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# Posts Tagged ‘state vector’

## PHY456H1F: Quantum Mechanics II. Lecture 1 (Taught by Prof J.E. Sipe). Review: Composite systems

Posted by peeterjoot on September 15, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Peeter’s lecture notes from class. May not be entirely coherent.

# Composite systems.

This is apparently covered as a side effect in the text [1] in one of the advanced material sections. FIXME: what section?

Example, one spin one half particle and one spin one particle. We can describe either quantum mechanically, described by a pair of Hilbert spaces

\begin{aligned}H_1,\end{aligned} \hspace{\stretch{1}}(1.1)

of dimension $D_1$

\begin{aligned}H_2,\end{aligned} \hspace{\stretch{1}}(1.2)

of dimension $D_2$

Recall that a Hilbert space (finite or infinite dimensional) is the set of states that describe the system. There were some additional details (completeness, normalizable, $L2$ integrable, …) not really covered in the physics curriculum, but available in mathematical descriptions.

We form the composite (Hilbert) space

\begin{aligned}H = H_1 \otimes H_2\end{aligned} \hspace{\stretch{1}}(1.3)

\begin{aligned}H_1 : { {\lvert {\phi_1^{(i)}} \rangle} }\end{aligned} \hspace{\stretch{1}}(1.4)

for any ket in $H_1$

\begin{aligned}{\lvert {I} \rangle} = \sum_{i=1}^{D_1} c_i {\lvert {\phi_1^{(i)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.5)

where

\begin{aligned}{\langle { \phi_1^{(i)}} \rvert}{\lvert { \phi_1^{(j)}} \rangle} = \delta^{i j}\end{aligned} \hspace{\stretch{1}}(1.6)

Similarly

\begin{aligned}H_2 : { {\lvert {\phi_2^{(i)}} \rangle} }\end{aligned} \hspace{\stretch{1}}(1.7)

for any ket in $H_2$

\begin{aligned}{\lvert {II} \rangle} = \sum_{i=1}^{D_2} d_i {\lvert {\phi_2^{(i)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.8)

where

\begin{aligned}{\langle { \phi_2^{(i)}} \rvert}{\lvert { \phi_2^{(j)}} \rangle} = \delta^{i j}\end{aligned} \hspace{\stretch{1}}(1.9)

The composite Hilbert space has dimension $D_1 D_2$

basis kets:

\begin{aligned}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} = {\lvert { \phi^{(ij)}} \rangle},\end{aligned} \hspace{\stretch{1}}(1.10)

where

\begin{aligned}{\langle { \phi^{(ij)}} \rvert}{\lvert { \phi^{(kl)}} \rangle} = \delta^{ik} \delta^{jl}.\end{aligned} \hspace{\stretch{1}}(1.11)

Any ket in $H$ can be written

\begin{aligned}{\lvert {\psi} \rangle} &= \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \\ &= \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi^{(ij)}} \rangle}.\end{aligned}

Direct product of kets:

\begin{aligned}{\lvert {I} \rangle} \otimes {\lvert {II} \rangle} &\equiv\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}c_i d_j{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \\ &=\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}c_i d_j{\lvert { \phi^{(ij)}} \rangle} \end{aligned}

If ${\lvert {\psi} \rangle}$ in $H$ cannot be written as ${\lvert {I} \rangle} \otimes {\lvert {II} \rangle}$, then ${\lvert {\psi} \rangle}$ is said to be “entangled”.

FIXME: insert a concrete example of this, with some low dimension.

## Operators.

With operators $\mathcal{O}_1$ and $\mathcal{O}_2$ on the respective Hilbert spaces. We’d now like to build

\begin{aligned}\mathcal{O}_1 \otimes \mathcal{O}_2\end{aligned} \hspace{\stretch{1}}(1.12)

If one defines

\begin{aligned}\mathcal{O}_1 \otimes \mathcal{O}_2\equiv\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.13)

Q:Can every operator that can be defined on the composite space have a representation of this form?

