week 2

indentical particle states (with spin!)

Q1: What is the hilbert space of a single particle with spin $s$?
A1: A single particle lives in the space $F \otimes \mathbb{C}^{2s +1}$, where:

$F$ is the space of functions $f: \R^3 \to \C$, with an inner product $\ip{\phi}{\psi}$ given by the familiar $$ \ip{\phi}{\psi} = \int dx \int dy \int dz \phi^*(x,y,z)\psi(x,y,z) $$ and
$\C^{2s+1}$ is the space of $(2s+1)$-tuples of complex numbers $\bm{c} = (c_1,c_2,\ldots,c_{2s+1})$, with the natural inner product $\ip{\bm{c'}}{\bm{c}}$ $$ \ip{\bm{c'}}{\bm{c}} = \sum_j^{2s+1} c'^*_j c_j $$

(we learned last week how a tensor product space $V_A \otimes V_B$ is formed from two hilbert spaces $V_A$ and $V_B$)

Q2:
a) What is the hilbert space of two distinguishable particles, each with spin $s = \frac{1}{2}$?
b) For two identical fermions with spin $s = \frac{1}{2}$?
c) For two identical bosons spin $s = \frac{1}{2}$? Why is this an oxymoron?
A2:
a) From last week last week we learned that if given two distinguishable particles, one living in a hilbert space $V_A$ and the other in $V_B$, then their space of joint states is $V_A \otimes V_B$. In our case each particle lives in the space $F \otimes \C^2$, so their joint space is $\S$.
b) The fermions we restrict ourselves to the antisymmetric subspace of $\S$. This is the subset of tensors $t \in \S$ satisfying the condition $Pt = -t$, where $P$ is the permutation operator, defined for any tensor product space $V \otimes V$, which acts on tensor products $v \otimes w$, $v, w \in V$ according to the rule: $$ P( v \otimes w ) = w \otimes v $$ ¹. Check for yourself that tensors satisfying this condition indeed constitute a linear subspace of $\S$.
c) Here we require $Pt = +t$. It is a fact of nature that bosons have integer spin, so it is somewhat meaningless to speak of a spin $\frac{1}{2}$ boson.

Now and in general we will find it useful to rearrange our tensor product space from $\S$ to $\T$. We go from one a tensor $\S$ to its equivalent tensor on $\T$ using the map $\eta$ which uses the simple rule: $$ \eta\left( \left( f_1 \otimes \chi_1 \right) \otimes \left( f_2 \otimes \chi_2 \right) \right) = \left( f_1 \otimes f_2 \right) \otimes \left( \chi_1 \otimes \chi_2 \right) $$ Q3: Show that the condition for bosons / fermions $Pt = \pm t$ on $\S$ can be translated to a equivalent condition on $\T$: $$ P' t' = \pm t' $$ where $t' \in \T$ and

$P' \equiv P_F \otimes P_{\C^2}$, where
$P_F$ and $P_{\C^2}$ are the permutation operators on $F \otimes F$ and $\C^2 \otimes \C^2$, respectively, and
we've hijacked the $\otimes$ notation (again!), this time to define a tensor product of operators, which takes an operator $A$ that acts on $V_A$ and operator $B$ that acts on $V_B$ and returns an operator $A \otimes B$ on $V_A \otimes V_B$ that acts on tensor products $a \otimes b$, $a \in V_A$, $b \in V_B$, according to the rule: $$ \left( A \otimes B \right) a \otimes b := Aa \otimes Bb $$

