Consider the integrand function in the second set of brackets. In particular,  $i\hbar\frac{\partial }{\partial p} e^{-ipx/\hbar}$.

Using the fact that $\frac{\mathrm{d} }{\mathrm{d} p}\left ( e^{-ipx/\hbar} \right )= -\frac{ix}{\hbar} e^{-ipx/\hbar}$ we can rewrite (2) using $x e^{-ipx/\hbar}$ in place of $i\hbar \frac{\partial }{\partial p}\left ( e^{-ipx/\hbar} \right )$ ...

$$\int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left [ \frac{1}{\sqrt{2\pi \hbar}}\int x e^{-ipx/\hbar} \Psi dx \right ]dp$$

Simplifying further and using Fubini's Theorem gives

$$\frac{1}{2\pi \hbar}\int \int \int \left [ e^{ipx^{'}/\hbar}\Psi^{*} x e^{-ipx/\hbar}\Psi \right ]dx^{'}dxdp     (3)$$

Letting $y = \frac{p}{\hbar}$ and simplifying the integrand further gives

$$\frac{1}{2\pi }\int \int \int \left [ e^{-i(x-x^{'})/y} \Psi^{*} x \Psi \right ]dx^{'}dxdy     (4)$$

Evaluating the y-integral first...

$$\frac{1}{2\pi }\int e^{-i(x-x^{'})/y}dy = \delta (x-x^{'})$$

If you're not comfortable with this, check it for yourself by using Plancherel's Theorem to determine the Fourier transform of $\delta(x)$.

Now we have

$$\int \int \delta(x-x^{'})  \Psi^{*}(x^{'},t) x \Psi (x,t) dx^{'}dx     (5)$$

But $ \int \delta(x-x^{'}) \Psi^{*}(x^{'},t) = \Psi^{*} (x,t)$  (see Eq. 3.129 in the first edition and Eq. 3.52 in the second edition), so we have

$$\int \Psi^{*}x \Psi dx = \int x |\Psi (x,t)|^{2}dx$$

QED

Two Linear Algebra Proofs - Hermitian and Transpose: Quantum Mechanics (Griffiths)

Problem A.11 from D.J. Griffiths' "Introduction to Quantum Mechanics: 2nd Edition" asks the student to prove a few properties from linear algebra. The properties aren't all that difficult to prove, but give some students trouble because these properties are generally used by students to prove other relationships in linear algebra.

The first property: The transpose of a product is the product of the transposes in reverse order...

$$\widetilde{\mathbf{ST}}=\widetilde{\mathbf{T}}\widetilde{\mathbf{S}}$$

$$\widetilde{\mathbf{ST}}$$

$$=(ST)_{ji}$$

$$=\sum_{k=1}^{n}S_{jk}T_{ki}$$

$$=\sum_{k=1}^{n}\widetilde{S}_{kj}\widetilde{T}_{ik}$$

$$=\sum_{k=1}^{n}\widetilde{T}_{ik}\widetilde{S}_{kj}$$

$$=\mathbf{\widetilde{T}}\mathbf{\widetilde{S}}$$

QED

The second property: The Hermitian conjugate of a product is the product of the Hermitian conjugates in reverse order...

$$(\mathbf{ST})^{\dagger }=\mathbf{T}^{\dagger }\mathbf{S}^{\dagger }$$

$$(\mathbf{ST})^{\dagger }$$

$$=(\widetilde{\mathbf{ST}})^{*}$$

$$=(\mathbf{\widetilde{T}}\mathbf{\widetilde{S}})^{*}$$

$$=\mathbf{\widetilde{T}}^{*}\mathbf{\widetilde{S}}^{*}$$

$$=\mathbf{T}^{\dagger }\mathbf{S}^{\dagger }$$

QED