No.

Special cases. The identity operators. Suppose that

\begin{aligned}{\lvert {\psi} \rangle}=\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.14)

then

\begin{aligned}(\mathcal{O}_1 \otimes \mathcal{I}_2) {\lvert {\psi} \rangle}=\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.15)

### Example commutator.

Can do other operations. Example:

\begin{aligned}\left[{ \mathcal{O}_1 \otimes \mathcal{I}_2 },{ \mathcal{I}_1 \otimes \mathcal{O}_2 }\right] = 0\end{aligned} \hspace{\stretch{1}}(1.16)

Let’s verify this one. Suppose that our state has the representation

\begin{aligned}{\lvert {\psi} \rangle} = \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.17)

so that the action on this ket from the composite operations are

\begin{aligned}(\mathcal{O}_1 \otimes \mathcal{I}_2){\lvert {\psi} \rangle} &= \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \\ (\mathcal{I}_1 \otimes \mathcal{O}_2){\lvert {\psi} \rangle} &= \sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.18)

Our commutator is

\begin{aligned}\left[{(\mathcal{O}_1 \otimes \mathcal{I}_2)},{(\mathcal{I}_1 \otimes \mathcal{O}_2)}\right]{\lvert {\psi} \rangle} &=(\mathcal{O}_1 \otimes \mathcal{I}_2)(\mathcal{I}_1 \otimes \mathcal{O}_2) {\lvert {\psi} \rangle} -(\mathcal{I}_1 \otimes \mathcal{O}_2)(\mathcal{O}_1 \otimes \mathcal{I}_2){\lvert {\psi} \rangle} \\ &=(\mathcal{O}_1 \otimes \mathcal{I}_2)\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle}-(\mathcal{I}_1 \otimes \mathcal{O}_2)\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \phi_2^{(j)}} \rangle} \\ &=\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle}-\sum_{i = 1}^{D_1}\sum_{j = 1}^{D_2}f_{ij}{\lvert { \mathcal{O}_1 \phi_1^{(i)}} \rangle} \otimes {\lvert { \mathcal{O}_2 \phi_2^{(j)}} \rangle} \\ &=0 \qquad \square\end{aligned}

### Generalizations.

Can generalize to

\begin{aligned}H_1 \otimes H_2 \otimes H_3 \otimes \cdots\end{aligned} \hspace{\stretch{1}}(1.20)

Can also start with $H$ and seek factor spaces. If $H$ is not prime there are, in general, many ways to find factor spaces

\begin{aligned}H = H_1 \otimes H_2 =H_1' \otimes H_2'\end{aligned} \hspace{\stretch{1}}(1.21)

A ket ${\lvert {\psi} \rangle}$, if unentangled in the first factor space, then it will be in general entangled in a second space. Thus ket entanglement is not a property of the ket itself, but instead is intrinsically related to the space in which it is represented.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

## PHY356F: Quantum Mechanics I. Lecture 11 notes. Harmonic Oscillator.

Posted by peeterjoot on November 30, 2010

# Setup.

Why study this problem?

It is relevant to describing the oscillation of molecules, quantum states of light, vibrations of the lattice structure of a solid, and so on.

FIXME: projected picture of masses on springs, with a ladle shaped well, approximately Harmonic about the minimum of the bucket.

The problem to solve is the one dimensional Hamiltonian

\begin{aligned}V(X) &= \frac{1}{{2}} K X^2 \\ K &= m \omega^2 \\ H &= \frac{P^2}{2m} + V(X)\end{aligned} \hspace{\stretch{1}}(8.168)

where $m$ is the mass, $\omega$ is the frequency, $X$ is the position operator, and $P$ is the momentum operator. Of these quantities, $\omega$ and $m$ are classical quantities.

This problem can be used to illustrate some of the reasons why we study the different pictures (Heisenberg, Interaction and Schr\”{o}dinger). This is a problem well suited to all of these (FIXME: lookup an example of this with the interaction picture. The book covers H and S methods.