A3: We need to show that a tensor $t \in \S$ is bosonic / fermionic if and only if $\eta(t)$ is bosonic / fermionic on $\T$. To do this we first show that $P' = \eta \circ P \circ \eta^{-1}$, where $\circ$ takes two function $f$ and $g$ and creates another function $g \circ f$ according to the rule "apply $f$ then apply $g$". Since $P'$ and $\eta \circ P \circ \eta^{-1}$ are both linear operators (prove this to yourself sometime if it is not clear why), it suffices to prove their equality of their action on a general tensor product $\left(f_1 \otimes f_2\right) \otimes (\chi_1 \otimes \chi_2)$ ². To this end: \begin{align} & \left( \eta \circ P \circ \eta^{-1} \right) \left( \left(f_1 \otimes f_2\right) \otimes \left(\chi_1 \otimes \chi_2\right) \right) \\ = & \left( \eta \circ P \right) \left( \left(f_1 \otimes \chi_1\right) \otimes \left(f_2 \otimes \chi_2\right) \right) \\ = & \eta \left( \left(f_2 \otimes \chi_2\right) \otimes \left(f_1 \otimes \chi_1\right) \right) \\ = & \left(f_2 \otimes f_1\right) \otimes \left(\chi_2 \otimes \chi_1\right) \\ = & P_F\left(f_2 \otimes f_1\right) \otimes P_{\C^2} \left(\chi_2 \otimes \chi_1\right) \\ = & P' \left( \left(f_1 \otimes f_2\right) \otimes \left(\chi_1 \otimes \chi_2\right) \right) \end{align} Having established this we simply note that if $P t = \pm t$ then $$ P'\left( \eta(t) \right) = \left( P' \circ \eta \right) (t) = \left( \eta \circ P \circ \eta^{-1} \circ \eta \right) (t) = \left( \eta \circ P \right)(t) = \eta ( \pm t ) = \pm \eta(t) $$ the proof of the converse (i.e. that $P' t' = \pm t'$ implies $P(\eta^{-1}(t'))=\pm \eta^{-1}(t')$) works in exactly the same way.

Q4: From a permutation operator $P$ we can generate projection operators³ $\Pi_{\pm} := \frac{1}{2} \left( 1 \pm P \right)$. Check the action of $\Pi_{+(-)}$ on any tensor will product a(n) (anti)symmetric tensor⁴.

Q5: Diagonalize $\C^2 \otimes \C^2$ with respect to $P_{\C^2}$.
A5: Define an orthonormal basis on $\C^2$ consisting of $\Up \equiv (1,0) \in \C^2$ and $\Down \equiv (0,1) \in \C^2$. We can construct a basis for $\C^2 \otimes \C^2$ from all possible pairs $\Up \otimes \Up, \Up \otimes \Down, \Down \otimes \Up, \Down \otimes \Down$. We act on these basis vectors with the projection operators just defined. The action of $\Pi_+$ on the basis generates three orthogonal vectors:

$\Pi_+ \Up \otimes \Up = \Up \otimes \Up \equiv \ket{11}$
$\Pi_+ \Down \otimes \Down = \Down \otimes \Down \equiv \ket{1-1}$
$\Pi_+ \Up \otimes \Down = \Pi_+ \Down \otimes \Up = \frac{1}{2} \left( \Up \otimes \Down + \Down \otimes \Up \right) \equiv \frac{1}{\sqrt{2}}\ket{10}$

The action of $\Pi_-$ on the basis generates the remaining vector, which is required to be orthogonal to other three (why?)

$\Pi_- \Up \otimes \Up = \Pi_- \Down \otimes \Down = 0$
$\Pi_- \Up \otimes \Down = - \Pi_- \Down \otimes \Up = \frac{1}{2} \left( \Up \otimes \Down - \Down \otimes \Up \right) = \frac{1}{\sqrt{2}}\ket{00}$

On the far right we've written the states in terms of the familiar spin triplet and singlet states, i.e. states of definite total spin angular momentum quantum number $S$ (and z-projection $S_z$, for that matter). It is not a coincidence that states of definite permutation symmetry can also be written as states of definite total spin angular momentum. We'll prove in a bit that this can be done for any collection of mutally commuting operators.

It turns out, however, that the connection between permutation symmetry and spin angular momentum in particular goes even deeper: the permutation symmetry of a spin state actually determines its total spin, and vice versa! This is true for any number of spin $\frac{1}{2}$ particles. This result is much more difficult to prove. It is a special case of a more general result known as the Schur-Weyl duality.

Last week we showed that for each tensor in $F \otimes F$ there is an equivalent vector in what I called $\tilde{F}$, the space of complex functions $\Psi$ that act on pairs of points $\bm{x_1},\bm{x_2} \in \R^3$, i.e. $\Psi(\bm{x_1},\bm{x_2}) \in \C$. The linear map $\tilde{\eta}: F \otimes F \to \tilde{F}$ relating equivalent vectors is defined by the rule: $$ \tilde{\eta}(\psi \otimes \phi)(\bm{x_1},\bm{x_2}) := \psi(\bm{x_1}) \phi(\bm{x_2}) $$ Q6: Show that the permutation operator $P_F$ acting on $F \otimes F$ is equivalent to the operator $P_{\tilde{F}}$ acting on $\tilde{F}$ which acts according to the rule: $$ \left(P_{\tilde{F}}\Psi\right)(\bm{x_1},\bm{x_2}):=\Psi(\bm{x_2},\bm{x_1}) $$ (i.e. show that $P_{\tilde{F}} = \tilde{\eta} \circ P \circ \tilde{\eta}^{-1}$)

Note that the hamiltonian $H$ for the helium atom from equation 5.38 in Griffiths is an operator on the space $\tilde{F}$.

