## PHY2403 (QFT I). Pondering the ground state bra formula.

November 5, 2018 phy2403 No comments , ,

In lecture 14 we found

\label{eqn:qftLecture15:60}
\ket{\Omega}
=
\evalbar{
\frac{ U(t_0, -T) \ket{0} }
{
e^{-i E_0(T – t_0)} \braket{\Omega}{0}
}
}{T \rightarrow \infty(1 – i \epsilon)},

and it was stated that we can also show that
\label{eqn:qftLecture15:80}
\bra{\Omega}
=
\evalbar{
\frac{ \bra{0} U(T, t_0) }
{
e^{-i E_0(T – t_0)} \braket{0}{\Omega}
}
}{T \rightarrow \infty(1 – i \epsilon)}.

This second statement is actually not obvious since
\label{eqn:braOmega:100}
\lr{
\frac{ U(t_0, -T) \ket{0} }
{
e^{-i E_0(T – t_0)} \braket{\Omega}{0}
}
}^\dagger
=
\frac{ \bra{0} U(-T, t_0) \ket{0} }
{
e^{+i E_0(T – t_0)} \braket{0}{\Omega}
}

My first thought was that I’d written down \ref{eqn:qftLecture15:80} in my notes wrong, but this is actually also consistent with [1], which our Prof is following loosely (i.e. he is explicitly filling in many of the holes in that dense little 800 page book).

The resolution of this inconsistency is that the limit point $$\infty(1 – i \epsilon)$$ doesn’t work if you just conjugate, and you’d also have to conjugate that limit, so while
\ref{eqn:braOmega:100} is correct, it is only part of the story, and should really be stated as
\label{eqn:braOmega:120}
\bra{\Omega}
=
\evalbar{
\frac{ \bra{0} U(-T, t_0) \ket{0} }
{
e^{+i E_0(T – t_0)} \braket{0}{\Omega}
}
}{T \rightarrow \infty(1 + i \epsilon)}.

This is awkward because now our expressions for $$\bra{\Omega}$$ and $$\ket{\Omega}$$ approach $$T$$ from different directions, and we want to evaluate both with a single limiting argument.

To resolve this, we really have to start back with the identity expansion we used in lecture 14
\label{eqn:braOmega:140}
\begin{aligned}
\bra{0} e^{-i H T}
&=
\lr{
\braket{0}{\Omega}\bra{\Omega}
+ {\int\kern-1em\sum}_n \braket{0}{n} \bra{n}
}
e^{-i H T} \\
&=
\braket{0}{\Omega}\bra{\Omega}
e^{-i E_0 T}
+ {\int\kern-1em\sum}_n \braket{0}{n} \bra{n} e^{-i E_n T}.
\end{aligned}

We argued (as does the text) that approaching to as $$T( 1 – i \epsilon)$$ kills off the energetic states since
\label{eqn:braOmega:160}
\bra{n} e^{-i E_n T}
\rightarrow
\bra{n} e^{-i E_n T} e^{-E_n T \epsilon}

and the exponential damping factor is smaller for each $$E_n > E_0$$, so it can be neglected in the large $$T$$ limit, leaving
\label{eqn:braOmega:180}
\bra{0} e^{-i H T}
=
\lim_{T \rightarrow \infty(1 – i \epsilon)}
\braket{0}{\Omega}\bra{\Omega}.

As we did for $$\ket{\Omega}$$ we can shift the large time $$T$$ by a small constant (this time $$-t_0$$ instead of $$t_0$$), to give
\label{eqn:braOmega:200}
\begin{aligned}
\bra{\Omega}
&=
\lim_{T \rightarrow \infty(1 – i \epsilon)}
\frac{ \bra{0} e^{-i H T} }
{
\braket{0}{\Omega} e^{-i E_0 T}
} \\
&\approx
\lim_{T \rightarrow \infty(1 – i \epsilon)}
\frac{ \bra{0} e^{-i H (T – t_0)} }
{
\braket{0}{\Omega} e^{-i E_0 (T – t_0)}
} \\
&=
\lim_{T \rightarrow \infty(1 – i \epsilon)}
\frac{ \bra{0} e^{i H_0( T – t_0)} e^{-i H (T – t_0)} }
{
\braket{0}{\Omega} e^{-i E_0 (T – t_0)}
} \\
&=
\lim_{T \rightarrow \infty(1 – i \epsilon)}
\frac{ \bra{0} U(T, t_0) }
{
\braket{0}{\Omega} e^{-i E_0 (T – t_0)}
},
\end{aligned}

where the projective property $$\bra{0} e^{i H_0 \alpha} = \bra{0}$$ has been used to insert a no-op (i.e. $$\bra{0} H_0 = 0$$). This recovers the result stated in class (also: eq. (4.29) in the text.)

# References

[1] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

## Plane wave ground state expectation for SHO

Problem [1] 2.18 is, for a 1D SHO, show that

\label{eqn:exponentialExpectationGroundState:20}
\bra{0} e^{i k x} \ket{0} = \exp\lr{ -k^2 \bra{0} x^2 \ket{0}/2 }.

Despite the simple appearance of this problem, I found this quite involved to show. To do so, start with a series expansion of the expectation

\label{eqn:exponentialExpectationGroundState:40}
\bra{0} e^{i k x} \ket{0}
=
\sum_{m=0}^\infty \frac{(i k)^m}{m!} \bra{0} x^m \ket{0}.

Let

\label{eqn:exponentialExpectationGroundState:60}
X = \lr{ a + a^\dagger },

so that

\label{eqn:exponentialExpectationGroundState:80}
x
= \sqrt{\frac{\Hbar}{2 \omega m}} X
= \frac{x_0}{\sqrt{2}} X.

Consider the first few values of $$\bra{0} X^n \ket{0}$$

\label{eqn:exponentialExpectationGroundState:100}
\begin{aligned}
\bra{0} X \ket{0}
&=
\bra{0} \lr{ a + a^\dagger } \ket{0} \\
&=
\braket{0}{1} \\
&=
0,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:120}
\begin{aligned}
\bra{0} X^2 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^2 \ket{0} \\
&=
\braket{1}{1} \\
&=
1,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:140}
\begin{aligned}
\bra{0} X^3 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^3 \ket{0} \\
&=
\bra{1} \lr{ \sqrt{2} \ket{2} + \ket{0} } \\
&=
0.
\end{aligned}

Whenever the power $$n$$ in $$X^n$$ is even, the braket can be split into a bra that has only contributions from odd eigenstates and a ket with even eigenstates. We conclude that $$\bra{0} X^n \ket{0} = 0$$ when $$n$$ is odd.

Noting that $$\bra{0} x^2 \ket{0} = \ifrac{x_0^2}{2}$$, this leaves

\label{eqn:exponentialExpectationGroundState:160}
\begin{aligned}
\bra{0} e^{i k x} \ket{0}
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \bra{0} x^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \lr{ \frac{x_0^2}{2} }^m \bra{0} X^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{1}{(2 m)!} \lr{ -k^2 \bra{0} x^2 \ket{0} }^m \bra{0} X^{2m} \ket{0}.
\end{aligned}

This problem is now reduced to showing that

\label{eqn:exponentialExpectationGroundState:180}
\frac{1}{(2 m)!} \bra{0} X^{2m} \ket{0} = \inv{m! 2^m},

or

\label{eqn:exponentialExpectationGroundState:200}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&= \frac{(2m)!}{m! 2^m} \\
&= \frac{ (2m)(2m-1)(2m-2) \cdots (2)(1) }{2^m m!} \\
&= \frac{ 2^m (m)(2m-1)(m-1)(2m-3)(m-2) \cdots (2)(3)(1)(1) }{2^m m!} \\
&= (2m-1)!!,
\end{aligned}

where $$n!! = n(n-2)(n-4)\cdots$$.

It looks like $$\bra{0} X^{2m} \ket{0}$$ can be expanded by inserting an identity operator and proceeding recursively, like

\label{eqn:exponentialExpectationGroundState:220}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&=
\bra{0} X^2 \lr{ \sum_{n=0}^\infty \ket{n}\bra{n} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^2 \lr{ \ket{0}\bra{0} + \ket{2}\bra{2} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^{2m-2} \ket{0} + \bra{0} X^2 \ket{2} \bra{2} X^{2m-2} \ket{0}.
\end{aligned}

This has made use of the observation that $$\bra{0} X^2 \ket{n} = 0$$ for all $$n \ne 0,2$$. The remaining term includes the factor

\label{eqn:exponentialExpectationGroundState:240}
\begin{aligned}
\bra{0} X^2 \ket{2}
&=
\bra{0} \lr{a + a^\dagger}^2 \ket{2} \\
&=
\lr{ \bra{0} + \sqrt{2} \bra{2} } \ket{2} \\
&=
\sqrt{2},
\end{aligned}

Since $$\sqrt{2} \ket{2} = \lr{a^\dagger}^2 \ket{0}$$, the expectation of interest can be written

\label{eqn:exponentialExpectationGroundState:260}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + \bra{0} a^2 X^{2m-2} \ket{0}.

How do we expand the second term. Let’s look at how $$a$$ and $$X$$ commute

\label{eqn:exponentialExpectationGroundState:280}
\begin{aligned}
a X
&=
\antisymmetric{a}{X} + X a \\
&=
\antisymmetric{a}{a + a^\dagger} + X a \\
&=
\antisymmetric{a}{a^\dagger} + X a \\
&=
1 + X a,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:300}
\begin{aligned}
a^2 X
&=
a \lr{ a X } \\
&=
a \lr{ 1 + X a } \\
&=
a + a X a \\
&=
a + \lr{ 1 + X a } a \\
&=
2 a + X a^2.
\end{aligned}

Proceeding to expand $$a^2 X^n$$ we find
\label{eqn:exponentialExpectationGroundState:320}
\begin{aligned}
a^2 X^3 &= 6 X + 6 X^2 a + X^3 a^2 \\
a^2 X^4 &= 12 X^2 + 8 X^3 a + X^4 a^2 \\
a^2 X^5 &= 20 X^3 + 10 X^4 a + X^5 a^2 \\
a^2 X^6 &= 30 X^4 + 12 X^5 a + X^6 a^2.
\end{aligned}

It appears that we have
\label{eqn:exponentialExpectationGroundState:340}
\antisymmetric{a^2 X^n}{X^n a^2} = \beta_n X^{n-2} + 2 n X^{n-1} a,

where

\label{eqn:exponentialExpectationGroundState:360}
\beta_n = \beta_{n-1} + 2 (n-1),

and $$\beta_2 = 2$$. Some goofing around shows that $$\beta_n = n(n-1)$$, so the induction hypothesis is

\label{eqn:exponentialExpectationGroundState:380}
\antisymmetric{a^2 X^n}{X^n a^2} = n(n-1) X^{n-2} + 2 n X^{n-1} a.

Let’s check the induction
\label{eqn:exponentialExpectationGroundState:400}
\begin{aligned}
a^2 X^{n+1}
&=
a^2 X^{n} X \\
&=
\lr{ n(n-1) X^{n-2} + 2 n X^{n-1} a + X^n a^2 } X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} a X + X^n a^2 X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} \lr{ 1 + X a } + X^n \lr{ 2 a + X a^2 } \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} + 2 n X^{n} a
+ 2 X^n a
+ X^{n+1} a^2 \\
&=
X^{n+1} a^2 + (2 + 2 n) X^{n} a + \lr{ 2 n + n(n-1) } X^{n-1} \\
&=
X^{n+1} a^2 + 2(n + 1) X^{n} a + (n+1) n X^{n-1},
\end{aligned}

which concludes the induction, giving

\label{eqn:exponentialExpectationGroundState:420}
\bra{ 0 } a^2 X^{n} \ket{0 } = n(n-1) \bra{0} X^{n-2} \ket{0},

and

\label{eqn:exponentialExpectationGroundState:440}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + (2m-2)(2m-3) \bra{0} X^{2m-4} \ket{0}.

