MHB Relation between matrix elements of momentum and position operators

Click For Summary
The discussion revolves around finding the relationship between the matrix elements of the momentum operator (p) and the position operator (x) in the context of a Hamiltonian for a single particle. The key equation derived is $$\frac{ \langle i \, | \, p \, | j \rangle}{\langle i \, | \, x \, | j \rangle} = \frac{iM(E_i - E_j)}{\hbar},$$ which connects the eigenvalues of the Hamiltonian to the matrix elements of the operators. The calculations involve using canonical commutation relations and the commutator of x with the Hamiltonian to derive necessary expressions. The approach highlights the importance of understanding how to compute matrix elements and the role of energy eigenvalues in these calculations. Overall, the discussion emphasizes the mathematical relationships that can be established through operator algebra in quantum mechanics.
Fantini
Gold Member
MHB
Messages
267
Reaction score
0
Hello. I'm having trouble understanding what is required in the following problem:

Find the relation between the matrix elements of the operators $\widehat{p}$ and $\widehat{x}$ in the base of eigenvectors of the Hamiltonian for one particle, that is, $$\widehat{H} = \frac{1}{2M} \widehat{p}^2 + V(\widehat{x}).$$

I don't understand what kind of relation he is asking for. The eigenvectors for the Hamiltonian satisfy the equation $\widehat{H} \varphi_i = E_i \varphi_i$, but I don't know how to use that. The answer is $$\frac{ \langle i \, | \, p \, | j \rangle}{\langle i \, | \, x \, | j \rangle} = \frac{iM(E_i - E_j)}{\hbar},$$ but it doesn't enlighten me.

How do I find an arbitrary matrix element of an operator?
 
Mathematics news on Phys.org
Just thinking out loud here. First off, we can tidy up the notation so that everything matches: $\hat{H} \, |i\rangle =E_i \, |i\rangle$, and the same for $j$. Let's remember some of the canonical commutation relations, in case they prove useful:
\begin{align*}
[x,p]&=i\hbar \\
[x,x]&=0 \\
[p,p]&=0 \\
[AB,C]&=A[B,C]+[A,C]B.
\end{align*}
The last allows us to compute
\begin{align*}
[x,H]&=\left[x,\frac{p^2}{2M}+V(x)\right] \\
&=\frac{1}{2M} \, [x,p^2]+\underbrace{[x,V(x)]}_{=0} \\
&=-\frac{1}{2M} \, [p^2,x] \\
&=-\frac{1}{2M} \, \left( p[p,x]+[p,x]p \right) \\
&=\frac{i \hbar}{M} \, p.
\end{align*}
The reason I computed this commutator is because the answer has $E_i$ and $E_j$ in it, leading me to think that we're going to have to calculate something with an $H$ in it. If you calculate $[x,H] \, |j\rangle$ two different ways, you get
$$[x,H] \, |j\rangle = \frac{i\hbar}{M} \, p \, |j\rangle = (E_j-H) \, x \, |j\rangle.$$
Here, I have done
\begin{align*}
[x,H] \, |j\rangle&=(xH-Hx) \, |j\rangle \\
&=xH \, |j\rangle - Hx \, |j\rangle \\
&=x E_j \, |j\rangle-Hx \, |j\rangle \\
&=E_j \, x \, |j\rangle -Hx \, |j\rangle \\
&=(E_j-H) \, x \, |j\rangle.
\end{align*}

It seems to me that these computations might help. Does this give you any ideas?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 1 ·
Replies
1
Views
935
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K