Q7: Show that $P_{\tilde{F}}$ commutes with

a) any potential energy operator $\hat{V}$ that has the property $\hat{V}(\bm{x_1},\bm{x_2}) = \hat{V}(\bm{x_2},\bm{x_1})$.
b) the kinetic energy
c) the hamiltonian for the helium atom

A7:

a) \begin{align} & \left( P_{\tilde{F}} \hat{V} \Psi \right) (\bm{x_1},\bm{x_2}) \\ = & \left( \hat{V} \Psi\right) (\bm{x_2},\bm{x_1}) \\ = & V(\bm{x_2},\bm{x_1}) \Psi(\bm{x_2},\bm{x_1}) \\ = & V(\bm{x_1},\bm{x_2}) \Psi(\bm{x_2},\bm{x_1}) \\ = & V(\bm{x_1},\bm{x_2}) \left( P_{\tilde{F}} \Psi \right)(\bm{x_1},\bm{x_2}) \\ = & \left( \hat{V} P_{\tilde{F}} \Psi \right)(\bm{x_1},\bm{x_2}) \end{align}
b) We start by showing \begin{align} & \left(P_{\tilde{F}} \frac{\partial}{\partial x_1} \Psi \right)(\bm{x_1},\bm{x_2}) \\ = & \left( \frac{\partial}{\partial x_1} \Psi \right)(\bm{x_2},\bm{x_1}) \\ = & \lim_{dx \to 0} \frac{ \Psi\left(\left( x_2 + dx,y_2,z_2 \right),\left(x_1,y_1,z_1\right)\right) - \Psi\left(\left( x_2,y_2,z_2 \right),\left(x_1,y_1,z_1\right)\right) }{dx} \\ = & \lim_{dx \to 0} \frac{ \left( P_{\tilde{F}} \Psi \right)\left(\left( x_1,y_1,z_1 \right),\left(x_2 + dx,y_2,z_2\right)\right) - \left( P_{\tilde{F}} \Psi \right)\left(\left( x_1,y_1,z_1 \right),\left(x_2,y_2,z_2\right)\right) }{dx} \\ = & \left(\frac{\partial}{\partial x_2}P_{\tilde{F}} \Psi \right)(\bm{x_1},\bm{x_2}) \end{align} i.e. $P_{\tilde{F}}\frac{\partial}{\partial x_1}=\frac{\partial}{\partial x_2} P_{\tilde{F}}$. The same holds for $y$ and $z$, and vice versa for $1 \leftrightarrow 2$. Using these relations it is simple to show that $P_{\tilde{F}}$ commutes with $\nabla^2_1 + \nabla^2_2$ and thus the kinetic energy.
c) We just need to check that the potential energy satisfies the condition $V(\bm{x_1},\bm{x_2})=V(\bm{x_2},\bm{x_1})$.

Q8: Let $H$ be a hamiltonian and $A$ some operator, both acting on some general hilbert space $V$. Let $[A,H]=0$. Check that:
a) $A$ acting on a state of definite energy produces another state of the same energy.
b) $H$ acting on an eigenstate of $A$ produces another eigenstate of $A$ with the same eigenvalue.
A8:
a) This manipulation underlies all the relationships between symmetry and the structure of the energy eigenstates. Let $v \in V$ be an energy eigenstate with energy $E$, i.e. $Hv = Ev$. Then $$ H(Av) = (HA)v = (AH)v = A(Hv) = A(Ev) = E(Av) $$ b) This proof is identical to a). The result tells us that if we want to interpret an operator $A$ (e.g. a hamiltonian) defined on a space of distinguishable particles $V \otimes V$ as an operator on the bosonic (symmetric) or fermionic (antisymmetric) subspaces of $V \otimes V$, then the property $[A,P]=0$ is required. Otherwise there is no gaurantee that $H$ acting on a (anti)symmetric will produce another (anti)symmetric state, which is required if $H$ is to satisfy the definition of being an operator on these subspaces.