Let

\label{eqn:exponentialExpectationGroundState:460}
\sigma_{n} = \bra{0} X^n \ket{0},

so that the recurrence relation, for $$2n \ge 4$$ is

\label{eqn:exponentialExpectationGroundState:480}
\sigma_{2n} = \sigma_{2n -2} + (2n-2)(2n-3) \sigma_{2n -4}

We want to show that this simplifies to

\label{eqn:exponentialExpectationGroundState:500}
\sigma_{2n} = (2n-1)!!

The first values are

\label{eqn:exponentialExpectationGroundState:540}
\sigma_0 = \bra{0} X^0 \ket{0} = 1

\label{eqn:exponentialExpectationGroundState:560}
\sigma_2 = \bra{0} X^2 \ket{0} = 1

which gives us the right result for the first term in the induction

\label{eqn:exponentialExpectationGroundState:580}
\begin{aligned}
\sigma_4
&= \sigma_2 + 2 \times 1 \times \sigma_0 \\
&= 1 + 2 \\
&= 3!!
\end{aligned}

For the general induction term, consider

\label{eqn:exponentialExpectationGroundState:600}
\begin{aligned}
\sigma_{2n + 2}
&= \sigma_{2n} + 2 n (2n – 1) \sigma_{2n -2} \\
&= (2n-1)!! + 2n ( 2n – 1) (2n -3)!! \\
&= (2n + 1) (2n -1)!! \\
&= (2n + 1)!!,
\end{aligned}

which completes the final induction. That was also the last thing required to complete the proof, so we are done!

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## PHY1520H Graduate Quantum Mechanics. Lecture 3: Density matrix (cont.). Taught by Prof. Arun Paramekanti

September 24, 2015 phy1520 No comments , , , , , , , , ,

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 3 content.

### Density matrix (cont.)

An example of a partitioned system with four total states (two spin 1/2 particles) is sketched in fig. 1.

fig. 1. Two spins

An example of a partitioned system with eight total states (three spin 1/2 particles) is sketched in fig. 2.

fig. 2. Three spins

The density matrix

\label{eqn:qmLecture3:20}
\hat{\rho} = \ket{\Psi}\bra{\Psi}

is clearly an operator as can be seen by applying it to a state

\label{eqn:qmLecture3:40}
\hat{\rho} \ket{\phi} = \ket{\Psi} \lr{ \braket{ \Psi }{\phi} }.

The quantity in braces is just a complex number.

After expanding the pure state $$\ket{\Psi}$$ in terms of basis states for each of the two partitions

\label{eqn:qmLecture3:60}
\ket{\Psi}
= \sum_{m,n} C_{m, n} \ket{m}_{\textrm{L}} \ket{n}_{\textrm{R}},

With $$\textrm{L}$$ and $$\textrm{R}$$ implied for $$\ket{m}, \ket{n}$$ indexed states respectively, this can be written

\label{eqn:qmLecture3:460}
\ket{\Psi}
= \sum_{m,n} C_{m, n} \ket{m} \ket{n}.

The density operator is

\label{eqn:qmLecture3:80}
\hat{\rho} =
\sum_{m,n}
C_{m, n}
C_{m’, n’}^\conj
\ket{m} \ket{n}
\sum_{m’,n’}
\bra{m’} \bra{n’}.

Suppose we trace over the right partition of the state space, defining such a trace as the reduced density operator $$\hat{\rho}_{\textrm{red}}$$

\label{eqn:qmLecture3:100}
\begin{aligned}
\hat{\rho}_{\textrm{red}}
&\equiv
\textrm{Tr}_{\textrm{R}}(\hat{\rho}) \\
&= \sum_{\tilde{n}} \bra{\tilde{n}} \hat{\rho} \ket{ \tilde{n}} \\
&= \sum_{\tilde{n}}
\bra{\tilde{n} }
\lr{
\sum_{m,n}
C_{m, n}
\ket{m} \ket{n}
}
\lr{
\sum_{m’,n’}
C_{m’, n’}^\conj
\bra{m’} \bra{n’}
}
\ket{ \tilde{n} } \\
&=
\sum_{\tilde{n}}
\sum_{m,n}
\sum_{m’,n’}
C_{m, n}
C_{m’, n’}^\conj
\ket{m} \delta_{\tilde{n} n}
\bra{m’ }
\delta_{ \tilde{n} n’ } \\
&=
\sum_{\tilde{n}, m, m’}
C_{m, \tilde{n}}
C_{m’, \tilde{n}}^\conj
\ket{m} \bra{m’ }
\end{aligned}

Computing the matrix element of $$\hat{\rho}_{\textrm{red}}$$, we have

\label{eqn:qmLecture3:120}
\begin{aligned}
\bra{\tilde{m}} \hat{\rho}_{\textrm{red}} \ket{\tilde{m}}
&=
\sum_{m, m’, \tilde{n}} C_{m, \tilde{n}} C_{m’, \tilde{n}}^\conj \braket{ \tilde{m}}{m} \braket{m’}{\tilde{m}} \\
&=
\sum_{\tilde{n}} \Abs{C_{\tilde{m}, \tilde{n}} }^2.
\end{aligned}

This is the probability that the left partition is in state $$\tilde{m}$$.

### Average of an observable

Suppose we have two spin half particles. For such a system the total magnetization is

\label{eqn:qmLecture3:140}
S_{\textrm{Total}} =
S_1^z
+
S_1^z,

as sketched in fig. 3.

fig. 3. Magnetic moments from two spins.

The average of some observable is

\label{eqn:qmLecture3:160}
\expectation{\hatA}
= \sum_{m, n, m’, n’} C_{m, n}^\conj C_{m’, n’}
\bra{m}\bra{n} \hatA \ket{n’} \ket{m’}.

Consider the trace of the density operator observable product

\label{eqn:qmLecture3:180}
\textrm{Tr}( \hat{\rho} \hatA )
= \sum_{m, n} \braket{m n}{\Psi} \bra{\Psi} \hatA \ket{m, n}.

Let

\label{eqn:qmLecture3:200}
\ket{\Psi} = \sum_{m, n} C_{m n} \ket{m, n},

so that

\label{eqn:qmLecture3:220}
\begin{aligned}
\textrm{Tr}( \hat{\rho} \hatA )
&= \sum_{m, n, m’, n’, m”, n”} C_{m’, n’} C_{m”, n”}^\conj
\braket{m n}{m’, n’} \bra{m”, n”} \hatA \ket{m, n} \\
&= \sum_{m, n, m”, n”} C_{m, n} C_{m”, n”}^\conj
\bra{m”, n”} \hatA \ket{m, n}.
\end{aligned}

This is just

\label{eqn:qmLecture3:240}
\boxed{
\bra{\Psi} \hatA \ket{\Psi} = \textrm{Tr}( \hat{\rho} \hatA ).
}

### Left observables

Consider

\label{eqn:qmLecture3:260}
\begin{aligned}
\bra{\Psi} \hatA_{\textrm{L}} \ket{\Psi}
&= \textrm{Tr}(\hat{\rho} \hatA_{\textrm{L}}) \\
&=
\textrm{Tr}_{\textrm{L}}
\textrm{Tr}_{\textrm{R}}
(\hat{\rho} \hatA_{\textrm{L}}) \\
&=
\textrm{Tr}_{\textrm{L}}
\lr{
\lr{
\textrm{Tr}_{\textrm{R}} \hat{\rho}
}
\hatA_{\textrm{L}})
} \\
&=
\textrm{Tr}_{\textrm{L}}
\lr{
\hat{\rho}_{\textrm{red}}
\hatA_{\textrm{L}})
}.
\end{aligned}

We see

\label{eqn:qmLecture3:280}
\bra{\Psi} \hatA_{\textrm{L}} \ket{\Psi}
=
\textrm{Tr}_{\textrm{L}} \lr{ \hat{\rho}_{\textrm{red}, \textrm{L}} \hatA_{\textrm{L}} }.

We find that we don’t need to know the state of the complete system to answer questions about portions of the system, but instead just need $$\hat{\rho}$$, a “probability operator” that provides all the required information about the partitioning of the system.

### Pure states vs. mixed states

For pure states we can assign a state vector and talk about reduced scenarios. For mixed states we must work with reduced density matrix.

## Example: Two particle spin half pure states

Consider

\label{eqn:qmLecture3:300}
\ket{\psi_1} = \inv{\sqrt{2}} \lr{ \ket{ \uparrow \downarrow } – \ket{ \downarrow \uparrow } }

\label{eqn:qmLecture3:320}
\ket{\psi_2} = \inv{\sqrt{2}} \lr{ \ket{ \uparrow \downarrow } + \ket{ \uparrow \uparrow } }.

For the first pure state the density operator is
\label{eqn:qmLecture3:360}
\hat{\rho} = \inv{2}
\lr{ \ket{ \uparrow \downarrow } – \ket{ \downarrow \uparrow } }
\lr{ \bra{ \uparrow \downarrow } – \bra{ \downarrow \uparrow } }

What are the reduced density matrices?

\label{eqn:qmLecture3:340}
\begin{aligned}
\hat{\rho}_{\textrm{L}}
&= \textrm{Tr}_{\textrm{R}} \lr{ \hat{\rho} } \\
&=
\inv{2} (-1)(-1) \ket{\downarrow}\bra{\downarrow}
+\inv{2} (+1)(+1) \ket{\uparrow}\bra{\uparrow},
\end{aligned}

so the matrix representation of this reduced density operator is

\label{eqn:qmLecture3:380}
\hat{\rho}_{\textrm{L}}
=
\inv{2}
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.

For the second pure state the density operator is
\label{eqn:qmLecture3:400}
\hat{\rho} = \inv{2}
\lr{ \ket{ \uparrow \downarrow } + \ket{ \uparrow \uparrow } }
\lr{ \bra{ \uparrow \downarrow } + \bra{ \uparrow \uparrow } }.

This has a reduced density matrice

\label{eqn:qmLecture3:420}
\begin{aligned}
\hat{\rho}_{\textrm{L}}
&= \textrm{Tr}_{\textrm{R}} \lr{ \hat{\rho} } \\
&=
\inv{2} \ket{\uparrow}\bra{\uparrow}
+\inv{2} \ket{\uparrow}\bra{\uparrow} \\
&=
\ket{\uparrow}\bra{\uparrow} .
\end{aligned}

This has a matrix representation

\label{eqn:qmLecture3:440}
\hat{\rho}_{\textrm{L}}
=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.

In this second example, we have more information about the left partition. That will be seen as a zero entanglement entropy in the problem set. In contrast we have less information about the first state, and will find a non-zero positive entanglement entropy in that case.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## PHY1520H Graduate Quantum Mechanics. Lecture 1: Lighting review. Taught by Prof. Arun Paramekanti

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 1 content.

