week 1
multi particle qm crash course
Problem 5.5a asks us to write down (i.e. define) the hamiltonian for two noninteracting identical particles.
Let's dive a little bit into the mathematical details implicit in this question.
The effort will pay off later when spin is added into the mix.
Q: What is a hamiltonian mathematically?
A: A hamiltonian is a hermitian operator.
Q: Ok, but what is an operator?
A: An operator takes a vector $\ket{\psi}$ that lives in some hilbert space $V$ and returns another vector $\ket{\phi}$ that also lives in $V$.
So before we can begin to answer 5.5a, we must first define what vectors the hamiltonian acts on.
If particle $A$ lives in some hilbert space $V_A$ and another particle $B$ lives in some hilbert space $V_B$, then what is the space $V_{AB}$ that describes both particles collectively?
The answer is that the particles collectively live in a hilbert space $V_A \otimes V_B$ called the tensor product space of $V_A$ and $V_B$.
1
This term doesn't even appear in the index of Griffiths so I will define it for you here.
The tensor product space of two hilbert spaces $V_A$ and $V_B$ is the set of all antilinear functions $t$ which take a vector $a$ from $V_A$ and a vector $b$ from $V_B$ and return some complex number $c$, i.e.
$$a,b \to t(a,b) = c$$
By antilinear we mean that for any vector $a$ from $V_A$, any two vectors $b,b'$ from $V_B$, and any complex number $c$ we require that
$$t(a,b+cb') = t(a,b) + c^*t(a,b')$$
and vice versa, i.e.
$$t(a+ca',b) = t(a,b) + c^*t(a',b)$$
We can make $V_A \otimes V_B$ a vector space by defining addition and scalar multiplication in the following way:
$$(t+s)(a,b) := t(a,b) + s(a,b)$$
and
$$(ct)(a,b) := c(t(a,b))$$
2
, where $t$ and $s$ are elements of $V_A \otimes V_B$ and $c$ is a complex number. Dwell on the distinction between the left- and right-hand-sides of both of these definitions.
We can further make $V_A \otimes V_B$ a hilbert space by defining the inner product:
$$\ip{s}{t} := \sum_{i,j=1}^{N,M} s^*(e_i,f_j) t(e_i,f_j)$$
where $\{e_1,\ldots,e_i,\ldots,e_N\}$ and $\{f_1,\ldots,f_j,\ldots,f_M\}$ are orthonormal bases for $V_A$ and $V_B$, respectively.
3
Q: Let's get some practice working with vectors in $V_A \otimes V_B$. Show that the inner product just defined satisfies the necessary requirements:
-
$\ip{s}{t} = \ip{t}{s}^*$
-
$\ip{s}{t + ct'} = \ip{s}{t} + c\ip{s}{t'}$
-
$\ip{t}{t} > 0$ if $t \neq 0$ and $\ip{0}{0} = 0$.
Q: Show that the inner product is well-defined in the sense that it is independent of which pair of orthonormal bases you use.
(hint: remember that two orthonormal bases $\{e_1,\ldots,e_i,\ldots,e_N\}$ and $\{e'_1,\ldots,e'_j,\ldots,e'_N\}$ are related by a unitary matrix $u_{ij}$, i.e. $e_i = \sum_{j=1}^N u_{ij} e'_j$)
A:
\begin{align}& \sum_{ij} s^*(e_i,f_j) t(e_i,f_j) \\
= & \sum_{ij} s^*(\sum_k u_{ik} e'_k,\sum_l \tilde{u}_{jl} f'_l) t(\sum_m u_{im} e'_m,\sum_n \tilde{u}_{jn} f'_n) \\
= & \sum_{ijklmn} u^*_{im} u_{ik} \tilde{u}^*_{jn} \tilde{u}_{jl} s^*(e'_k,f'_l) t(e'_m,f'_n) \\
= & \sum_{ijklmn} u^{-1}_{mi} u_{ik} \tilde{u}^{-1}_{nj} \tilde{u}_{jl} s^*(e'_k,f'_l) t(e'_m,f'_n) \\
= & \sum_{klmn} \delta_{km} \delta_{ln} s^*(e'_k,f'_l) t(e'_m,f'_n) \\
= & \sum_{kl} s^*(e'_k,f'_l) t(e'_k,f'_l)\end{align}
We can also define an operation called the tensor product that takes a vector $a$ from $V_A$ and a vector $b$ from $V_B$ and returns a vector $a \otimes b$ that lives in $V_A \otimes V_B$ according to the following rule:
$$(a \otimes b)(c,d) := \ip{c}{a} \ip{d}{b}$$
Q: Verify that the tensor product is indeed antilinear, as required for membership in $V_A \otimes V_B$.
