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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] 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
\begin{equation}\label{eqn:qmLecture1:20} \begin{aligned} \ddt{\Br} &= \inv{m} \Bp \\ \ddt{\Bp} &= \spacegrad V \end{aligned} \end{equation}
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
\begin{equation}\label{eqn:qmLecture1:480} \boxed{ \Hbar = 1. } \end{equation}
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
\begin{equation}\label{eqn:qmLecture1:40} \braket{ x }{\Psi } = \Psi(x), \end{equation}
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
\begin{equation}\label{eqn:qmLecture1:60} \braket{ p }{\Psi } = \Psi(p), \end{equation}
a probability amplitude with respect to momentum p .
More precisely,
\begin{equation}\label{eqn:qmLecture1:80} \Abs{\Psi(x)}^2 dx \ge 0 \end{equation}
is the probability of finding the particle in the range (x, x + dx ) . To have meaning as a probability, we require
\begin{equation}\label{eqn:qmLecture1:100} \int_{-\infty}^\infty \Abs{\Psi(x)}^2 dx = 1. \end{equation}
The average position can be calculated using this probability density function. For example
\begin{equation}\label{eqn:qmLecture1:120} \expectation{x} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 x dx, \end{equation}
or
\begin{equation}\label{eqn:qmLecture1:140}
\expectation{f(x)} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 f(x) dx.
\end{equation}
Similarly, calculation of an average of a function of momentum can be expressed as
\begin{equation}\label{eqn:qmLecture1:160} \expectation{f(p)} = \int_{-\infty}^\infty \Abs{\Psi(p)}^2 f(p) dp. \end{equation}
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
\begin{equation}\label{eqn:qmLecture1:180} \Psi(p) \stackrel{?}{\leftrightarrow} \Psi(x), \end{equation}
or equivalently
\begin{equation}\label{eqn:qmLecture1:200}
\braket{p}{\Psi}
\stackrel{?}{\leftrightarrow}
\braket{x}{\Psi}.
\end{equation}
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
\begin{equation}\label{eqn:qmLecture1:220} \int_{-\infty}^\infty dp \ket{p}\bra{p} = 1, \end{equation}
or
\begin{equation}\label{eqn:qmLecture1:240}
\int_{-\infty}^\infty dx \ket{x}\bra{x} = 1
\end{equation}
Some interpretations:
- \ket{x_0} \leftrightarrow \text{sits at} x = x_0
- \braket{x}{x’} \leftrightarrow \delta(x – x’)
- \braket{p}{p’} \leftrightarrow \delta(p – p’)
- \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
\begin{equation}\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} \end{equation}
This also the determination of an integral operator representation for the delta function
\begin{equation}\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} \end{equation}
or
\begin{equation}\label{eqn:qmLecture1:520} \delta(x – x’) = \inv{V} \int dp e^{i p (x- x’)}. \end{equation}
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
\begin{equation}\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} \end{equation}
The momentum space to position space conversion can be written as
\begin{equation}\label{eqn:qmLecture1:300} \Psi(x) = \int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi(p) dp. \end{equation}
Now we can go back and figure out the an expectation
\begin{equation}\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} \end{equation}
Here we’ve essentially calculated the position space representation of the momentum operator, allowing identifications of the following form
\begin{equation}\label{eqn:qmLecture1:380}
p \leftrightarrow -i \PD{x}{}
\end{equation}
\begin{equation}\label{eqn:qmLecture1:400}
p^2 \leftrightarrow – \PDSq{x}{}.
\end{equation}
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
\begin{equation}\label{eqn:qmLecture1:540} \bra{x} P \ket{p} = -i \Hbar \PD{x}{} \braket{x}{p}, \end{equation}
but \bra{x} P \ket{p} = p \braket{x}{p} , providing a differential equation for \braket{x}{p}
\begin{equation}\label{eqn:qmLecture1:560} p \braket{x}{p} = -i \Hbar \PD{x}{} \braket{x}{p}, \end{equation}
with solution
\begin{equation}\label{eqn:qmLecture1:580} i p x/\Hbar = \ln \braket{x}{p} + \text{const}, \end{equation}
or
\begin{equation}\label{eqn:qmLecture1:600}
\braket{x}{p} \propto e^{i p x/\Hbar}.
\end{equation}
Matrix interpretation
- Ket’s \ket{\Psi} \leftrightarrow \text{column vector}
- Bra’s \bra{\Psi} \leftrightarrow {(\text{row vector})}^\conj
- Operators \leftrightarrow matrices that act on vectors.
\begin{equation}\label{eqn:qmLecture1:420} \hat{p} \ket{\Psi} \rightarrow \ket{\Psi’} \end{equation}
Time evolution
For a state subject to the equations of motion given by the Hamiltonian operator \hat{H}
\begin{equation}\label{eqn:qmLecture1:440} i \PD{t}{} \ket{\Psi} = \hat{H} \ket{\Psi}, \end{equation}
the time evolution is given by
\begin{equation}\label{eqn:qmLecture1:460}
\ket{\Psi(t)} = e^{-i \hat{H} t} \ket{\Psi(0)}.
\end{equation}
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.