### Classical mechanics

We’ll be talking about one body physics for most of this course. In classical mechanics we can figure out the particle trajectories using both of $$(\Br, \Bp$$, where

\label{eqn:qmLecture1:20}
\begin{aligned}
\ddt{\Br} &= \inv{m} \Bp \\
\end{aligned}

A two dimensional phase space as sketched in fig. 1 shows the trajectory of a point particle subject to some equations of motion

fig. 1. One dimensional classical phase space example

### Quantum mechanics

For this lecture, we’ll work with natural units, setting

\label{eqn:qmLecture1:480}
\boxed{
\Hbar = 1.
}

In QM we are no longer allowed to think of position and momentum, but have to start asking about state vectors $$\ket{\Psi}$$.

We’ll consider the state vector with respect to some basis, for example, in a position basis, we write

\label{eqn:qmLecture1:40}
\braket{ x }{\Psi } = \Psi(x),

a complex numbered “wave function”, the probability amplitude for a particle in $$\ket{\Psi}$$ to be in the vicinity of $$x$$.

We could also consider the state in a momentum basis

\label{eqn:qmLecture1:60}
\braket{ p }{\Psi } = \Psi(p),

a probability amplitude with respect to momentum $$p$$.

More precisely,

\label{eqn:qmLecture1:80}
\Abs{\Psi(x)}^2 dx \ge 0

is the probability of finding the particle in the range $$(x, x + dx )$$. To have meaning as a probability, we require

\label{eqn:qmLecture1:100}
\int_{-\infty}^\infty \Abs{\Psi(x)}^2 dx = 1.

The average position can be calculated using this probability density function. For example

\label{eqn:qmLecture1:120}
\expectation{x} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 x dx,

or
\label{eqn:qmLecture1:140}
\expectation{f(x)} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 f(x) dx.

Similarly, calculation of an average of a function of momentum can be expressed as

\label{eqn:qmLecture1:160}
\expectation{f(p)} = \int_{-\infty}^\infty \Abs{\Psi(p)}^2 f(p) dp.

### Transformation from a position to momentum basis

We have a problem, if we which to compute an average in momentum space such as $$\expectation{p}$$, when given a wavefunction $$\Psi(x)$$.

How do we convert

\label{eqn:qmLecture1:180}
\Psi(p)
\stackrel{?}{\leftrightarrow}
\Psi(x),

or equivalently
\label{eqn:qmLecture1:200}
\braket{p}{\Psi}
\stackrel{?}{\leftrightarrow}
\braket{x}{\Psi}.

Such a conversion can be performed by virtue of an the assumption that we have a complete orthonormal basis, for which we can introduce identity operations such as

\label{eqn:qmLecture1:220}
\int_{-\infty}^\infty dp \ket{p}\bra{p} = 1,

or
\label{eqn:qmLecture1:240}
\int_{-\infty}^\infty dx \ket{x}\bra{x} = 1

Some interpretations:

1. $$\ket{x_0} \leftrightarrow \text{sits at} x = x_0$$
2. $$\braket{x}{x’} \leftrightarrow \delta(x – x’)$$
3. $$\braket{p}{p’} \leftrightarrow \delta(p – p’)$$
4. $$\braket{x}{p’} = \frac{e^{i p x}}{\sqrt{V}}$$, where $$V$$ is the volume of the box containing the particle. We’ll define the appropriate normalization for an infinite box volume later.

The delta function interpretation of the braket $$\braket{p}{p’}$$ justifies the identity operator, since we recover any state in the basis when operating with it. For example, in momentum space

\label{eqn:qmLecture1:260}
\begin{aligned}
1 \ket{p}
&=
\lr{ \int_{-\infty}^\infty dp’
\ket{p’}\bra{p’} }
\ket{p} \\
&=
\int_{-\infty}^\infty dp’
\ket{p’}
\braket{p’}{p} \\
&=
\int_{-\infty}^\infty dp’
\ket{p’}
\delta(p – p’) \\
&=
\ket{p}.
\end{aligned}

This also the determination of an integral operator representation for the delta function

\label{eqn:qmLecture1:500}
\begin{aligned}
\delta(x – x’)
&=
\braket{x}{x’} \\
&=
\int dp \braket{x}{p} \braket{p}{x’} \\
&=
\inv{V} \int dp e^{i p x} e^{-i p x’},
\end{aligned}

or

\label{eqn:qmLecture1:520}
\delta(x – x’)
=
\inv{V} \int dp e^{i p (x- x’)}.

Here we used the fact that $$\braket{p}{x} = \braket{x}{p}^\conj$$.

FIXME: do we have a justification for that conjugation with what was defined here so far?

The conversion from a position basis to momentum space is now possible

\label{eqn:qmLecture1:280}
\begin{aligned}
\braket{p}{\Psi}
&= \Psi(p) \\
&= \int_{-\infty}^\infty \braket{p}{x} \braket{x}{\Psi} dx \\
&= \int_{-\infty}^\infty \frac{e^{-ip x}}{\sqrt{V}} \Psi(x) dx.
\end{aligned}

The momentum space to position space conversion can be written as

\label{eqn:qmLecture1:300}
\Psi(x)
= \int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi(p) dp.

Now we can go back and figure out the an expectation

\label{eqn:qmLecture1:320}
\begin{aligned}
\expectation{p}
&=
\int \Psi^\conj(p) \Psi(p) p d p \\
&=
\int dp
\lr{
\int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi^\conj(x) dx
}
\lr{
\int_{-\infty}^\infty \frac{e^{-ip x’}}{\sqrt{V}} \Psi(x’) dx’
}
p \\
&=\int dp dx dx’
\Psi^\conj(x)
\inv{V} e^{ip (x-x’)} \Psi(x’) p \\
&=
\int dp dx dx’
\Psi^\conj(x)
\inv{V} \lr{ -i\PD{x}{e^{ip (x-x’)}} }\Psi(x’) \\
&=
\int dp dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\inv{V} \int dx’ e^{ip (x-x’)} \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\int dx’ \lr{ \inv{V} \int dp e^{ip (x-x’)} } \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\int dx’ \delta(x – x’) \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\Psi(x)
\end{aligned}

Here we’ve essentially calculated the position space representation of the momentum operator, allowing identifications of the following form

\label{eqn:qmLecture1:380}
p \leftrightarrow -i \PD{x}{}

\label{eqn:qmLecture1:400}
p^2 \leftrightarrow – \PDSq{x}{}.

### Alternate starting point.

Most of the above results followed from the claim that $$\braket{x}{p} = e^{i p x}$$. Note that this position space representation of the momentum operator can also be taken as the starting point. Given that, the exponential representation of the position-momentum braket follows

\label{eqn:qmLecture1:540}
\bra{x} P \ket{p}
=
-i \Hbar \PD{x}{} \braket{x}{p},

but $$\bra{x} P \ket{p} = p \braket{x}{p}$$, providing a differential equation for $$\braket{x}{p}$$

\label{eqn:qmLecture1:560}
p \braket{x}{p} = -i \Hbar \PD{x}{} \braket{x}{p},

with solution

\label{eqn:qmLecture1:580}
i p x/\Hbar = \ln \braket{x}{p} + \text{const},

or
\label{eqn:qmLecture1:600}
\braket{x}{p} \propto e^{i p x/\Hbar}.

### Matrix interpretation

1. Ket’s $$\ket{\Psi} \leftrightarrow \text{column vector}$$
2. Bra’s $$\bra{\Psi} \leftrightarrow {(\text{row vector})}^\conj$$
3. Operators $$\leftrightarrow$$ matrices that act on vectors.

\label{eqn:qmLecture1:420}
\hat{p} \ket{\Psi} \rightarrow \ket{\Psi’}

### Time evolution

For a state subject to the equations of motion given by the Hamiltonian operator $$\hat{H}$$

\label{eqn:qmLecture1:440}
i \PD{t}{} \ket{\Psi} = \hat{H} \ket{\Psi},

the time evolution is given by
\label{eqn:qmLecture1:460}
\ket{\Psi(t)} = e^{-i \hat{H} t} \ket{\Psi(0)}.

### Incomplete information

We’ll need to introduce the concept of Density matrices. This will bring us to concepts like entanglement.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Free particle propagator

September 7, 2015 phy1520 6 comments , , , , , ,

## Question: Free particle propagator ([1] pr. 2.31)

Derive the free particle propagator in one and three dimensions.

I found the description in the text confusing, so let’s start from scratch with the definition of the propagator. This is the kernel of the spatial convolution integral that encodes time evolution, and can be expressed by expanding a general time state with two sets of identity operators. Let the position relative state at time $$t$$, relative to an initial time $$t_0$$ be given by $$\braket{\Bx}{\alpha, t ; t_0 }$$, and expand this in terms of a complete basis of energy eigenstates $$| a’ >$$ and the time evolution operator

\label{eqn:freeParticlePropagator:20}
\begin{aligned}
\braket{\Bx”}{\alpha, t ; t_0 }
&= \bra{\Bx”} U \ket{\alpha, t_0 } \\
&= \bra{\Bx”} e^{-i H (t -t_0)/\Hbar} \ket{\alpha, t_0 } \\
&= \bra{\Bx”} e^{-i H (t -t_0)/\Hbar} \lr{ \sum_{a’} \ket{a’} \bra{a’ }} \ket{\alpha, t_0 } \\
&= \bra{\Bx”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \braket{a’ }{\alpha, t_0 } \\
&=
\bra{\Bx”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \bra{a’ }
\lr{ \int d^3 \Bx’
\ket{\Bx’}\bra{\Bx’}
}
\ket{\alpha, t_0 } \\
&=
\int d^3 \Bx’
\lr{
\bra{\Bx”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \braket{a’ }{\Bx’}
}
\braket{\Bx’}{\alpha, t_0 } \\
&=
\int d^3 \Bx’ K(\Bx”, t ; \Bx’, t_0) \braket{\Bx’}{\alpha, t_0 },
\end{aligned}

where

\label{eqn:freeParticlePropagator:40}
K(\Bx”, t ; \Bx’, t_0) =
\sum_{a’}
\braket{\Bx”}{a’}\braket{a’ }{\Bx’}
e^{-i E_{a’} (t -t_0)/\Hbar},

the propagator, is the kernel of the convolution integral that takes the state $$\ket{\alpha, t_0}$$ to state $$\ket{\alpha, t ; t_0}$$. Evaluating this over the momentum states (where integration and not plain summation is required), we have

\label{eqn:freeParticlePropagator:60}
\begin{aligned}
K(\Bx”, t ; \Bx’, t_0)
&=
\int d^3 \Bp’
\braket{\Bx”}{\Bp’}\braket{\Bp’ }{\Bx’}
e^{-i E_{\Bp’} (t -t_0)/\Hbar} \\
&=
\int d^3 \Bp’
\braket{\Bx”}{\Bp’}\braket{\Bp’ }{\Bx’}
\exp\lr{-i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}} \\
&=
\int d^3 \Bp’
\frac{e^{i \Bx” \cdot \Bp’/\Hbar}}{(\sqrt{2 \pi \Hbar})^3}
\frac{e^{-i \Bx’ \cdot \Bp’/\Hbar}}{(\sqrt{2 \pi \Hbar})^3}
\exp\lr{-i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}} \\
&=
\inv{(2 \pi \Hbar)^3}
\int d^3 \Bp’
e^{i (\Bx” -\Bx’) \cdot \Bp’/\Hbar}
\exp\lr{-i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}} \\
&=
\inv{ 2 \pi \Hbar }
\int_{-\infty}^\infty dp_1′
e^{i (x_1” -x_1′) p_1’/\Hbar}
\exp\lr{-i \frac{(p_1′)^2 (t -t_0)}{2 m \Hbar}} \times \\
&\quad \inv{ 2 \pi \Hbar }
\int_{-\infty}^\infty dp_2′
e^{i (x_2” -x_2′) p_2’/\Hbar}
\exp\lr{-i \frac{(p_2′)^2 (t -t_0)}{2 m \Hbar}} \times \\
&\quad \inv{ 2 \pi \Hbar }
\int_{-\infty}^\infty dp_3′
e^{i (x_3” -x_3′) p_3’/\Hbar}
\exp\lr{-i \frac{(p_3′)^2 (t -t_0)}{2 m \Hbar}}
\end{aligned}