Q: Prove the following useful relationship:
$$(a + ca') \otimes b = a \otimes b + c (a' \otimes b)$$
where $a,a' \in V_A$, $b \in V_B$, and $c$ is a complex number. By the same token:
$$a \otimes (b + cb') = a \otimes b + c(a \otimes b')$$
Q: Show that
$$t(a,b) = \ip{a \otimes b}{t}$$
where $t \in V_A \otimes V_B$, $a \in V_A$, and $b \in V_B$.
Q:
a) Show that the full set of pairs $e_i \otimes f_j$, where $\{e_1,\ldots,e_i,\ldots,e_N\}$ and $\{f_1,\ldots,f_j,\ldots,f_M\}$ are respective orthonormal bases for $V_A$ and $V_B$, constitute an orthonormal basis for $V_A \otimes V_B$.
This implies that any vector in $V_A \otimes V_B$ can be written in the form $\sum_{ij} t_{ij} e_i \otimes f_j$, where the $t_{ij}$ are coefficients (i.e. complex numbers).
(The proof is somewhat difficult if you've never done one like it before. Feel free to jump straight to my answer, or just take it on faith.)
b) Use this result to determine the dimension of the space $V_A \otimes V_B$ in terms of $N$ and $M$, the respective dimensions of $V_A$ and $V_B$.
c) What are the coefficients $t'_{kl}$ of a vector $t \in V_A \otimes V_B$ in the basis $\{e'_k \otimes f'_l\}$ in terms of $t_{ij}$, its coefficients in a different basis $\{e_i \otimes f_j\}$? Assume that the $e_i,f_j,e'_k,f'_l$ are all orthonormal.
A:
a)
In general, to show that some set of vectors $\{e_1,\ldots,e_N\}$ consitute a basis of some vector space $V$ we must show that:
-
Any vector $v \in V$ can be written as a linear combination of the $\{e_1,\ldots,e_N\}$.
-
The $\{e_1,\ldots,e_N\}$ must be linearly independent so that the only solution to the equation $\sum_i^N c_i e_i = 0$ is the one where all the coefficients $c_i$ are zero.
Take some vector $t \in V_A \otimes V_B$. We generate some coordinates $t_{ij} = t(e_i,f_j)$. I claim that the vector $\tilde{t} \equiv \sum_{ij} t_{ij} e_i \otimes f_j$ is equal to $t$.
Remember that $t$ and $\tilde{t}$ are elements of $V_A \otimes V_B$, which is the set of (antilinear) functions which take a vector from $V_A$ and one from $V_B$ and return a complex number.
So to prove that they are equal we must show that they act the same on any pair of vectors $a$ and $b$, where $a \in V_A$ and $b \in V_B$.
To this end:
\begin{align}
& \tilde{t}(a,b) \\
= & \sum_{ij} t_{ij} e_i \otimes f_j (a,b) \\
= & \sum_{ij} t_{ij} \ip{a}{e_i} \ip{b}{f_j} \\
= & \sum_{ij} t_{ij} (\ip{e_i}{a} \ip{f_j}{b})^* \\
= & \sum_{ij} t_{ij} (a \otimes b (e_i,f_j))^* \\
= & \sum_{ij} t(e_i,f_j) (a \otimes b (e_i,f_j))^* \\
= & \ip{a \otimes b}{t} \\
= & t(a,b) \\
\end{align}
To show that the basis is linearly independent, we suppose there is some set of coefficients $c_{ij}$ where $\sum_{ij} c_{ij} e_i \otimes f_j = 0$. Plugging in $e_a$ and $f_b$ into the left hand side (which is a function rememmber?), we get:
$$
(\sum_{ij} c_{ij} e_i \otimes f_j)(e_a,e_b)
= \sum_{ij} c_{ij} e_i \otimes f_j(e_a,e_b)
= \sum_{ij} c_{ij} \ip{e_a}{e_i} \ip{f_b}{f_j}
= \sum_{ij} c_{ij} \delta_{ai} \delta_{bj}
= c_{ab}
$$
If we compare this with what we get when we feed $e_a$ and $f_b$ into the right hand side, which is the zero function, defined to return zero for any pair of arguments, we find that $c_{ab}=0$.