We attack this with a non-intuitive, but cool technique. Introduce the raising $a^\dagger$ and lowering $a$ operators:

\begin{aligned}a &= \sqrt{\frac{m \omega}{2 \hbar}} \left( X + i \frac{P}{m\omega} \right) \\ a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} \left( X - i \frac{P}{m\omega} \right)\end{aligned} \hspace{\stretch{1}}(8.171)

\paragraph{Question:} are we using the dagger for more than Hermitian conjugation in this case.
\paragraph{Answer:} No, this is precisely the Hermitian conjugation operation.

Solving for $X$ and $P$ in terms of $a$ and $a^\dagger$, we have

\begin{aligned}a + a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} 2 X \\ a - a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} 2 i \frac{P }{m \omega}\end{aligned}

or

\begin{aligned}X &= \sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) \\ P &= i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)\end{aligned} \hspace{\stretch{1}}(8.173)

Express $H$ in terms of $a$ and $a^\dagger$

\begin{aligned}H &= \frac{P^2}{2m} + \frac{1}{{2}} K X^2 \\ &= \frac{1}{2m} \left(i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)\right)^2+ \frac{1}{{2}} m \omega^2\left(\sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) \right)^2 \\ &= \frac{-\hbar \omega}{4} \left(a^\dagger a^\dagger + a^2 - a a^\dagger - a^\dagger a\right)+ \frac{\hbar \omega}{4}\left(a^\dagger a^\dagger + a^2 + a a^\dagger + a^\dagger a\right) \\ \end{aligned}

\begin{aligned}H= \frac{\hbar \omega}{2} \left(a a^\dagger + a^\dagger a\right) = \frac{\hbar \omega}{2} \left(2 a^\dagger a + \left[{a},{a^\dagger}\right]\right) \end{aligned} \hspace{\stretch{1}}(8.175)

Since $\left[{X},{P}\right] = i \hbar \mathbf{1}$ then we can show that $\left[{a},{a^\dagger}\right] = \mathbf{1}$. Solve for $\left[{a},{a^\dagger}\right]$ as follows

\begin{aligned}i \hbar &=\left[{X},{P}\right] \\ &=\left[{\sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) },{i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)}\right] \\ &=\sqrt{\frac{\hbar}{2 m \omega}} i \sqrt{\frac{\hbar m \omega}{2}} \left[{a^\dagger + a},{a^\dagger -a}\right] \\ &= \frac{i \hbar}{2}\left(\left[{a^\dagger},{a^\dagger}\right] -\left[{a^\dagger},{a}\right] +\left[{a},{a^\dagger}\right] -\left[{a},{a}\right] \right) \\ &= \frac{i \hbar}{2}\left(0+2 \left[{a},{a^\dagger}\right] -0\right)\end{aligned}

Comparing LHS and RHS we have as stated

\begin{aligned}\left[{a},{a^\dagger}\right] = \mathbf{1}\end{aligned} \hspace{\stretch{1}}(8.176)

and thus from 8.175 we have

\begin{aligned}H = \hbar \omega \left( a^\dagger a + \frac{\mathbf{1}}{2} \right)\end{aligned} \hspace{\stretch{1}}(8.177)

Let ${\lvert {n} \rangle}$ be the eigenstate of $H$ so that $H{\lvert {n} \rangle} = E_n {\lvert {n} \rangle}$. From 8.177 we have

\begin{aligned}H {\lvert {n} \rangle} =\hbar \omega \left( a^\dagger a + \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.178)

or

\begin{aligned}a^\dagger a {\lvert {n} \rangle} + \frac{{\lvert {n} \rangle}}{2} = \frac{E_n}{\hbar \omega} {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.179)

\begin{aligned}a^\dagger a {\lvert {n} \rangle} = \left( \frac{E_n}{\hbar \omega} - \frac{1}{{2}} \right) {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.180)

We wish now to find the eigenstates of the “Number” operator $a^\dagger a$, which are simultaneously eigenstates of the Hamiltonian operator.