With $$a = \ifrac{(t -t_0)}{2 m \Hbar}$$, each of these three integral factors is of the form

\label{eqn:freeParticlePropagator:80}
\begin{aligned}
\inv{ 2 \pi \Hbar }
\int_{-\infty}^\infty dp
e^{i \Delta x p/\Hbar }
\exp\lr{-i a p^2}
&=
\inv{2 \pi \Hbar \sqrt{a}}
\int_{-\infty}^\infty du
e^{i \Delta x u/(\sqrt{a}\Hbar) }
\exp\lr{-i u^2} \\
&=
\inv{2 \pi \Hbar \sqrt{a}}
\int_{-\infty}^\infty du
e^{i \Delta x u/(\sqrt{a} \Hbar) }
\exp\lr{-i (u – \Delta x /(2\sqrt{a}\Hbar))^2 + i(\Delta x/(2\sqrt{a}\Hbar))^2} \\
&=
\inv{2 \pi \Hbar \sqrt{a}}
\exp\lr{ \frac{i(\Delta x)^2 2 m \Hbar}{4 (t -t_0) \Hbar^2} }
\int_{-\infty}^\infty dz
e^{-i z^2} \\
&= \sqrt{ \frac{ -i \pi 2 m \Hbar}{ 4 \pi^2 \Hbar^2 (t -t_0)} }
\exp\lr{ \frac{i(\Delta x)^2 m}{2 (t -t_0) \Hbar} } \\
&= \sqrt{ \frac{ m }{ 2 \pi i \Hbar (t -t_0)} }
\exp\lr{ \frac{i(\Delta x)^2 m}{2 (t -t_0) \Hbar} }.
\end{aligned}

Note that the integral above has value $$\sqrt{-i\pi}$$ which can be found by integrating over the contour of fig. 1, letting $$R \rightarrow \infty$$.

fig. 1. Integration contour for $$\int e^{-i z^2}$$

Multiplying out each of the spatial direction factors gives the propagator in its closed form
\label{eqn:freeParticlePropagator:120}
\boxed{
K(\Bx”, t ; \Bx’, t_0)
= \lr{ \sqrt{ \frac{ m }{ 2 \pi i \Hbar (t -t_0)} } }^3
\exp\lr{ \frac{i(\Bx” – \Bx’)^2 m}{2 (t -t_0) \Hbar} }.
}

In one or two dimensions the exponential power $$3$$ need only be adjusted appropriately.

## Question: Momentum space free particle propagator ([1] pr. 2.33)

Derive the free particle propagator in momentum space.

The momentum space propagator follows in the same fashion as the spatial propagator

\label{eqn:freeParticlePropagator:140}
\begin{aligned}
\braket{\Bp”}{\alpha, t ; t_0 }
&= \bra{\Bp”} U \ket{\alpha, t_0 } \\
&= \bra{\Bp”} e^{-i H (t -t_0)/\Hbar} \ket{\alpha, t_0 } \\
&= \bra{\Bp”} e^{-i H (t -t_0)/\Hbar} \lr{ \sum_{a’} \ket{a’} \bra{a’ }} \ket{\alpha, t_0 } \\
&= \bra{\Bp”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \braket{a’ }{\alpha, t_0 } \\
&=
\bra{\Bp”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \bra{a’ }
\lr{ \int d^3 \Bp’
\ket{\Bp’}\bra{\Bp’}
}
\ket{\alpha, t_0 } \\
&=
\int d^3 \Bp’
\lr{
\bra{\Bp”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \braket{a’ }{\Bp’}
}
\braket{\Bp’}{\alpha, t_0 } \\
&=
\int d^3 \Bp’ K(\Bp”, t ; \Bp’, t_0) \braket{\Bp’}{\alpha, t_0 },
\end{aligned}

so

\label{eqn:freeParticlePropagator:160}
K(\Bp”, t ; \Bp’, t_0)
=
\sum_{a’}
\braket{\Bp”}{a’}
\braket{a’ }{\Bp’}
e^{-i E_{a’} (t -t_0)/\Hbar}.

For the free particle Hamiltonian, this can be evaluated over a momentum space basis

\label{eqn:freeParticlePropagator:170}
\begin{aligned}
K(\Bp”, t ; \Bp’, t_0)
&=
\int d^3 \Bp”’
\braket{\Bp”}{\Bp”’}
\braket{\Bp”’ }{\Bp’}
e^{-i E_{\Bp”’} (t -t_0)/\Hbar} \\
&=
\int d^3 \Bp”’
\braket{\Bp”}{\Bp”’}
\delta(\Bp”’ – \Bp’)
\exp\lr{ -i \frac{(\Bp”’)^2 (t -t_0)}{2 m \Hbar}} \\
&=
\braket{\Bp”}{\Bp’}
\exp\lr{ -i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}}
\end{aligned}

or

\label{eqn:freeParticlePropagator:200}
\boxed{
K(\Bp”, t ; \Bp’, t_0)
=
\delta( \Bp” – \Bp’ )
\exp\lr{ -i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}}.
}

This is what we expect since the time evolution is given by just this exponential factor

\label{eqn:freeParticlePropagator:220}
\begin{aligned}
\braket{\Bp’}{\alpha, t_0 ; t}
&= \bra{\Bp’} \exp\lr{ -i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}} \ket{\alpha, t_0} \\
&=
\exp\lr{ -i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}}
\braket{\Bp’}
{\alpha, t_0}.
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Weird dreams

I woke up today having a dream still in my head from the night, but it was a strange one. I was expanding out the Dirac notation representation of an operator in matrix form, but the symbols in the kets were elaborate pictures of Disney princesses that I was drawing with forestry scenery in the background, including little bears. At the point that I woke up from the dream, I noticed that I’d gotten the proportion of the bears wrong in one of the pictures, and they looked like they were ready to eat one of the princess characters.

## Guts

As a side effect of this weird dream I actually started thinking about matrix element representation of operators.

When forming the matrix element of an operator using Dirac notation the elements are of the form $$\bra{\textrm{row}} A \ket{\textrm{column}}$$. I’ve gotten that mixed up a couple of times, so I thought it would be helpful to write this out explicitly for a $$2 \times 2$$ operator representation for clarity.

To start, consider a change of basis for a single matrix element from basis $$\setlr{\ket{q}, \ket{r} }$$, to basis $$\setlr{\ket{a}, \ket{b} }$$

\label{eqn:operatorMatrixElement:20}
\begin{aligned}
\bra{q} A \ket{r}
&=
\braket{q}{a} \bra{a} A \ket{r}
+
\braket{q}{b} \bra{b} A \ket{r} \\
&=
\braket{q}{a} \bra{a} A \ket{a}\braket{a}{r}
+ \braket{q}{a} \bra{a} A \ket{b}\braket{b}{r} \\
&+ \braket{q}{b} \bra{b} A \ket{a}\braket{a}{r}
+ \braket{q}{b} \bra{b} A \ket{b}\braket{b}{r} \\
&=
\braket{q}{a}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix}
+
\braket{q}{b}
\begin{bmatrix}
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix} \\
&=
\begin{bmatrix}
\braket{q}{a} &
\braket{q}{b}
\end{bmatrix}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix}.
\end{aligned}

Suppose the matrix representation of $$\ket{q}, \ket{r}$$ are respectively

\label{eqn:operatorMatrixElement:40}
\begin{aligned}
\ket{q} &\sim
\begin{bmatrix}
\braket{a}{q} \\
\braket{b}{q} \\
\end{bmatrix} \\
\ket{r} &\sim
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r} \\
\end{bmatrix} \\
\end{aligned},

then

\label{eqn:operatorMatrixElement:60}
\bra{q} \sim
{\begin{bmatrix}
\braket{a}{q} \\
\braket{b}{q} \\
\end{bmatrix}}^\dagger
=
\begin{bmatrix}
\braket{q}{a} &
\braket{q}{b}
\end{bmatrix}.

The matrix element is then

\label{eqn:operatorMatrixElement:80}
\bra{q} A \ket{r}
\sim
\bra{q}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\ket{r},

and the corresponding matrix representation of the operator is

\label{eqn:operatorMatrixElement:100}
A \sim
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}.

## Question: Uncertainty relation. ([1] pr. 1.20)

Find the ket that maximizes the uncertainty product

\label{eqn:moreKet:140}
\expectation{\lr{\Delta S_x}^2}
\expectation{\lr{\Delta S_y}^2},

and compare to the uncertainty bound $$\inv{4} \Abs{ \expectation{\antisymmetric{S_x}{S_y}}}^2$$.

To parameterize the ket space, consider first the kets that where both components are both not zero, where a single complex number can parameterize the ket

\label{eqn:moreKet:160}
\ket{s} =
\begin{bmatrix}
\beta’ e^{i\phi’} \\
\alpha’ e^{i\theta’} \\
\end{bmatrix}
\propto
\begin{bmatrix}
1 \\
\alpha e^{i\theta} \\
\end{bmatrix}

The expectation values with respect to this ket are
\label{eqn:moreKet:180}
\begin{aligned}
\expectation{S_x}
&=
\frac{\Hbar}{2}
\begin{bmatrix}
1 & \alpha e^{-i\theta} \\
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
1 \\
\alpha e^{i\theta} \\
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
1 &
\alpha e^{-i\theta} \\
\end{bmatrix}
\begin{bmatrix}
\alpha e^{i\theta} \\
1 \\
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\alpha e^{i\theta} + \alpha e^{-i\theta} \\
&=
\frac{\Hbar}{2}
2 \alpha \cos\theta \\
&=
\Hbar \alpha \cos\theta.
\end{aligned}

\label{eqn:moreKet:200}
\begin{aligned}
\expectation{S_y}
&=
\frac{\Hbar}{2}
\begin{bmatrix}
1 & \alpha e^{-i\theta} \\
\end{bmatrix}
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
1 \\
\alpha e^{i\theta} \\
\end{bmatrix} \\
&=
\frac{i\Hbar}{2}
\begin{bmatrix}
1 & \alpha e^{-i\theta} \\
\end{bmatrix}
\begin{bmatrix}
-\alpha e^{i\theta} \\
1 \\
\end{bmatrix} \\
&=
\frac{-i \alpha \Hbar}{2} 2 i \sin\theta \\
&=
\alpha \Hbar \sin\theta.
\end{aligned}

