projection operator

PHY1520H Graduate Quantum Mechanics. Lecture 20: Perturbation theory. Taught by Prof. Arun Paramekanti

December 3, 2015 phy1520 No comments , , , , , , , , , ,

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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] ch. 5 content.

Perturbation theory

Given a \( 2 \times 2 \) Hamiltonian \( H = H_0 + V \), where

\begin{equation}\label{eqn:qmLecture20:20}
H =
\begin{bmatrix}
a & c \\
c^\conj & b
\end{bmatrix}
\end{equation}

which has eigenvalues

\begin{equation}\label{eqn:qmLecture20:40}
\lambda_\pm = \frac{a + b}{2} \pm \sqrt{ \lr{ \frac{a – b}{2}}^2 + \Abs{c}^2 }.
\end{equation}

If \( c = 0 \),

\begin{equation}\label{eqn:qmLecture20:60}
H_0 =
\begin{bmatrix}
a & 0 \\
0 & b
\end{bmatrix},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:80}
V =
\begin{bmatrix}
0 & c \\
c^\conj & 0
\end{bmatrix}.
\end{equation}

Suppose that \( \Abs{c} \ll \Abs{a – b} \), then

\begin{equation}\label{eqn:qmLecture20:100}
\lambda_\pm \approx \frac{a + b}{2} \pm \Abs{ \frac{a – b}{2} } \lr{ 1 + 2 \frac{\Abs{c}^2}{\Abs{a – b}^2} }.
\end{equation}

If \( a > b \), then

\begin{equation}\label{eqn:qmLecture20:120}
\lambda_\pm \approx \frac{a + b}{2} \pm \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} }.
\end{equation}

\begin{equation}\label{eqn:qmLecture20:140}
\begin{aligned}
\lambda_{+}
&= \frac{a + b}{2} + \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&= a + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= a + \frac{\Abs{c}^2}{a – b},
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:qmLecture20:680}
\begin{aligned}
\lambda_{-}
&= \frac{a + b}{2} – \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&=
b + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= b + \frac{\Abs{c}^2}{a – b}.
\end{aligned}
\end{equation}

This adiabatic evolution displays a “level repulsion”, quadradic in \( \Abs{c} \) as sketched in fig. 1, and is described as a non-degenerate perbutation.

fig. 1.  Adiabatic (non-degenerate) perturbation

fig. 1. Adiabatic (non-degenerate) perturbation

If \( \Abs{c} \gg \Abs{a -b} \), then

\begin{equation}\label{eqn:qmLecture20:160}
\begin{aligned}
\lambda_\pm
&= \frac{a + b}{2} \pm \Abs{c} \sqrt{ 1 + \inv{\Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&\approx \frac{a + b}{2} \pm \Abs{c} \lr{ 1 + \inv{2 \Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&= \frac{a + b}{2} \pm \Abs{c} \pm \frac{\lr{a – b}^2}{8 \Abs{c}}.
\end{aligned}
\end{equation}

Here we loose the adiabaticity, and have “level repulsion” that is linear in \( \Abs{c} \), as sketched in fig. 2. We no longer have the sign of \( a – b \) in the expansion. This is described as a degenerate perbutation.

fig. 2.  Degenerate perbutation

fig. 2. Degenerate perbutation

General non-degenerate perturbation

Given an unperturbed system with solutions of the form

\begin{equation}\label{eqn:qmLecture20:180}
H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}},
\end{equation}

we want to solve the perturbed Hamiltonian equation

\begin{equation}\label{eqn:qmLecture20:200}
\lr{ H_0 + \lambda V } \ket{ n } = \lr{ E_n^{(0)} + \Delta n } \ket{n}.
\end{equation}

Here \( \Delta n \) is an energy shift as that goes to zero as \( \lambda \rightarrow 0 \). We can write this as

\begin{equation}\label{eqn:qmLecture20:220}
\lr{ E_n^{(0)} – H_0 } \ket{ n } = \lr{ \lambda V – \Delta_n } \ket{n}.
\end{equation}

We are hoping to iterate with application of the inverse to an initial estimate of \( \ket{n} \)

\begin{equation}\label{eqn:qmLecture20:240}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n}.
\end{equation}