Q9 (difficult!): Show that two hermitean operators ⁵ $A$ and $B$ commute if and only if they can be simultaneously diagonalized, i.e. there exists an orthonormal basis where each basis vector is an eigenvector of both $A$ and $B$.
A9: The easy part is proving that simultaneous diagonalizability implies commutivity. Suppose there is an orthonormal basis $\{\ket{e_n}\}$ simultaneously diagonalizing $A$ and $B$, so that $A\ket{e_n} = a_n\ket{e_n}$ and $B\ket{e_n} = b_n \ket{e_n}$. Then \begin{align} AB = & AB \sum_n \ket{e_n}\bra{e_n} = \sum_n AB \ket{e_n}\bra{e_n} \\ = & \sum_n A b_n \ket{e_n}\bra{e_n} = \sum_n a_n b_n \ket{e_n}\bra{e_n} \\ = & \sum_n b_n a_n \ket{e_n}\bra{e_n} = \sum_n B a_n \ket{e_n}\bra{e_n} = \\ = & \sum_n B A \ket{e_n}\bra{e_n} = BA \sum_n \ket{e_n}\bra{e_n} = BA \end{align} To prove that commutivity implies simultaneous diagonalizability, we first construct an orthonormal basis $\{\ket{a_{ij}}\}$ that diagonalizes $A$ (i.e. an eigenbasis of $A$), so that $A\ket{a_{ij}} = a_i \ket{a_{ij}}$. In other words $\ket{a_{ij}}$ is the $j^{\text{th}}$ eigenvector with eigenvalue $a_i$. We showed in the previous problem that the action of $B$ on any $\ket{a_{ij}}$ will produce another eigenvector of $A$ with eigenvalue $a_i$. This means for any $i$ we can define a hermitean operator $B_i$ by restricting $B$ to act only on the subspace spanned by $\{\ket{a_{ij}}\}$ with $i$ fixed, i.e. $B_i: \text{span}(\{\ket{a_{i1}},\ldots,\ket{a_{ij}},\ldots\}) \to \text{span}(\{\ket{a_{i1}},\ldots,\ket{a_{ij}},\ldots\})$. Since $B_i$ is hermitean it possesses an eigenbasis $\{\ket{b_{ij}}\}$, i.e. $B_i \ket{b_{ij}} = b_{ij} \ket{b_{ij}}$. By combining the $\{ \ket{b_{ij}} \}$ for every $i$, we obtain an orthonormal basis which simultaneously diagonalizes $A$ and $B$.

To see that the $\ket{b_{ij}}$ diagonalize $B$, we just note that $$ B\ket{b_{ij}} = B_i \ket{b_{ij}} = b_{ij} \ket{b_{ij}} $$ To see that the $\ket{b_{ij}}$ diagonalize $A$, we note that the $\ket{b_{ij}} \in \text{span}(\{\ket{a_{i1}},\ldots,\ket{a_{ij}},\ldots\})$ and so are linear combinations of eigenvectors of $A$ all with eigenvalues $a_i$. We showed last week that a linear combination of eigenvectors with the same eigenvalue is also an eigenvector with the same eigenvalue.

So we see that we can construct a basis $\Psi_{n}(\bm{x_1},\bm{x_2})$ of $\tilde{F}$ whose basis vectors are simultaneously eigenstates of $H$ and $P_{\tilde{F}}$.

Q10:
a) Show that the basis $\tilde{\eta}^{-1}(\Phi_n)\otimes\ket{S M_S}$ diagonalizes the hamiltonian $H'$ which operates on $\T$ according to the rule: $$ H'(t_F \otimes t_{\C^2}):= H_F t_{F} \otimes t_{\C^2} $$ Where

$t_F \in F \otimes F$ and $t_{\C^2} \in \C^2 \otimes \C^2$ and
$H_F$ operates on $t \in F \otimes F$ using the rule $H_F(t) := \tilde{\eta}^{-1} \circ H \circ \tilde{\eta}$.