The variances are
\label{eqn:moreKet:220}
\begin{aligned}
\lr{ \Delta S_x }^2
&=
\lr{
\frac{\Hbar}{2}
\begin{bmatrix}
-2 \alpha \cos\theta & 1 \\
1 & -2 \alpha \cos\theta \\
\end{bmatrix}
}^2 \\
&=
\frac{\Hbar^2}{4}
\begin{bmatrix}
-2 \alpha \cos\theta & 1 \\
1 & -2 \alpha \cos\theta \\
\end{bmatrix}
\begin{bmatrix}
-2 \alpha \cos\theta & 1 \\
1 & -2 \alpha \cos\theta \\
\end{bmatrix} \\
&=
\frac{\Hbar^2}{4}
\begin{bmatrix}
4 \alpha^2 \cos^2\theta + 1 & -4 \alpha \cos\theta \\
-4 \alpha \cos\theta & 4 \alpha^2 \cos^2\theta + 1 \\
\end{bmatrix},
\end{aligned}

and

\label{eqn:moreKet:240}
\begin{aligned}
\lr{ \Delta S_y }^2
&=
\lr{
\frac{\Hbar}{2}
\begin{bmatrix}
-2 \alpha \sin\theta & -i \\
i & -2 \alpha \sin\theta \\
\end{bmatrix}
}^2 \\
&=
\frac{\Hbar^2}{4}
\begin{bmatrix}
-2 \alpha \sin\theta & -i \\
i & -2 \alpha \sin\theta \\
\end{bmatrix}
\begin{bmatrix}
-2 \alpha \sin\theta & -i \\
i & -2 \alpha \sin\theta \\
\end{bmatrix} \\
&=
\frac{\Hbar^2}{4}
\begin{bmatrix}
4 \alpha^2 \sin^2\theta + 1 & 4 \alpha i \sin\theta \\
-4 \alpha i \sin\theta & 4 \alpha^2 \sin^2\theta + 1 \\
\end{bmatrix}.
\end{aligned}

The uncertainty factors are

\label{eqn:moreKet:260}
\begin{aligned}
\expectation{\lr{\Delta S_x}^2}
&=
\frac{\Hbar^2}{4}
\begin{bmatrix}
1 & \alpha e^{-i\theta}
\end{bmatrix}
\begin{bmatrix}
4 \alpha^2 \cos^2\theta + 1 & -4 \alpha \cos\theta \\
-4 \alpha \cos\theta & 4 \alpha^2 \cos^2\theta + 1 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
\alpha e^{i\theta}
\end{bmatrix} \\
&=
\frac{\Hbar^2}{4}
\begin{bmatrix}
1 & \alpha e^{-i\theta}
\end{bmatrix}
\begin{bmatrix}
4 \alpha^2 \cos^2\theta + 1 -4 \alpha^2 \cos\theta e^{i\theta} \\
-4 \alpha \cos\theta + 4 \alpha^3 \cos^2\theta e^{i\theta} + \alpha e^{i\theta} \\
\end{bmatrix} \\
&=
\frac{\Hbar^2}{4}
\lr{
4 \alpha^2 \cos^2\theta + 1 -4 \alpha^2 \cos\theta e^{i\theta}
-4 \alpha^2 \cos\theta e^{-i\theta} + 4 \alpha^4 \cos^2\theta + \alpha^2
} \\
&=
\frac{\Hbar^2}{4}
\lr{
4 \alpha^2 \cos^2\theta + 1 -8 \alpha^2 \cos^2\theta
+ 4 \alpha^4 \cos^2\theta + \alpha^2
} \\
&=
\frac{\Hbar^2}{4}
\lr{
-4 \alpha^2 \cos^2\theta + 1
+ 4 \alpha^4 \cos^2\theta + \alpha^2
} \\
&=
\frac{\Hbar^2}{4}
\lr{
4 \alpha^2 \cos^2\theta \lr{ \alpha^2 – 1 }
+ \alpha^2 + 1
}
,
\end{aligned}

and

\label{eqn:moreKet:280}
\begin{aligned}
\expectation{ \lr{ \Delta S_y }^2 }
&=
\frac{\Hbar^2}{4}
\begin{bmatrix}
1 & \alpha e^{-i\theta}
\end{bmatrix}
\begin{bmatrix}
4 \alpha^2 \sin^2\theta + 1 & 4 \alpha i \sin\theta \\
-4 \alpha i \sin\theta & 4 \alpha^2 \sin^2\theta + 1 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
\alpha e^{i\theta}
\end{bmatrix} \\
&=
\frac{\Hbar^2}{4}
\begin{bmatrix}
1 & \alpha e^{-i\theta}
\end{bmatrix}
\begin{bmatrix}
4 \alpha^2 \sin^2\theta + 1 + 4 \alpha^2 i \sin\theta e^{i\theta} \\
-4 \alpha i \sin\theta + 4 \alpha^3 \sin^2\theta e^{i\theta} + \alpha e^{i\theta} \\
\end{bmatrix} \\
&=
\frac{\Hbar^2}{4}
\lr{
4 \alpha^2 \sin^2\theta + 1 + 4 \alpha^2 i \sin\theta e^{i\theta}
-4 \alpha^2 i \sin\theta e^{-i\theta} + 4 \alpha^4 \sin^2\theta + \alpha^2
} \\
&=
\frac{\Hbar^2}{4}
\lr{
-4 \alpha^2 \sin^2\theta + 1
+ 4 \alpha^4 \sin^2\theta + \alpha^2
} \\
&=
\frac{\Hbar^2}{4}
\lr{
4 \alpha^2 \sin^2\theta \lr{ \alpha^2 – 1}
+ \alpha^2
+ 1
}
.
\end{aligned}

The uncertainty product can finally be calculated

\label{eqn:moreKet:300}
\begin{aligned}
\expectation{\lr{\Delta S_x}^2}
\expectation{\lr{\Delta S_y}^2}
&=
\lr{\frac{\Hbar}{2} }^4
\lr{
4 \alpha^2 \cos^2\theta \lr{ \alpha^2 – 1 }
+ \alpha^2 + 1
}
\lr{
4 \alpha^2 \sin^2\theta \lr{ \alpha^2 – 1}
+ \alpha^2
+ 1
} \\
&=
\lr{\frac{\Hbar}{2} }^4
\lr{
4 \alpha^4 \sin^2 \lr{ 2\theta } \lr{ \alpha^2 – 1 }
+ 4 \alpha^2 \lr{ \alpha^4 – 1 }
+ \lr{\alpha^2 + 1 }^2
}
\end{aligned}

The maximum occurs when $$f = \sin^2 2 \theta$$ is extremized. Those points are
\label{eqn:moreKet:320}
\begin{aligned}
0
&= \PD{\theta}{f} \\
&= 2 \sin 2 \theta \cos 2\theta \\
&= 4 \sin 4 \theta.
\end{aligned}

Those points are at $$4 \theta = \pi n$$, for integer $$n$$, or

\label{eqn:moreKet:340}
\theta = \frac{\pi}{4} n, n \in [0, 7],

Minimums will occur when

\label{eqn:moreKet:360}
0 < \PDSq{\theta}{f} = 8 \cos 4\theta, or $$\label{eqn:moreKet:380} n = 0, 2, 4, 6.$$ At these points $$\sin^2 2\theta$$ takes the values $$\label{eqn:moreKet:400} \sin^2 \lr{ 2 \frac{\pi}{4} \setlr{ 0, 2, 4, 6 } } = \sin^2 \lr{ \pi \setlr{ 0, 1, 2, 3 } } \in \setlr{ 0 },$$ so the maximumization of the uncertainty product can be reduced to that of $$\label{eqn:moreKet:420} \expectation{\lr{\Delta S_x}^2} \expectation{\lr{\Delta S_y}^2} = \lr{\frac{\Hbar}{2} }^4 \lr{ 4 \alpha^2 \lr{ \alpha^4 - 1 } + \lr{\alpha^2 + 1 }^2 }$$ We seek \label{eqn:moreKet:440} \begin{aligned} 0 &= \PD{\alpha}{} \lr{ 4 \alpha^2 \lr{ \alpha^4 - 1 } + \lr{\alpha^2 + 1 }^2 } \\ &= \lr{ 8 \alpha \lr{ \alpha^4 - 1 } +16 \alpha^5 + 4 \lr{\alpha^2 + 1 } \alpha } \\ &= 4 \alpha \lr{ 2 \alpha^4 - 2 +4 \alpha^4 + 4 \alpha^2 + 4 } \\ &= 8 \alpha \lr{ 3 \alpha^4 + 2 \alpha^2 + 1 }. \end{aligned} The only real root of this polynomial is $$\alpha = 0$$, so the ket where both $$\ket{+}$$ and $$\ket{-}$$ are not zero that maximizes the uncertainty product is $$\label{eqn:moreKet:460} \ket{s} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \ket{+}.$$ The search for this maximizing value excluded those kets proportional to $$\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \ket{-}$$. Let's see the values of this uncertainty product at both $$\ket{\pm}$$, and compare to the uncertainty commutator. First $$\ket{s} = \ket{+}$$ $$\label{eqn:moreKet:480} \expectation{S_x} = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = 0.$$ $$\label{eqn:moreKet:500} \expectation{S_y} = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = 0.$$ so $$\label{eqn:moreKet:520} \expectation{ \lr{ \Delta S_x }^2 } = \lr{\frac{\Hbar}{2}}^2 \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \lr{\frac{\Hbar}{2}}^2$$ $$\label{eqn:moreKet:540} \expectation{ \lr{ \Delta S_y }^2 } = \lr{\frac{\Hbar}{2}}^2 \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \lr{\frac{\Hbar}{2}}^2.$$ For the commutator side of the uncertainty relation we have \label{eqn:moreKet:560} \begin{aligned} \inv{4} \Abs{ \expectation{ \antisymmetric{ S_x}{ S_y } } }^2 &= \inv{4} \Abs{ \expectation{ i \hbar S_z } }^2 \\ &= \lr{ \frac{\Hbar}{2} }^4 \Abs{ \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} }^2, \end{aligned} so for the $$\ket{+}$$ state we have an equality condition for the uncertainty relation $$\label{eqn:moreKet:580} \expectation{\lr{\Delta S_x}^2} \expectation{\lr{\Delta S_y}^2} = \inv{4} \Abs{ \expectation{\antisymmetric{S_x}{S_y}}}^2 = \lr{ \frac{\Hbar}{2} }^4.$$ It's reasonable to guess that the $$\ket{-}$$ state also matches the equality condition. Let's check $$\label{eqn:moreKet:600} \expectation{S_x} = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = 0.$$ $$\label{eqn:moreKet:620} \expectation{S_y} = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = 0.$$ so $$\expectation{ \lr{ \Delta S_x }^2 } = \expectation{ \lr{ \Delta S_y }^2 } = \lr{\frac{\Hbar}{2}}^2$$. For the commutator side of the uncertainty relation will be identical, so the equality of \ref{eqn:moreKet:580} is satisfied for both $$\ket{\pm}$$. Note that it wasn't explicitly verified that $$\ket{-}$$ maximized the uncertainty product, but I don't feel like working through that second set of algebraic mess. We can see by example that equality does not mean that the equality condition means that the product is maximized. For example, it is straightforward to show that $$\ket{ S_x ; \pm }$$ also satisfy the equality condition of the uncertainty relation. However, in that case the product is not maximized, but is zero.

## Question: Degenerate ket space example. ([1] pr. 1.23)

Consider operators with representation

\label{eqn:moreKet:20}
A =
\begin{bmatrix}
a & 0 & 0 \\
0 & -a & 0 \\
0 & 0 & -a
\end{bmatrix}
,
B =
\begin{bmatrix}
b & 0 & 0 \\
0 & 0 & -ib \\
0 & ib & 0
\end{bmatrix}.