Observe that we have

\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) &= a^\dagger ( a a^\dagger {\lvert {n} \rangle} ) \\ &= a^\dagger ( \mathbf{1} + a^\dagger a ) {\lvert {n} \rangle}\end{aligned}

where we used $\left[{a},{a^\dagger}\right] = a a^\dagger - a^\dagger a = \mathbf{1}$.

\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) &= a^\dagger \left( \mathbf{1} + \frac{E_n}{\hbar\omega} - \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle} \\ &= a^\dagger \left( \frac{E_n}{\hbar\omega} + \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle},\end{aligned}

or

\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) = (\lambda_n + 1) (a^\dagger {\lvert {n} \rangle} )\end{aligned} \hspace{\stretch{1}}(8.181)

The new state $a^\dagger {\lvert {n} \rangle}$ is presumed to lie in the same space, expressible as a linear combination of the basis states in this space. We can see the effect of the operator $a a^\dagger$ on this new state, we find that the energy is changed, but the state is otherwise unchanged. Any state $a^\dagger {\lvert {n} \rangle}$ is an eigenstate of $a^\dagger a$, and therefore also an eigenstate of the Hamiltonian.

Play the same game and win big by discovering that

\begin{aligned}a^\dagger a ( a {\lvert {n} \rangle} ) = (\lambda_n -1) (a {\lvert {n} \rangle} )\end{aligned} \hspace{\stretch{1}}(8.182)

There will be some state ${\lvert {0} \rangle}$ such that

\begin{aligned}a {\lvert {0} \rangle} = 0 {\lvert {0} \rangle}\end{aligned} \hspace{\stretch{1}}(8.183)

which implies

\begin{aligned}a^\dagger (a {\lvert {0} \rangle}) = (a^\dagger a) {\lvert {0} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(8.184)

so from 8.180 we have

\begin{aligned}\lambda_0 = 0\end{aligned} \hspace{\stretch{1}}(8.185)

Observe that we can identify $\lambda_n = n$ for

\begin{aligned}\lambda_n = \left( \frac{E_n}{\hbar\omega} - \frac{1}{{2}} \right) = n,\end{aligned} \hspace{\stretch{1}}(8.186)

or

\begin{aligned}\frac{E_n}{\hbar\omega} = n + \frac{1}{{2}}\end{aligned} \hspace{\stretch{1}}(8.187)

or

\begin{aligned}E_n = \hbar \omega \left( n + \frac{1}{{2}} \right)\end{aligned} \hspace{\stretch{1}}(8.188)

where $n = 0, 1, 2, \cdots$.

We can write

\begin{aligned}\hbar \omega \left( a^\dagger a + \frac{1}{{2}} \mathbf{1} \right) {\lvert {n} \rangle} &= E_n {\lvert {n} \rangle} \\ a^\dagger a {\lvert {n} \rangle} + \frac{1}{{2}} {\lvert {n} \rangle} &= \frac{E_n}{\hbar \omega} {\lvert {n} \rangle} \\ \end{aligned}

or

\begin{aligned}a^\dagger a {\lvert {n} \rangle} = \left( \frac{E_n}{\hbar \omega} - \frac{1}{{2}} \right) {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle} = n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.189)

We call this operator $a^\dagger a = N$, the number operator, so that

\begin{aligned}N {\lvert {n} \rangle} = n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.190)

# Relating states.

Recall the calculation we performed for

\begin{aligned}L_{+} {\lvert {lm} \rangle} &= C_{+} {\lvert {l, m+1} \rangle} \\ L_{-} {\lvert {lm} \rangle} &= C_{+} {\lvert {l, m-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.191)

Where $C_{+}$, and $C_{+}$ are constants. The next game we are going to play is to work out $C_n$ for the lowering operation

\begin{aligned}a{\lvert {n} \rangle} = C_n {\lvert {n-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.193)

and the raising operation

\begin{aligned}a^\dagger {\lvert {n} \rangle} = B_n {\lvert {n+1} \rangle}.\end{aligned} \hspace{\stretch{1}}(9.194)