Since this argument applies for any pair $e_a$ and $f_b$, we conclude that all the $c_{ij}$ equal zero.
b)
The dimension of a space is the size of a basis. Here we have a basis
$$
\begin{bmatrix}
e_1 \otimes f_1 , & \cdots & , e_1 \otimes f_M, \\
\vdots & \ddots & \\
e_N \otimes f_1, & & e_N \otimes f_M
\end{bmatrix}
$$
which has a size $NM$.
c)
Since the $e'_k \otimes f'_l$ basis is orthonormal (since the $e'_k$ basis and $f'_l$ basis are both assumed orthonormal), we can write the coefficient $t'_{kl}$ as $\ip{e'_k \otimes f'_l}{t}$. Changing basis
$$e_i \leftarrow e'_k = \sum_i u_{ik} e_i$$
and
$$f_j \leftarrow f'_l = \sum_j \tilde{u}_{jl} f_j$$
for some (unitary) matrices $u_{ik}$ and $\tilde{u}_{jl}$ we get
$$t'_{kl} = \ip{e'_k \otimes f'_l}{t}=\ip{\sum_i u_{ik} e_i \otimes \sum_j \tilde{u}_{jl} f_j}{t} = \sum_{ij} u^*_{ik} \tilde{u}^*_{jl} \ip{e_i \otimes f_j}{t} = \sum_{ij} u^*_{ik} \tilde{u}^*_{jl} t_{ij}$$
Q:(bonus) Can any vector in $V_A \otimes V_B$ be written as a (single) tensor product?
A: No! You can check this for some simple cases or note that the space of vector pairs is $N+M$-dimensional, while $V_A \otimes V_B$ is $NM$-dimensional. The fact that the latter space can be larger than the former opens up the possibility of entanglement.
Q:
Describe the state space of two distinguishable particles in the same box (infinite square well).
A:
The particles individually live in the (hilbert) space of (square integrable) (wave) functions that vanish everywhere outside the region $(0,a)$, where $a$ is the width of the box. Let's call this space $F$. The state space is then the space $F \otimes F$ of antilinear functions $f$ that take a pair of functions $\phi,\psi \in F$ and return a complex number $c$, i.e.
$$
\phi, \psi \to f(\phi,\psi) = c
$$
That is the definition according to the principles we've so far discussed. In example 5.1, however, Griffiths presents multiparticle states as wave functions $\Psi(x_1,x_2)$, nonlinear, in general, that take in two numbers (not functions!) $x_1,y_1 \in (0,a)$, and return a complex number.
These objects do not live in $F \otimes F$, but instead in the space (let's call it $\tilde{F}$) of (square integrable) functions on coordinate pairs $(x_1,x_2)$, where $x_1,x_2 \in (0,a)$.
So we appear to have a contradiction!
In fact $\tilde{F}$ and $F \otimes F$ are in one-to-one linear correspondence, so we can think of them as equivalent.
The correspondence $\eta: \tilde{F} \to F \otimes F$ that allows us to convert from one to the other is the following:
$$
\eta(\Psi)(\phi,\psi) := \int_0^a dx_1 \int_0^a dx_2 \phi^*(x_1) \psi^*(x_2) \Psi(x_1,x_2)
$$
Q:
a) Verify that $\eta(\Psi)$ is indeed an element of $F \otimes F$.
b) What is the function $\eta^{-1}$, that takes us from $F \otimes F$ to $\tilde{F}$?
A:
b) $\eta^{-1}(f)(x_1,x_2) = f(\delta(x - x_1),\delta(x - x_2))$
Knowing the hamiltonian $h$ for a single particle in a square well, we can define the hamiltonian for two particles by transforming $h$ from acting on $F$ to acting on $F \otimes F$ using the following recipe: $H = \tilde{h}_1 + \tilde{h}_2$, where $\tilde{h}_1 \phi \otimes \psi := h\phi \otimes \psi$ and $\tilde{h}_2 = \phi \otimes h\psi$.
(Why is it sufficient to specify only how $H$ acts on tensor products?)
To see how $H$ acts on the wave function
$$
\Psi^-_{12}(x_1,x_2) = \frac{1}{\sqrt{2}}( \psi_1(x_1)\psi_2(x_2) - \psi_2(x_1) \psi_1(x_2) )
$$
which lives in $\tilde{F}$, we must first convert the wave function to its corresponding vector in $F \otimes F$.
Then, since $h$ acts very simply on the energy eigenstates $\psi_n$, i.e. $h\psi_n = E_n\psi_n$, where $E_n=\frac{\pi^2 \hbar^2 n^2}{2 m a^2}$, we will want to then expand the state in terms of the $\psi_m \otimes \psi_n$ basis.