Show that these both have degeneracies, commute, and compute a simultaneous ket space for both operators.

The eigenvalues and eigenvectors for $$A$$ can be read off by inspection, with values of $$a, -a, -a$$, and kets

\label{eqn:moreKet:40}
\ket{a_1} =
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix},
\ket{a_2} =
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
\ket{a_3} =
\begin{bmatrix}
0 \\
0 \\
1 \\
\end{bmatrix}

Notice that the lower-right $$2 \times 2$$ submatrix of $$B$$ is proportional to $$\sigma_y$$, so it’s eigenvalues can be formed by inspection

\label{eqn:moreKet:60}
\ket{b_1} =
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix},
\ket{b_2} =
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
1 \\
i
\end{bmatrix},
\ket{b_3} =
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
1 \\
-i \\
\end{bmatrix}.

Computing $$B \ket{b_i}$$ shows that the eigenvalues are $$b, b, -b$$ respectively.

Because of the two-fold degeneracy in the $$-a$$ eigenvalues of $$A$$, any linear combination of $$\ket{a_2}, \ket{a_3}$$ will also be an eigenket. In particular,

\label{eqn:moreKet:80}
\begin{aligned}
\ket{a_2} + i \ket{a_3} &= \ket{b_2} \\
\ket{a_2} – i \ket{a_3} &= \ket{b_3},
\end{aligned}

so the basis $$\setlr{ \ket{b_i}}$$ is a simulaneous eigenspace for both $$A$$ and $$B$$. Because there is a simulaneous eigenspace, the matrices must commute. This can be confirmed with direct computation

\label{eqn:moreKet:100}
\begin{aligned}
A B
&= a b
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0
\end{bmatrix} \\
&=
a b
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & i \\
0 & -i & 0
\end{bmatrix},
\end{aligned}

and

\label{eqn:moreKet:120}
\begin{aligned}
B A
&= a b
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix} \\
&=
a b
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & i \\
0 & -i & 0
\end{bmatrix}.
\end{aligned}

## Question: Unitary transformation. ([1] pr. 1.26)

Construct the transformation matrix that maps between the $$S_z$$ diagonal basis, to the $$S_x$$ diagonal basis.

Based on the definition

\label{eqn:moreKet:640}
U \ket{a^{(r)}} = \ket{b^{(r)}},

the matrix elements can be computed

\label{eqn:moreKet:660}
\bra{a^{(s)}} U \ket{a^{(r)}} = \braket{a^{(s)}}{b^{(r)}},

that is

\label{eqn:moreKet:680}
\begin{aligned}
U
&=
\begin{bmatrix}
\bra{a^{(1)}} U \ket{a^{(1)}} & \bra{a^{(1)}} U \ket{a^{(2)}} \\
\bra{a^{(2)}} U \ket{a^{(1)}} & \bra{a^{(2)}} U \ket{a^{(2)}}
\end{bmatrix} \\
&=
\begin{bmatrix}
\braket{a^{(1)}}{b^{(1)}} & \braket{a^{(1)}}{b^{(2)}} \\
\braket{a^{(2)}}{b^{(1)}} & \braket{a^{(2)}}{b^{(2)}}
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 \\ 1
\end{bmatrix} &
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 \\ -1
\end{bmatrix} \\
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\ 1
\end{bmatrix} &
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\ -1
\end{bmatrix} \\
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.
\end{aligned}

As a similarity transformation, we have

\label{eqn:moreKet:700}
\begin{aligned}
\bra{b^{(r)}} S_z \ket{b^{(s)}}
&=
\braket{b^{(r)}}{a^{(t)}}\bra{a^{(t)}} S_z \ket{a^{(u)}}\braket{a^{(u)}}{b^{(s)}} \\
&=
\braket{a^{(r)}}U^\dagger {a^{(t)}}\bra{a^{(t)}} S_z \ket{a^{(u)}}\bra{a^{(u)}}U \ket{a^{(s)}},
\end{aligned}

or

\label{eqn:moreKet:720}
S_z’ = U^\dagger S_z U.

Let’s check that the computed similarity transformation does it’s job.
\label{eqn:moreKet:740}
\begin{aligned}
\sigma_z’
&= U^\dagger \sigma_z U \\
&= \inv{2}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix} \\
&=
\inv{2}
\begin{bmatrix}
1 & -1 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix} \\
&=
\inv{2}
\begin{bmatrix}
0 & 2 \\
2 & 0
\end{bmatrix} \\
&= \sigma_x.
\end{aligned}

The transformation matrix can also be computed more directly

\label{eqn:moreKet:760}
\begin{aligned}
U
&= U \ket{a^{(r)}} \bra{a^{(r)}} \\
&= \ket{b^{(r)}}\bra{a^{(r)}} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
\begin{bmatrix}
1 & 0
\end{bmatrix}
+
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-1
\end{bmatrix}
\begin{bmatrix}
0 & 1
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 0 \\
1 & 0
\end{bmatrix}
+
\inv{\sqrt{2}}
\begin{bmatrix}
0 & 1 \\
0 & -1
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## (a)

Determine the matrix representation of $$\ket{\alpha}\bra{\beta}$$ given a complete set of eigenvectors $$\ket{a^r}$$.

## (b)

Verify with $$\ket{\alpha} = \ket{s_z = \Hbar/2}, \ket{s_x = \Hbar/2}$$.

## (a)

Forming the matrix element

\label{eqn:moreBraKetProblems:20}
\begin{aligned}
\bra{a^r} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^s}
&=
\braket{a^r}{\alpha}\braket{\beta}{a^s} \\
&=
\braket{a^r}{\alpha}
\braket{a^s}{\beta}^\conj,
\end{aligned}

the matrix representation is seen to be

\label{eqn:moreBraKetProblems:40}
\ket{\alpha}\bra{\beta}
\sim
\begin{bmatrix}
\bra{a^1} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^1} & \bra{a^1} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^2} & \cdots \\
\bra{a^2} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^1} & \bra{a^2} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^2} & \cdots \\
\vdots & \vdots & \ddots \\
\end{bmatrix}
=
\begin{bmatrix}
\braket{a^1}{\alpha} \braket{a^1}{\beta}^\conj & \braket{a^1}{\alpha} \braket{a^2}{\beta}^\conj & \cdots \\
\braket{a^2}{\alpha} \braket{a^1}{\beta}^\conj & \braket{a^2}{\alpha} \braket{a^2}{\beta}^\conj & \cdots \\
\vdots & \vdots & \ddots \\
\end{bmatrix}.

## (b)

First compute the spin-z representation of $$\ket{s_x = \Hbar/2 }$$.

\label{eqn:moreBraKetProblems:60}
\begin{aligned}
\lr{ S_x – \Hbar/2 I }
\begin{bmatrix}
a \\
b
\end{bmatrix}
&=
\lr{
\begin{bmatrix}
0 & \Hbar/2 \\
\Hbar/2 & 0 \\
\end{bmatrix}

\begin{bmatrix}
\Hbar/2 & 0 \\
0 & \Hbar/2 \\
\end{bmatrix}
} \\
&=
\begin{bmatrix}
a \\
b
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
-1 & 1 \\
1 & -1 \\
\end{bmatrix}
\begin{bmatrix}
a \\
b
\end{bmatrix},
\end{aligned}

so $$\ket{s_x = \Hbar/2 } \propto (1,1)$$.

Normalized we have

\label{eqn:moreBraKetProblems:80}
\begin{aligned}
\ket{\alpha} &= \ket{s_z = \Hbar/2 } =
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
\ket{\beta} &= \ket{s_z = \Hbar/2 }
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1
\end{bmatrix}.
\end{aligned}

Using \ref{eqn:moreBraKetProblems:40} the matrix representation is

\label{eqn:moreBraKetProblems:100}
\ket{\alpha}\bra{\beta}
\sim
\begin{bmatrix}
(1) (1/\sqrt{2})^\conj & (1) (1/\sqrt{2})^\conj \\
(0) (1/\sqrt{2})^\conj & (0) (1/\sqrt{2})^\conj \\
\end{bmatrix}
=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}.

This can be confirmed with direct computation
\label{eqn:moreBraKetProblems:120}
\begin{aligned}
\ket{\alpha}\bra{\beta}
&=
\begin{bmatrix}
1 \\
0
\end{bmatrix}
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}.
\end{aligned}

## Question: eigenvalue of sum of kets ([1] pr. 1.6)

Given eigenkets $$\ket{i}, \ket{j}$$ of an operator $$A$$, what are the conditions that $$\ket{i} + \ket{j}$$ is also an eigenvector?

Let $$A \ket{i} = i \ket{i}, A \ket{j} = j \ket{j}$$, and suppose that the sum is an eigenket. Then there must be a value $$a$$ such that

\label{eqn:moreBraKetProblems:140}
A \lr{ \ket{i} + \ket{j} } = a \lr{ \ket{i} + \ket{j} },

so

\label{eqn:moreBraKetProblems:160}
i \ket{i} + j \ket{j} = a \lr{ \ket{i} + \ket{j} }.

Operating with $$\bra{i}, \bra{j}$$ respectively, gives

\label{eqn:moreBraKetProblems:180}
\begin{aligned}
i &= a \\
j &= a,
\end{aligned}

so for the sum to be an eigenket, both of the corresponding energy eigenvalues must be identical (i.e. linear combinations of degenerate eigenkets are also eigenkets).

## Question: Null operator ([1] pr. 1.7)

Given eigenkets $$\ket{a’}$$ of operator $$A$$

## (a)

show that

\label{eqn:moreBraKetProblems:200}
\prod_{a’} \lr{ A – a’ }

is the null operator.

## (b)

\label{eqn:moreBraKetProblems:220}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”}

## (c)

Illustrate using $$S_z$$ for a spin 1/2 system.

## (a)

Application of $$\ket{a}$$, the eigenket of $$A$$ with eigenvalue $$a$$ to any term $$A – a’$$ scales $$\ket{a}$$ by $$a – a’$$, so the product operating on $$\ket{a}$$ is

\label{eqn:moreBraKetProblems:240}
\prod_{a’} \lr{ A – a’ } \ket{a} = \prod_{a’} \lr{ a – a’ } \ket{a}.

Since $$\ket{a}$$ is one of the $$\setlr{\ket{a’}}$$ eigenkets of $$A$$, one of these terms must be zero.

## (b)

Again, consider the action of the operator on $$\ket{a}$$,

\label{eqn:moreBraKetProblems:260}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a}
=
\prod_{a” \ne a’} \frac{\lr{ a – a” }}{a’ – a”} \ket{a}.

