2
$\begingroup$

I was reading Feynman's Lectures III's Hamiltonian Matrix. There I found this property of Hamiltonian Matrix:

The Hamiltonian has one property that can be deduced right away, namely, that $$H^*_{ij} = H_{ji} $$This follows from the condition that the total probability that the system is in some state does not change. If you start with a particle—an object or the world—then you’ve still got it as time goes on. The total probability of finding it somewhere is $$\sum_i |C_i(t)|^2$$which must not vary with time. If this is to be true for any starting condition $ϕ$, then Eq. (8.40) must also be true.

I haven't understood Feynman's reasoning for proving the law; how "the total probability that the system is in some state does not change" assure the validity of the property? Can anyone explain me Feynman's reasoning?

$\endgroup$
  • $\begingroup$ I think he is just arguing that, given that the time evolution is $U(t) = e^{iHt}$, by requiring the probability to be conserved one is requiring $U$ to be unitary, hence $H$ to be self-adjoint. $\endgroup$ – Phoenix87 Jul 31 '15 at 19:04
3
$\begingroup$

Okay, so, let's get this thing going right.

The property you're talking about is called by different names that you can google: e.g. that the Hamiltonian matrix must be self-adjoint (where the adjoint is the conjugate transpose) or Hermitian.

The easiest property to prove about Hermitian matrices is that they have real eigenvalues, for their defining characteristic $\langle H \psi| \psi\rangle = \langle \psi| H \psi\rangle$ combines with the property $\langle a | b\rangle = \left(\langle b | a\rangle\right)^*$ to say that a certain number (an "expectation value") is equal to its own conjugate, which is only true for real numbers.

How does this connect to probability conservation? The total probability is$$P = \sum_i C_i^* ~ C_i$$and the Schrodinger equation 8.39 that he presents is:$$i\hbar\,\frac{\rm dC_i}{\rm dt}=\sum_jH_{ij}(t)C_j(t).$$The conjugate expression must therefore be: $$-i\hbar\,\frac{\rm dC_i^*}{\rm dt}=\sum_jH_{ij}^*(t)C_j^*(t).$$ Now let us just take a derivative of $P$ with the product rule, as:$$\begin{align}i\hbar \frac{dP}{dt} &= i\hbar \sum_i \left({\rm dC_i^*\over \rm dt} ~ C_i + C_i^* ~ {\rm dC_i\over \rm dt}\right)\\ &= \sum_{ij}\left(C_i^* ~H_{ij} ~C_j - H_{ij}^* ~C_j^*~ C_i\right)\end{align}$$We now perform a dirty trick: we split this sum into its two terms and we relabel each part of the sum differently. In the first term we take $i \mapsto m$ and $j \mapsto n$. But in the second term we take $i \mapsto n$ and $j \mapsto m$. We then combine both terms together to find:$$i\hbar \frac{\rm dP}{\rm dt} = \sum_{mn}\left(C_m^* ~H_{mn} ~C_n - H_{nm}^* ~C_m^*~ C_n\right),$$and, combining like terms:$$i\hbar \frac{\rm dP}{\rm dt} = \sum_{mn}C^*_m~C_n \left(H_{mn} - H_{nm}^*\right).$$The implication is that if we want $\rm dP/\rm dt = 0$ for all $C_n$ (total probability does not escape the system) then we must have that the parenthesized matrix is the 0-matrix.

Why? It's actually not too complicated to see. If we steal the $i$ from the left to define $$D_{mn} = -i \left(H_{mn} - H_{nm}^*\right),$$then we know for sure that $D_{mn}$ is Hermitian by construction. (Just take its conjugate transpose, I dare you.) Because it's Hermitian, it is never defective and we can diagonalize it with real eigenvalues down the diagonal. Now either one of these eigenvalues $\lambda$ is nonzero, in which case we can use the corresponding eigenvector as $C_n$ above and we get $\hbar \frac{\rm dP}{\rm dt} = \lambda \sum_m C_m^* C_m = \lambda P$, violating probability conservation, or else all of the eigenvalues are 0 and this expression is the 0-matrix. But the 0 matrix has all-zero components in all bases, not just its eigenbasis, so therefore $H_{mn} = H^*_{nm}$ in general.