For the Hermitian conjugate of $a {\lvert {n} \rangle}$ we have

\begin{aligned}(a {\lvert {n} \rangle})^\dagger = ( C_n {\lvert {n-1} \rangle} )^\dagger = C_n^{*} {\lvert {n-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.195)

So

\begin{aligned}({\langle {n} \rvert} a^\dagger) (a {\lvert {n} \rangle}) = C_n C_n^{*} \left\langle{{n-1}} \vert {{n-1}}\right\rangle = {\left\lvert{C_n}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(9.196)

Expanding the LHS we have

\begin{aligned}{\left\lvert{C_n}\right\rvert}^2 &={\langle {n} \rvert} a^\dagger a {\lvert {n} \rangle} \\ &={\langle {n} \rvert} n {\lvert {n} \rangle} \\ &=n \left\langle{{n}} \vert {{n}}\right\rangle \\ &=n \end{aligned}

For

\begin{aligned}C_n = \sqrt{n}\end{aligned} \hspace{\stretch{1}}(9.197)

Similarly

\begin{aligned}({\langle {n} \rvert} a^\dagger) (a {\lvert {n} \rangle}) = B_n B_n^{*} \left\langle{{n+1}} \vert {{n+1}}\right\rangle = {\left\lvert{B_n}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(9.198)

and

\begin{aligned}{\left\lvert{B_n}\right\rvert}^2 &={\langle {n} \rvert} \underbrace{a a^\dagger}_{a a^\dagger - a^\dagger a = \mathbf{1}} {\lvert {n} \rangle} \\ &={\langle {n} \rvert} \left( \mathbf{1} + a^\dagger a \right) {\lvert {n} \rangle} \\ &=(1 + n) \left\langle{{n}} \vert {{n}}\right\rangle \\ &=1 + n \end{aligned}

for

\begin{aligned}B_n = \sqrt{n + 1}\end{aligned} \hspace{\stretch{1}}(9.199)

# Heisenberg picture.

\paragraph{How does the lowering operator $a$ evolve in time?}

\paragraph{A:} Recall that for a general operator $A$, we have for the time evolution of that operator

\begin{aligned}i \hbar \frac{d A}{dt} = \left[{ A },{H}\right]\end{aligned} \hspace{\stretch{1}}(10.200)

Let’s solve this one.

\begin{aligned}i \hbar \frac{d a}{dt} &= \left[{ a },{H}\right] \\ &= \left[{ a },{ \hbar \omega (a^\dagger a + \mathbf{1}/2) }\right] \\ &= \hbar\omega \left[{ a },{ (a^\dagger a + \mathbf{1}/2) }\right] \\ &= \hbar\omega \left[{ a },{ a^\dagger a }\right] \\ &= \hbar\omega \left( a a^\dagger a - a^\dagger a a \right) \\ &= \hbar\omega \left( (a a^\dagger) a - a^\dagger a a \right) \\ &= \hbar\omega \left( (a^\dagger a + \mathbf{1}) a - a^\dagger a a \right) \\ &= \hbar\omega a \end{aligned}

Even though $a$ is an operator, it can undergo a time evolution and we can think of it as a function, and we can solve for $a$ in the differential equation

\begin{aligned}\frac{d a}{dt} = -i \omega a \end{aligned} \hspace{\stretch{1}}(10.201)

This has the solution

\begin{aligned}a = a(0) e^{-i \omega t}\end{aligned} \hspace{\stretch{1}}(10.202)

here $a(0)$ is an operator, the value of that operator at $t = 0$. The exponential here is just a scalar (not effected by the operator so we can put it on either side of the operator as desired).

\paragraph{CHECK:}

\begin{aligned}a' = a(0) \frac{d}{dt} e^{-i \omega t} = a(0) (-i \omega) e^{-i \omega t} = -i \omega a\end{aligned} \hspace{\stretch{1}}(10.203)

# A couple comments on the Schr\”{o}dinger picture.

We don’t do this in class, but it is very similar to the approach of the hydrogen atom. See the text for full details.