If $$\ket{a} = \ket{a’}$$, then $$\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a} = \ket{a}$$, whereas if it does not, then it equals one of the $$a”$$ energy eigenvalues. This is a representation of the Kronecker delta function

\label{eqn:moreBraKetProblems:300}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a} \equiv \delta_{a’, a} \ket{a}

## (c)

For operator $$S_z$$ the eigenvalues are $$\setlr{ \Hbar/2, -\Hbar/2 }$$, so the null operator must be

\label{eqn:moreBraKetProblems:280}
\begin{aligned}
\prod_{a’} \lr{ A – a’ }
&=
\lr{ \frac{\Hbar}{2} }^2 \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} – \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\begin{bmatrix}
0 & 0 \\
0 & -2
\end{bmatrix}
\begin{bmatrix}
2 & 0 \\
0 & 0 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 0 \\
0 & 0 \\
\end{bmatrix}
\end{aligned}

For the delta representation, consider the $$\ket{\pm}$$ states and their eigenvalue. The delta operators are

\label{eqn:moreBraKetProblems:320}
\begin{aligned}
\prod_{a” \ne \Hbar/2} \frac{\lr{ A – a” }}{\Hbar/2 – a”}
&=
\frac{S_z – (-\Hbar/2) I}{\Hbar/2 – (-\Hbar/2)} \\
&=
\inv{2} \lr{ \sigma_z + I } \\
&=
\inv{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\inv{2}
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix}
\\
&=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.
\end{aligned}

\label{eqn:moreBraKetProblems:340}
\begin{aligned}
\prod_{a” \ne -\Hbar/2} \frac{\lr{ A – a” }}{-\Hbar/2 – a”}
&=
\frac{S_z – (\Hbar/2) I}{-\Hbar/2 – \Hbar/2} \\
&=
\inv{2} \lr{ \sigma_z – I } \\
&=
\inv{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} – \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\inv{2}
\begin{bmatrix}
0 & 0 \\
0 & -2
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}.
\end{aligned}

These clearly have the expected delta function property acting on kets $$\ket{+} = (1,0), \ket{-} = (0, 1)$$.

## Question: Spin half general normal ([1] pr. 1.9)

Construct $$\ket{\BS \cdot \ncap ; + }$$, where $$\ncap = ( \cos\alpha \sin\beta, \sin\alpha \sin\beta, \cos\beta )$$ such that

\label{eqn:moreBraKetProblems:360}
\BS \cdot \ncap \ket{\BS \cdot \ncap ; + } =
\frac{\Hbar}{2} \ket{\BS \cdot \ncap ; + },

Solve this as an eigenvalue problem.

The spin operator for this direction is

\label{eqn:moreBraKetProblems:380}
\begin{aligned}
\BS \cdot \ncap
&= \frac{\Hbar}{2} \Bsigma \cdot \ncap \\
&= \frac{\Hbar}{2}
\lr{
\cos\alpha \sin\beta \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + \sin\alpha \sin\beta \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + \cos\beta \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta &
e^{-i\alpha}
\sin\beta
\\
e^{i\alpha}
\sin\beta
& -\cos\beta
\end{bmatrix}.
\end{aligned}

Observed that this is traceless and has a $$-\Hbar/2$$ determinant like any of the $$x,y,z$$ spin operators.

Assuming that this has an $$\Hbar/2$$ eigenvalue (to be verified later), the eigenvalue problem is

\label{eqn:moreBraKetProblems:400}
\begin{aligned}
0
&=
\BS \cdot \ncap – \Hbar/2 I \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta -1 &
e^{-i\alpha}
\sin\beta
\\
e^{i\alpha}
\sin\beta
& -\cos\beta -1
\end{bmatrix} \\
&=
\Hbar
\begin{bmatrix}
– \sin^2 \frac{\beta}{2} &
e^{-i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
\\
e^{i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
& -\cos^2 \frac{\beta}{2}
\end{bmatrix}
\end{aligned}

This has a zero determinant as expected, and the eigenvector $$(a,b)$$ will satisfy

\label{eqn:moreBraKetProblems:420}
\begin{aligned}
0
&= – \sin^2 \frac{\beta}{2} a +
e^{-i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
b \\
&= \sin\frac{\beta}{2} \lr{ – \sin \frac{\beta}{2} a +
e^{-i\alpha} b
\cos\frac{\beta}{2}
}
\end{aligned}

\label{eqn:moreBraKetProblems:440}
\begin{bmatrix}
a \\
b
\end{bmatrix}
\propto
\begin{bmatrix}
\cos\frac{\beta}{2} \\
e^{i\alpha}
\sin\frac{\beta}{2}
\end{bmatrix}.

This is appropriately normalized, so the ket for $$\BS \cdot \ncap$$ is

\label{eqn:moreBraKetProblems:460}
\ket{ \BS \cdot \ncap ; + } =
\cos\frac{\beta}{2} \ket{+} +
e^{i\alpha}
\sin\frac{\beta}{2}
\ket{-}.

Note that the other eigenvalue is

\label{eqn:moreBraKetProblems:480}
\ket{ \BS \cdot \ncap ; – } =
-\sin\frac{\beta}{2} \ket{+} +
e^{i\alpha}
\cos\frac{\beta}{2}
\ket{-}.

It is straightforward to show that these are orthogonal and that this has the $$-\Hbar/2$$ eigenvalue.

## Question: Two state Hamiltonian ([1] pr. 1.10)

Solve the eigenproblem for

\label{eqn:moreBraKetProblems:500}
H = a \biglr{
\ket{1}\bra{1}
-\ket{2}\bra{2}
+\ket{1}\bra{2}
+\ket{2}\bra{1}
}

In matrix form the Hamiltonian is

\label{eqn:moreBraKetProblems:520}
H = a
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.

The eigenvalue problem is

\label{eqn:moreBraKetProblems:540}
\begin{aligned}
0
&= \Abs{ H – \lambda I } \\
&= (a – \lambda)(-a – \lambda) – a^2 \\
&= (-a + \lambda)(a + \lambda) – a^2 \\
&= \lambda^2 – a^2 – a^2,
\end{aligned}

or

\label{eqn:moreBraKetProblems:560}
\lambda = \pm \sqrt{2} a.

An eigenket proportional to $$(\alpha,\beta)$$ must satisfy

\label{eqn:moreBraKetProblems:580}
0
= ( 1 \mp \sqrt{2} ) \alpha + \beta,

so

\label{eqn:moreBraKetProblems:600}
\ket{\pm} \propto
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix},

or

\label{eqn:moreBraKetProblems:620}
\begin{aligned}
\ket{\pm}
&=
\inv{2(2 – \sqrt{2})}
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix} \\
&=
\frac{2 + \sqrt{2}}{4}
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix}.
\end{aligned}

That is
\label{eqn:moreBraKetProblems:640}
\ket{\pm} =
\frac{2 + \sqrt{2}}{4} \lr{
-\ket{1} + (1 \mp \sqrt{2}) \ket{2}
}.

## Question: Spin half probability and dispersion ([1] pr. 1.12)

A spin $$1/2$$ system $$\BS \cdot \ncap$$, with $$\ncap = \sin \gamma \xcap + \cos\gamma \zcap$$, is in state with eigenvalue $$\Hbar/2$$.

## (a)

If $$S_x$$ is measured. What is the probability of getting $$+ \Hbar/2$$?

## (b)

Evaluate the dispersion in $$S_x$$, that is,

\label{eqn:moreBraKetProblems:660}
\expectation{\lr{ S_x – \expectation{S_x}}^2}.

## (a)

In matrix form the spin operator for the system is

\label{eqn:moreBraKetProblems:680}
\begin{aligned}
\BS \cdot \ncap
&= \frac{\Hbar}{2} \lr{ \cos\gamma \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \sin\gamma \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}} \\
&= \frac{\Hbar}{2}
\begin{bmatrix}
\cos\gamma & \sin\gamma \\
\sin\gamma & -\cos\gamma \\
\end{bmatrix}
\end{aligned}

An eigenket $$\ket{\BS \cdot \ncap ; + } = (a,b)$$ must satisfy

\label{eqn:moreBraKetProblems:700}
\begin{aligned}
0
&= \lr{ \cos \gamma – 1 } a + \sin\gamma b \\
&= \lr{ -2 \sin^2 \frac{\gamma}{2} } a + 2 \sin\frac{\gamma}{2} \cos\frac{\gamma}{2} b \\
&= -\sin \frac{\gamma}{2} a + \cos\frac{\gamma}{2} b,
\end{aligned}

so the eigenstate is
\label{eqn:moreBraKetProblems:720}
\ket{\BS \cdot \ncap ; + }
=
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}.

Pick $$\ket{S_x ; \pm } = \inv{\sqrt{2}} \begin{bmatrix} 1 \\ \pm 1 \end{bmatrix}$$ as the basis for the $$S_x$$ operator. Then, for the probability that the system will end up in the $$+ \Hbar/2$$ state of $$S_x$$, we have

\label{eqn:moreBraKetProblems:740}
\begin{aligned}
P
&= \Abs{\braket{ S_x ; + }{ \BS \cdot \ncap ; + } }^2 \\
&= \Abs{ \inv{\sqrt{2} }
{
\begin{bmatrix}
1 \\
1
\end{bmatrix}}^\dagger
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}
}^2 \\
&=\inv{2}
\Abs{
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}
}^2 \\
&=
\inv{2}
\lr{
\cos\frac{\gamma}{2} +
\sin\frac{\gamma}{2}
}^2 \\
&=
\inv{2}
\lr{ 1 + 2 \cos\frac{\gamma}{2} \sin\frac{\gamma}{2} } \\
&=
\inv{2}
\lr{ 1 + \sin\gamma }.
\end{aligned}

This is a reasonable seeming result, with $$P \in [0, 1]$$. Some special values also further validate this

\label{eqn:moreBraKetProblems:760}
\begin{aligned}
\gamma &= 0, \ket{\BS \cdot \ncap ; + } =
\begin{bmatrix}
1 \\
0
\end{bmatrix}
=
\ket{S_z ; +}
=
\inv{\sqrt{2}} \ket{S_x;+}
+\inv{\sqrt{2}} \ket{S_x;-}
\\
\gamma &= \pi/2, \ket{\BS \cdot \ncap ; + } =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\ket{S_x ; +}
\\
\gamma &= \pi, \ket{\BS \cdot \ncap ; + } =
\begin{bmatrix}
0 \\
1
\end{bmatrix}
=
\ket{S_z ; -}
=
\inv{\sqrt{2}} \ket{S_x;+}
-\inv{\sqrt{2}} \ket{S_x;-},
\end{aligned}

where we see that the probabilites are in proportion to the projection of the initial state onto the measured state $$\ket{S_x ; +}$$.

## (b)

The $$S_x$$ expectation is

\label{eqn:moreBraKetProblems:780}
\begin{aligned}
\expectation{S_x}
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix}
\sin\frac{\gamma}{2} \\
\cos\frac{\gamma}{2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2} 2 \sin\frac{\gamma}{2} \cos\frac{\gamma}{2} \\
&=
\frac{\Hbar}{2} \sin\gamma.
\end{aligned}

Note that $$S_x^2 = (\Hbar/2)^2I$$, so

\label{eqn:moreBraKetProblems:800}
\begin{aligned}
\expectation{S_x^2}
&=
\lr{\frac{\Hbar}{2}}^2
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix} \\
&=
\lr{ \frac{\Hbar}{2} }^2
\cos^2\frac{\gamma}{2} + \sin^2 \frac{\gamma}{2} \\
&=
\lr{ \frac{\Hbar}{2} }^2.
\end{aligned}

The dispersion is

\label{eqn:moreBraKetProblems:820}
\begin{aligned}
\expectation{\lr{ S_x – \expectation{S_x}}^2}
&=
\expectation{S_x^2} – \expectation{S_x}^2 \\
&=
\lr{ \frac{\Hbar}{2} }^2
\lr{1 – \sin^2 \gamma} \\
&=
\lr{ \frac{\Hbar}{2} }^2
\cos^2 \gamma.
\end{aligned}

At $$\gamma = \pi/2$$ the dispersion is 0, which is expected since $$\ket{\BS \cdot \ncap ; + } = \ket{ S_x ; + }$$ at that point. Similarily, the dispersion is maximized at $$\gamma = 0,\pi$$ where the $$\ket{\BS \cdot \ncap ; + }$$ component in the $$\ket{S_x ; + }$$ direction is minimized.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

# Some bra-ket manipulation problems.([1] pr. 1.4)

Using braket logic expand

## (a)

\label{eqn:braketManip:20}
\textrm{tr}{X Y}

## (b)

\label{eqn:braketManip:40}
(X Y)^\dagger

## (c)

\label{eqn:braketManip:60}
e^{i f(A)},

where $$A$$ is Hermitian with a complete set of eigenvalues.