b) Describe the energy levels.
c) Show that the basis from part a) also diagonalizes the permutation operator (which in this case would be the $P'$ operator defined earlier).
d) What are the bosonic and fermionic energy eigenstates?
A10:
a) \begin{align} & H'\tilde{\eta}^{-1}(\Phi_n)\otimes\ket{S M_S} \\ = & \left(H_F\tilde{\eta}^{-1}(\Phi_n)\right) \otimes \ket{S M_S} \\ = & \left( \left( \tilde{\eta}^{-1} \circ H \circ \tilde{\eta} \circ \tilde{\eta}^{-1} \right)\Phi_n \right) \otimes \ket{S M_S} \\ = & \left( \tilde{\eta}^{-1}\left( H \Phi_n \right) \right) \otimes \ket{S M_S} \\ = & \left( \tilde{\eta}^{-1}\left( E_n \Phi_n \right) \right) \otimes \ket{S M_S} \\ = & E_n \tilde{\eta}^{-1}(\Phi_n) \otimes \ket{S M_S} \end{align} b) Supposing all the $E_n$ are different (why might this be a bad assumption?), we find that the energy level with energy $E_n$ is spanned by the space $\{\Phi_n\} \otimes \left( \C^2 \otimes \C^2 \right)$. In other words we can tack onto $\Psi_n$ an arbitary tensor from $\C^2 \otimes \C^2$ to construct an energy eigenstate of $H'$ with energy $E_n$.
c) \begin{align} & P' \tilde{\eta}^{-1}(\Phi_n)\otimes\ket{S M_S} \\ = & \left( P_F \tilde{\eta}^{-1} \left(\Phi_n \right) \right) \otimes \left( P_{\C^2} \ket{S M_S} \right) \\ = & \left( \left( P_F \circ \tilde{\eta}^{-1} \right)\Phi_n \right) \otimes \left( P_{\C^2} \ket{S M_S} \right) \\ = & \left( \left( \tilde{\eta}^{-1} \circ P_{\tilde{F}} \circ \tilde{\eta} \circ \tilde{\eta}^{-1} \right) \Phi_n \right) \otimes \left( P_{\C^2} \ket{S M_S} \right) \\ = & \left( \tilde{\eta}^{-1} \left( P_{\tilde{F}} \Phi_n \right) \right) \otimes \left( P_{\C^2} \ket{S M_S} \right) \\ = & \left( \tilde{\eta}^{-1} \left( \sigma_{\tilde{F}}(n) \Phi_n \right) \right) \otimes \left( P_{\C^2} \ket{S M_S} \right) \\ = & \sigma_{\tilde{F}}(n) \tilde{\eta}^{-1} \left( \Phi_n \right) \otimes \left( P_{\C^2} \ket{S M_S} \right) \\ = & \sigma_{\tilde{F}}(n) \tilde{\eta}^{-1} \left( \Phi_n \right) \otimes \left( \sigma_{\C^2}(S) \ket{S M_S} \right) \\ = & \sigma_{\tilde{F}}(n) \sigma_{\C^2}(S) \tilde{\eta}^{-1} \left( \Phi_n \right) \otimes \ket{S M_S} \end{align} where

$\sigma_{\tilde{F}}(n)$ is $\pm1$ depending on the permutation symmetry of $\Phi_n$, and
$\sigma_{\C^2}(S)$ is $\pm1$ depending on the permutation symmetry of spin state $\ket{S M_S}$, i.e. +1 for $S=1$ and -1 for $S=0$

d) The bosonic (fermionic) energy eigenstates are obtained by taking the symmetric $\Phi_n$ with the $S=1$ ($S=0$) spin states and vice versa.

footnotes

¹ Note how we've stopped thinking of elements (i.e. tensors) of tensor product spaces $V_A \otimes V_B$ as "the set of antilinear functions that take a vector $a \in V_A$ and a vector $b \in V_B$ and return a complex number", and think of them as linear combinations of tensor products $a \otimes b$. This is fine and in fact much closer to how physicists typically use tensors, but you will want to keep in mind the properties tensor products possess, i.e.

For $a,a' \in V_A$, $b \in V_B$, and $c \in \C$, $(a + ca') \otimes b = a \otimes b + c a' \otimes b$, and vice versa.
$ \ip{c \otimes d}{a \otimes b} = \ip{c}{a} \ip{d}{b} $

All this won't tell you what $a \otimes b$ is. For that you have to go back to the antilinear mumbo jumbo.

² If it is not clear why both maps are linear operators or why it this implies that it sufficies to prove their equality for a general tensor product, try proving it yourself.

³ The defining property of projection operators is idempotence, or $P^2 = P$. Check that our operators satisfy this property.

⁴ Notice that these operators are just the special case of the general Slater permanent / determinant.

⁵ It is only important that each operator separately possess an orthonormal basis of eigenvectors. This is true not only of hermitean operators but also unitary operators and in general any operator $A$ satisfying the condition $AA^{\dagger}=A^{\dagger}A$. Such operators are called normal operators.