This gets us into trouble if \( \lambda \rightarrow 0 \), which can be fixed by using

\begin{equation}\label{eqn:qmLecture20:260}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

which can be seen to be a solution to \ref{eqn:qmLecture20:220}. We want to ask if

\begin{equation}\label{eqn:qmLecture20:280}
\lr{ \lambda V – \Delta_n } \ket{n} ,
\end{equation}

contains a bit of \( \ket{ n^{(0)} } \)? To determine this act with \( \bra{n^{(0)}} \) on the left

\begin{equation}\label{eqn:qmLecture20:300}
\begin{aligned}
\bra{ n^{(0)} } \lr{ \lambda V – \Delta_n } \ket{n}
&=
\bra{ n^{(0)} } \lr{ E_n^{(0)} – H_0 } \ket{n} \\
&=
\lr{ E_n^{(0)} – E_n^{(0)} } \braket{n^{(0)}}{n} \\
&=
0.
\end{aligned}
\end{equation}

This shows that \( \ket{n} \) is entirely orthogonal to \( \ket{n^{(0)}} \).

Define a projection operator

\begin{equation}\label{eqn:qmLecture20:320}
P_n = \ket{n^{(0)}}\bra{n^{(0)}},
\end{equation}

which has the idempotent property \( P_n^2 = P_n \) that we expect of a projection operator.

Define a rejection operator
\begin{equation}\label{eqn:qmLecture20:340}
\overline{{P}}_n
= 1 –
\ket{n^{(0)}}\bra{n^{(0)}}
= \sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}.
\end{equation}

Because \( \ket{n} \) has no component in the direction \( \ket{n^{(0)}} \), the rejection operator can be inserted much like we normally do with the identity operator, yielding

\begin{equation}\label{eqn:qmLecture20:360}
\ket{n}’ = \lr{ E_n^{(0)} – H_0 }^{-1} \overline{{P}}_n \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

valid for any initial \( \ket{n} \).

Power series perturbation expansion

Instead of iterating, suppose that the unknown state and unknown energy difference operator can be expanded in a \( \lambda \) power series, say

\begin{equation}\label{eqn:qmLecture20:380}
\ket{n}
=
\ket{n_0}
+ \lambda \ket{n_1}
+ \lambda^2 \ket{n_2}
+ \lambda^3 \ket{n_3} + \cdots
\end{equation}

and

\begin{equation}\label{eqn:qmLecture20:400}
\Delta_{n} = \Delta_{n_0}
+ \lambda \Delta_{n_1}
+ \lambda^2 \Delta_{n_2}
+ \lambda^3 \Delta_{n_3} + \cdots
\end{equation}

We usually interpret functions of operators in terms of power series expansions. In the case of \( \lr{ E_n^{(0)} – H_0 }^{-1} \), we have a concrete interpretation when acting on one of the unpertubed eigenstates

\begin{equation}\label{eqn:qmLecture20:420}
\inv{ E_n^{(0)} – H_0 } \ket{m^{(0)}} =
\inv{ E_n^{(0)} – E_m^0 } \ket{m^{(0)}}.
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture20:440}
\ket{n}
=
\inv{ E_n^{(0)} – H_0 }
\sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}
\lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

or

\begin{equation}\label{eqn:qmLecture20:460}
\boxed{
\ket{n}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_n } \ket{n}.
}
\end{equation}

From \ref{eqn:qmLecture20:220}, note that

\begin{equation}\label{eqn:qmLecture20:500}
\Delta_n =
\frac{\bra{n^{(0)}} \lambda V \ket{n}}{\braket{n^0}{n}},
\end{equation}

however, we will normalize by setting \( \braket{n^0}{n} = 1 \), so

\begin{equation}\label{eqn:qmLecture20:521}
\boxed{
\Delta_n =
\bra{n^{(0)}} \lambda V \ket{n}.
}
\end{equation}

to \( O(\lambda^0) \)

If all \( \lambda^n, n > 0 \) are zero, then we have

\label{eqn:qmLecture20:780}
\begin{equation}\label{eqn:qmLecture20:740}
\ket{n_0}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ – \Delta_{n_0} } \ket{n_0}
\end{equation}
\begin{equation}\label{eqn:qmLecture20:800}
\Delta_{n_0} \braket{n^{(0)}}{n_0} = 0
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:540}
\begin{aligned}
\ket{n_0} &= \ket{n^{(0)}} \\
\Delta_{n_0} &= 0.
\end{aligned}
\end{equation}

to \( O(\lambda^1) \)