Q:
Find the coefficients $\eta(\Psi^-_{12})_{mn}$
A:
We can do this in one fell swoop by:
\begin{align}
& \eta(\Psi^-_{12})_{mn} \\
= & \ip{\psi_m \otimes \psi_n}{\eta(\Psi^-_{12})} \\
= & \eta(\Psi^-_{12})(\psi_m,\psi_n) \\
= & \int_0^a dx_1 \int_0^a dx_2 \psi^*_m(x_1) \psi^*_n(x_2) \Psi^-_{12}(x_1,x_2) \\
\end{align}
By inspection we find that all the $\eta(\Psi^-_{12})_{mn}$ are zero except for $m=1,n=2$, where the coefficient is $\frac{1}{\sqrt{2}}$, and $m=2,n=1$, where the coefficient is $-\frac{1}{\sqrt{2}}$.
To see that $\Psi^-_{12}$ is an eigenstate of $H$ it is sufficient to show that the vector $\psi_1 \otimes \psi_2$ is an eigenvector.
Q:
Do this.
A:
$$
H \psi_1 \otimes \psi_2
= (\tilde{h}_1 + \tilde{h}_2) \psi_1 \otimes \psi_2
= h\psi_1 \otimes \psi_2 + \psi_1 \otimes h \psi_2
= E_1 \psi_1 \otimes \psi_2 + \psi_1 \otimes E_2 \psi_2
= E_1 \psi_1 \otimes \psi_2 + E_2 \psi_1 \otimes \psi_2
= (E_1 + E_2) \psi_1 \otimes \psi_2
$$
From here we define the exchange operator $P$ which acts on states according to the rule $P \phi \otimes \psi := \psi \otimes \phi$.
Q:
Show that $P^2 = 1$ and that this limits the possible eigenvalues of $P$ to $\pm 1$.
Q:
Show that:
-
$P\tilde{h}_1=\tilde{h}_2P$ and vice-versa
-
$[P,H]=0$
-
$\psi_2 \otimes \psi_1$ is also an eigenstate of $H$ with the same eigenvalue as $\psi_1 \otimes \psi_2$
-
any linear combination of $\psi_1 \otimes \psi_2$ and $\psi_2 \otimes \psi_1$ is an eigenstate of $H$.
A:
$$
P\tilde{h}_1 \phi \otimes \psi
= P h\phi \otimes \psi
= \psi \otimes h \phi
= \tilde{h}_2 \psi \otimes \phi
= \tilde{h}_2 P \phi \otimes \psi
$$
we use this to show
$$
PH
= P(\tilde{h}_1 + \tilde{h}_2)
= P\tilde{h}_1 + P\tilde{h}_2
= \tilde{h}_2 P + \tilde{h}_1 P
= ( \tilde{h}_2 + \tilde{h}_1 ) P
= ( \tilde{h}_1 + \tilde{h}_2 ) P
= HP
$$
next we see that
$$
H \psi_2 \otimes \psi_1
= H P \psi_1 \otimes \psi_2
= P H \psi_1 \otimes \psi_2
= P (E_1 + E_2) \psi_1 \otimes \psi_2
= (E_1 + E_2) P \psi_1 \otimes \psi_2
= (E_1 + E_2) \psi_2 \otimes \psi_1
$$
and then finally, we note that for any operator $A$ and any linear combination $c_1 \ket{a_1} + c_2 \ket{a_2}$ of two eigenstates $\ket{a_1}$, $\ket{a_2}$ of $A$ with the same eigenvalue $a$ we have
$$
A ( c_1 \ket{a_1} + c_2 \ket{a_2} )
= c_1 A \ket{a_1} + c_2 A \ket{a_2}
= c_1 a \ket{a_1} + c_2 a \ket{a_2}
= a ( c_1 \ket{a_1} + c_2 \ket{a_2} )
$$
or in otherwords that the set of eigenstates with the same eigenvalue form a vector subspace.
4
For 5.5b we have no work left to do.
The basis $\psi_m \otimes \psi_n$ diagonalizes $H$.
Two states $\psi_a \otimes \psi_b$ and $\psi_c \otimes \psi_d$ are degenerate if $E_a + E_b = E_c + E_d$.
This occurs for pairs $(m,n) \leftrightarrow (n,m)$, where $m \neq n$
5.
Since $P$ and $H$ commute, we know that we can "simultaneously diagonalize" their matrices, or that there exists some basis where each vector is an eigenstate of both $P$ and $H$.
Since $\psi_m \otimes \psi_n$ is not an eigenstate of $P$ when $m \neq n$, this means that the $(m,n),(n,m)$ degenerate subspaces must contain an odd and an even state, since $\pm1$ are the only eigenvalues of $P$.
6.