$\endgroup$
  • $\begingroup$ Hi, sir....couldn't contact(am suffering from measles!). Though most of the maths' operation I could understand, I couldn't comprehend why $\dfrac{dp}{dt}$ needs to be zero. Can you please provide a little explanation for it. Also, sir, can you please explain the line total probability does not escape the system? Why does probability need to be conserved?? Thanks in advance, sir:) $\endgroup$ – user36790 Aug 2 '15 at 15:38
  • 1
    $\begingroup$ @user36790 $dP/dt = 0$ because $P$ is the sum of probabilities for all possible outcomes, hence we want $P = 1$, constant over time. If an electron's total probability isn't conserved, then that means that sometimes, if we look everywhere for it, we don't find it -- which means other things like its mass cannot easily be conserved within the theory. It is effectively a commitment that, if we want to model electrons entering/leaving a system S, we're going to do quantum mechanics on a larger system consisting of S plus "electron reservoirs" that absorb/emit electrons. $\endgroup$ – CR Drost Aug 2 '15 at 18:39
  • $\begingroup$ I have queries, sir. Say, I have a state-vector: $|\psi\rangle = \sum_i c_i|a_i\rangle$. So, the total probability to find the particle in any of its base states is $\sum_i |c_i|^2$ which is always 1, right? Acc. to probability-conservation, the total probability remains the same i.e. 1, right? Now, suppose, the probability of finding the particle anywhere would be $\int_{-\infty}^{\infty}|\psi|^2 dx$ which would be 1 & again by probability-conservation, it would be same all the time. Just want to know, are $\sum_i |c_i|^2$ & $\int_{-\infty}^{\infty}|\psi|^2 dx$ describing same thing or not? $\endgroup$ – user36790 Aug 6 '15 at 7:18
  • $\begingroup$ Another one: I've understood from your explanation that in order to conserve mass, you need to have probability conserved. But I am not understanding why the concept of "probability-current" is needed; yes, we need probability to be 1 always & hence need to be conserved but how is it related to current? Why is the emergent of the concept of current?? $\endgroup$ – user36790 Aug 6 '15 at 7:27
  • $\begingroup$ In order to have a better grisp over the concept of probability-conservation, just I am conjuring up what would happen if the total probability decreased over time from 1 to, say, $1 \over 4$. What would be the scenario then? How would it affect the possibility of being in one of its base states $|a_i\rangle$? Thanks in advance. I'll be grateful if you help me, sir:) $\endgroup$ – user36790 Aug 6 '15 at 8:56
2
$\begingroup$

If that property doesn't hold then $H$ is not Hermitian. This means its eigenvalues could be complex and the time evolution operator won't be unitary: $U(t)U(t)^\dagger=e^\frac{iHt-itH^\dagger }{\hbar} \neq1$

If it's not unitary, then it will change the length of your state vectors which you're evolving. Since the length of these($\sum_i |C_i(t)|^2$) vectors are the sum of the probabilities, your total probability won't be conserved.

$\endgroup$
1
$\begingroup$

What he actually is referring to is the mathematical fact that in order for the standard norm (length) of $n$-tuple $[C_j]_{j=1,2,...n}$ to be preserved during evolution determined by the equation

$$ \frac{\rm dC_i}{\rm dt} = -\frac{i}{\hbar} \sum_j H_{ij} C_j $$ where $H_{ij}$ is time-independent complex matrix, this matrix needs to be hermitian, i.e. obey the condition for all pairs $i,j$,

$$ H_{ij}^* = H_{ji}. $$

This can be proven starting with the condition that norm squared is constant in time:

$$ \frac{\rm d}{\rm dt}\bigg(\sum_i C_i^*C_i\bigg)=0 $$

and using the above differential equation.

Feynman just rephrases this and uses term "total probability" instead of "norm squared", because it is common to interpret $|C_i|^2$ as probability that $i^{\textrm{th}}$ eigenvalue of $H$ will be measured.

$\endgroup$
  • 1
    $\begingroup$ Actually, the matrix that multiplies $C_j$ has to be unitary. But the unitary evolution operator is $e^{iHt}$, which is unitary if $iH$ is anti-Hermitian, which happens when $H$ is Hermitian. $\endgroup$ – knzhou Jul 31 '15 at 21:17
  • $\begingroup$ @KevinZhou, you're right! I need to edit this. $\endgroup$ – Ján Lalinský Jul 31 '15 at 23:30
  • $\begingroup$ Sir, you've forgotten to write the "-" sign in the equation; I've edited it. $\endgroup$ – user36790 Aug 1 '15 at 2:51

protected by Qmechanic Aug 6 '15 at 9:50

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?