## (d)

\label{eqn:braketManip:80}
\sum_{a’} \Psi_{a’}(\Bx’)^\conj \Psi_{a’}(\Bx”),

where $$\Psi_{a’}(\Bx”) = \braket{\Bx’}{a’}$$.

## (a)

\label{eqn:braketManip:100}
\begin{aligned}
\textrm{tr}{X Y}
&= \sum_a \bra{a} X Y \ket{a} \\
&= \sum_{a,b} \bra{a} X \ket{b}\bra{b} Y \ket{a} \\
&= \sum_{a,b}
\bra{b} Y \ket{a}
\bra{a} X \ket{b} \\
&= \sum_{a,b}
\bra{b} Y
X \ket{b} \\
&= \textrm{tr}{ Y X }.
\end{aligned}

## (b)

\label{eqn:braketManip:120}
\begin{aligned}
\bra{a} \lr{ X Y}^\dagger \ket{b}
&=
\lr{ \bra{b} X Y \ket{a} }^\conj \\
&=
\sum_c \lr{ \bra{b} X \ket{c}\bra{c} Y \ket{a} }^\conj \\
&=
\sum_c \lr{ \bra{b} X \ket{c} }^\conj \lr{ \bra{c} Y \ket{a} }^\conj \\
&=
\sum_c
\lr{ \bra{c} Y \ket{a} }^\conj
\lr{ \bra{b} X \ket{c} }^\conj \\
&=
\sum_c
\bra{a} Y^\dagger \ket{c}
\bra{c} X^\dagger \ket{b} \\
&=
\bra{a} Y^\dagger
X^\dagger \ket{b},
\end{aligned}

so $$\lr{ X Y }^\dagger = Y^\dagger X^\dagger$$.

## (c)

Let’s presume that the function $$f$$ has a Taylor series representation

\label{eqn:braketManip:140}
f(A) = \sum_r b_r A^r.

If the eigenvalues of $$A$$ are given by

\label{eqn:braketManip:160}
A \ket{a_s} = a_s \ket{a_s},

this operator can be expanded like

\label{eqn:braketManip:180}
\begin{aligned}
A
&= \sum_{a_s} A \ket{a_s} \bra{a_s} \\
&= \sum_{a_s} a_s \ket{a_s} \bra{a_s},
\end{aligned}

To compute powers of this operator, consider first the square

\label{eqn:braketManip:200}
\begin{aligned}
A^2 =
&=
\sum_{a_s} a_s \ket{a_s} \bra{a_s}
\sum_{a_r} a_r \ket{a_r} \bra{a_r} \\
&=
\sum_{a_s, a_r} a_s a_r \ket{a_s} \bra{a_s} \ket{a_r} \bra{a_r} \\
&=
\sum_{a_s, a_r} a_s a_r \ket{a_s} \delta_{s r} \bra{a_r} \\
&=
\sum_{a_s} a_s^2 \ket{a_s} \bra{a_s}.
\end{aligned}

The pattern for higher powers will clearly just be

\label{eqn:braketManip:220}
A^k =
\sum_{a_s} a_s^k \ket{a_s} \bra{a_s},

so the expansion of $$f(A)$$ will be

\label{eqn:braketManip:240}
\begin{aligned}
f(A)
&= \sum_r b_r A^r \\
&= \sum_r b_r
\sum_{a_s} a_s^r \ket{a_s} \bra{a_s} \\
&=
\sum_{a_s} \lr{ \sum_r b_r a_s^r } \ket{a_s} \bra{a_s} \\
&=
\sum_{a_s} f(a_s) \ket{a_s} \bra{a_s}.
\end{aligned}

The exponential expansion is

\label{eqn:braketManip:260}
\begin{aligned}
e^{i f(A)}
&=
\sum_t \frac{i^t}{t!} f^t(A) \\
&=
\sum_t \frac{i^t}{t!}
\lr{ \sum_{a_s} f(a_s) \ket{a_s} \bra{a_s} }^t \\
&=
\sum_t \frac{i^t}{t!}
\sum_{a_s} f^t(a_s) \ket{a_s} \bra{a_s} \\
&=
\sum_{a_s}
e^{i f(a_s) }
\ket{a_s} \bra{a_s}.
\end{aligned}

## (d)

\label{eqn:braketManip:n}
\begin{aligned}
\sum_{a’} \Psi_{a’}(\Bx’)^\conj \Psi_{a’}(\Bx”)
&=
\sum_{a’}
\braket{\Bx’}{a’}^\conj
\braket{\Bx”}{a’} \\
&=
\sum_{a’}
\braket{a’}{\Bx’}
\braket{\Bx”}{a’} \\
&=
\sum_{a’}
\braket{\Bx”}{a’}
\braket{a’}{\Bx’} \\
&=
\braket{\Bx”}{\Bx’} \\
&= \delta_{\Bx” – \Bx’}.
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Motivation

In [2] the Schwartz inequality

\label{eqn:qmSchwartz:20}
\boxed{
\braket{a}{a}
\braket{b}{b}
\ge \Abs{\braket{a}{b}}^2,
}

is used in the derivation of the uncertainty relation. The proof of the Schwartz inequality uses a sneaky substitution that doesn’t seem obvious, and is even less obvious since there is a typo in the value to be substituted. Let’s understand where that sneakiness is coming from.

## Without being sneaky

My ancient first year linear algebra text [1] contains a non-sneaky proof, but it only works for real vector spaces. Recast in bra-ket notation, this method examines the bounds of the norms of sums and differences of unit states (i.e. $$\braket{a}{a} = \braket{b}{b} = 1$$.)

\label{eqn:qmSchwartz:40}
\braket{a – b}{a – b}
= \braket{a}{a} + \braket{b}{b} – \braket{a}{b} – \braket{b}{a}
= 2 – 2 \textrm{Re} \braket{a}{b}
\ge 0,

so
\label{eqn:qmSchwartz:60}
1 \ge \textrm{Re} \braket{a}{b}.

Similarily

\label{eqn:qmSchwartz:80}
\braket{a + b}{a + b}
= \braket{a}{a} + \braket{b}{b} + \braket{a}{b} + \braket{b}{a}
= 2 + 2 \textrm{Re} \braket{a}{b}
\ge 0,

so
\label{eqn:qmSchwartz:100}
\textrm{Re} \braket{a}{b} \ge -1.

This means that for normalized state vectors

\label{eqn:qmSchwartz:120}
-1 \le \textrm{Re} \braket{a}{b} \le 1,

or
\label{eqn:qmSchwartz:140}
\Abs{\textrm{Re} \braket{a}{b}} \le 1.

Writing out the unit vectors explicitly, that last inequality is

\label{eqn:qmSchwartz:180}
\Abs{ \textrm{Re} \braket{ \frac{a}{\sqrt{\braket{a}{a}}} }{ \frac{b}{\sqrt{\braket{b}{b}}} } } \le 1,

squaring and rearranging gives

\label{eqn:qmSchwartz:200}
\Abs{\textrm{Re} \braket{a}{b}}^2 \le
\braket{a}{a}
\braket{b}{b}.

This is similar to, but not identical to the Schwartz inequality. Since $$\Abs{\textrm{Re} \braket{a}{b}} \le \Abs{\braket{a}{b}}$$ the Schwartz inequality cannot be demonstrated with this argument. This first year algebra method works nicely for demonstrating the inequality for real vector spaces, so a different argument is required for a complex vector space (i.e. quantum mechanics state space.)

## Arguing with projected and rejected components

Notice that the equality condition in the inequality holds when the vectors are colinear, and the largest inequality holds when the vectors are normal to each other. Given those geometrical observations, it seems reasonable to examine the norms of projected or rejected components of a vector. To do so in bra-ket notation, the correct form of a projection operation must be determined. Care is required to get the ordering of the bra-kets right when expressing such a projection.

Suppose we wish to calculation the rejection of $$\ket{a}$$ from $$\ket{b}$$, that is $$\ket{b – \alpha a}$$, such that

\label{eqn:qmSchwartz:220}
0
= \braket{a}{b – \alpha a}
= \braket{a}{b} – \alpha \braket{a}{a},

or
\label{eqn:qmSchwartz:240}
\alpha =
\frac{\braket{a}{b} }{ \braket{a}{a} }.

Therefore, the projection of $$\ket{b}$$ on $$\ket{a}$$ is

\label{eqn:qmSchwartz:260}
\textrm{Proj}_{\ket{a}} \ket{b}
= \frac{\braket{a}{b} }{ \braket{a}{a} } \ket{a}
= \frac{\braket{b}{a}^\conj }{ \braket{a}{a} } \ket{a}.

The conventional way to write this in QM is in the operator form

\label{eqn:qmSchwartz:300}
\textrm{Proj}_{\ket{a}} \ket{b}
= \frac{\ket{a}\bra{a}}{\braket{a}{a}} \ket{b}.

In this form the rejection of $$\ket{a}$$ from $$\ket{b}$$ can be expressed as

\label{eqn:qmSchwartz:280}
\textrm{Rej}_{\ket{a}} \ket{b} = \ket{b} – \frac{\ket{a}\bra{a}}{\braket{a}{a}} \ket{b}.

This state vector is normal to $$\ket{a}$$ as desired

\label{eqn:qmSchwartz:320}
\braket{a}{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a }
=
\braket{a}{ b} – \frac{ \braket{a}{b} }{ \braket{a}{a} } \braket{a}{a}
=
\braket{a}{ b} – \braket{a}{b}
= 0.

How about it’s length? That is

\label{eqn:qmSchwartz:340}
\begin{aligned}
\braket{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a}{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a }
&=
\braket{b}{b} – 2 \frac{\Abs{\braket{a}{b}}^2}{\braket{a}{a}} +\frac{\Abs{\braket{a}{b}}^2 }{ \braket{a}{a}^2 } \braket{a}{a} \\
&=
\braket{b}{b} – \frac{\Abs{\braket{a}{b}}^2}{\braket{a}{a}}.
\end{aligned}

Observe that this must be greater to or equal to zero, so

\label{eqn:qmSchwartz:360}
\braket{b}{b} \ge \frac{ \Abs{ \braket{a}{b} }^2 }{ \braket{a}{a} }.

Rearranging this gives \ref{eqn:qmSchwartz:20} as desired. The Schwartz proof in [2] obscures the geometry involved and starts with

\label{eqn:qmSchwartz:380}
\braket{b + \lambda a}{b + \lambda a} \ge 0,

where the “proof” is nothing more than a statement that one can “pick” $$\lambda = -\braket{b}{a}/\braket{a}{a}$$. The Pythagorean context of the Schwartz inequality is not mentioned, and without thinking about it, one is left wondering what sort of magic hat that $$\lambda$$ selection came from.

# References

[1] W Keith Nicholson. Elementary linear algebra, with applications. PWS-Kent Publishing Company, 1990.

[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.