In the Schr\”{o}dinger picture,

\begin{aligned}-\frac{\hbar^2}{2m} \frac{d^2 u}{dx^2} + \frac{1}{{2}} m \omega^2 x^2 u = E u\end{aligned} \hspace{\stretch{1}}(11.204)

This does directly to the wave function representation, but we can relate these by noting that we get this as a consequence of the identification $u = u(x) = \left\langle{{x}} \vert {{u}}\right\rangle$.

In 11.204, we can switch to dimensionless quantities with

\begin{aligned}\xi = \text{xi (z)''} = \alpha x\end{aligned} \hspace{\stretch{1}}(11.205)

with

\begin{aligned}\alpha = \sqrt{\frac{m \omega}{\hbar}}\end{aligned} \hspace{\stretch{1}}(11.206)

This gives, with $\lambda = 2E/\hbar\omega$,

\begin{aligned}\frac{d^2 u}{d\xi^2} + (\lambda - \xi^2) u = 0\end{aligned} \hspace{\stretch{1}}(11.207)

We can use polynomial series expansion methods to solve this, and find that we require a terminating expression, and write this in terms of the Hermite polynomials (courtesy of the clever French once again).

When all is said and done we will get the energy eigenvalues once again

\begin{aligned}E = E_n = \hbar \omega \left( n + \frac{1}{{2}} \right)\end{aligned} \hspace{\stretch{1}}(11.208)

# Back to the Heisenberg picture.

Let us express

\begin{aligned}\left\langle{{x}} \vert {{n}}\right\rangle = u_n(x)\end{aligned} \hspace{\stretch{1}}(12.209)

With

\begin{aligned}a {\lvert {0} \rangle} = 0,\end{aligned} \hspace{\stretch{1}}(12.210)

we have

\begin{aligned}0 =\left( X + i \frac{P}{m \omega} \right) {\lvert {0} \rangle},\end{aligned} \hspace{\stretch{1}}(12.211)

and

\begin{aligned}0 &= {\langle {x} \rvert} \left( X + i \frac{P}{m \omega} \right) {\lvert {0} \rangle} \\ &= {\langle {x} \rvert} X {\lvert {0 } \rangle} + i \frac{1}{m \omega} {\langle {x} \rvert} P {\lvert {0} \rangle} \\ &= x \left\langle{{x}} \vert {{0}}\right\rangle + i \frac{1}{m \omega} {\langle {x} \rvert} P {\lvert {0} \rangle} \\ \end{aligned}

Recall that our matrix operator is

\begin{aligned}{\langle {x'} \rvert} P {\lvert {x} \rangle} = \delta(x - x') \left( -i \hbar \frac{d}{dx} \right)\end{aligned} \hspace{\stretch{1}}(12.212)

\begin{aligned}{\langle {x} \rvert} P {\lvert {0} \rangle} &={\langle {x} \rvert} P \underbrace{\int {\lvert {x'} \rangle} {\langle {x'} \rvert} dx' }_{= \mathbf{1}}{\lvert {0} \rangle} \\ &=\int {\langle {x} \rvert} P {\lvert {x'} \rangle} \left\langle{{x'}} \vert {{0}}\right\rangle dx' \\ &=\int \delta(x - x') \left( -i \hbar \frac{d}{dx} \right)\left\langle{{x'}} \vert {{0}}\right\rangle dx' \\ &=\left( -i \hbar \frac{d}{dx} \right)\left\langle{{x}} \vert {{0}}\right\rangle\end{aligned}

We have then

\begin{aligned}0 =x u_0(x) + \frac{\hbar}{m \omega} \frac{d u_0(x)}{dx}\end{aligned} \hspace{\stretch{1}}(12.213)

NOTE: picture of the solution to this LDE on slide…. but I didn’t look closely enough.