Requiring identity for all \( \lambda^1 \) terms means

\begin{equation}\label{eqn:qmLecture20:760}
\ket{n_1} \lambda
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_{n_1} \lambda } \ket{n_0},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:560}
\ket{n_1}
=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}} \bra{ m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} } \ket{n_0}.
\end{equation}

With the assumption that \( \ket{n^{(0)}} \) is normalized, and with the shorthand

\begin{equation}\label{eqn:qmLecture20:600}
V_{m n} = \bra{ m^{(0)}} V \ket{n^{(0)}},
\end{equation}

that is

\begin{equation}\label{eqn:qmLecture20:580}
\begin{aligned}
\ket{n_1}
&=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}
\\
\Delta_{n_1} &= \bra{ n^{(0)} } V \ket{ n^0} = V_{nn}.
\end{aligned}
\end{equation}

to \( O(\lambda^2) \)

The second order perturbation states are found by selecting only the \( \lambda^2 \) contributions to

\begin{equation}\label{eqn:qmLecture20:820}
\lambda^2 \ket{n_2}
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – (\lambda \Delta_{n_1} + \lambda^2 \Delta_{n_2}) }
\lr{
\ket{n_0}
+ \lambda \ket{n_1}
}.
\end{equation}

Because \( \ket{n_0} = \ket{n^{(0)}} \), the \( \lambda^2 \Delta_{n_2} \) is killed, leaving

\begin{equation}\label{eqn:qmLecture20:840}
\begin{aligned}
\ket{n_2}
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\ket{n_1} \\
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\sum_{l \ne n}
\frac{
\ket{l^{(0)}}
}
{
E_n^{(0)} – E_l^{(0)}
}
V_{l n},
\end{aligned}
\end{equation}

which can be written as

\begin{equation}\label{eqn:qmLecture20:620}
\ket{n_2}
=
\sum_{l,m \ne n}
\ket{m^{(0)}}
\frac{V_{m l} V_{l n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}

\sum_{m \ne n}
\ket{m^{(0)}}
\frac{V_{n n} V_{m n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.
\end{equation}

For the second energy perturbation we have

\begin{equation}\label{eqn:qmLecture20:860}
\lambda^2 \Delta_{n_2} =
\bra{n^{(0)}} \lambda V \lr{ \lambda \ket{n_1} },
\end{equation}

or

\begin{equation}\label{eqn:qmLecture20:880}
\begin{aligned}
\Delta_{n_2}
&=
\bra{n^{(0)}} V \ket{n_1} \\
&=
\bra{n^{(0)}} V
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}.
\end{aligned}
\end{equation}

That is

\begin{equation}\label{eqn:qmLecture20:900}
\Delta_{n_2}
=
\sum_{m \ne n} \frac{V_{n m} V_{m n} }{E_n^{(0)} – E_m^{(0)}}.
\end{equation}

to \( O(\lambda^3) \)

Similarily, it can be shown that

\begin{equation}\label{eqn:qmLecture20:640}
\Delta_{n_3} =
\sum_{l, m \ne n} \frac{V_{n m} V_{m l} V_{l n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}

\sum_{ m \ne n} \frac{V_{n m} V_{n n} V_{m n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.
\end{equation}

In general, the energy perturbation is given by

\begin{equation}\label{eqn:qmLecture20:660}
\Delta_n^{(l)} = \bra{n^{(0)}} V \ket{n^{(l-1)}}.
\end{equation}

References

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

Schwartz inequality in bra-ket notation

July 6, 2015 phy1520 No comments , , , , , ,


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Motivation

In [2] the Schwartz inequality

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

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. \( \braket{a}{a} = \braket{b}{b} = 1 \).)

\begin{equation}\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,
\end{equation}

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

Similarily

\begin{equation}\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,
\end{equation}

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

This means that for normalized state vectors

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

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

Writing out the unit vectors explicitly, that last inequality is

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

squaring and rearranging gives

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

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

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

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

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

\begin{equation}\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}.
\end{equation}

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

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

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

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

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

\begin{equation}\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.
\end{equation}

How about it’s length? That is

\begin{equation}\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}
\end{equation}

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

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

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

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

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.