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Can anyone point me to a reference (preferably either something online or something a small liberal arts school would be likely to have in its library) that goes through a derivation of the minimum uncertainty wavefunction in more detail than in the Griffiths?

Edit I've moved the second part of my original question to a separate post: 3D Minimum uncertainty wavepackets

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    $\begingroup$ great question, but it would be much better if you split the two things you've asked into two separate posts. You could edit this one to remove one of the parts, and then ask a separate question for that part alone. $\endgroup$
    – David Z
    Mar 18, 2011 at 5:33
  • $\begingroup$ For the first question, do you want to derive equation on that minimizing wavefunction (i.e. annihilation of anticommutator and certain colinearity condition) or do you want the solution to that equation? $\endgroup$
    – Marek
    Mar 18, 2011 at 9:26
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    $\begingroup$ okay, I answered the first part. But I agree with David that you should ask the second question separately. Nevertheless, what do you mean w.r.t. $\mathbf r$ and $\mathbf p$? The Gaussian does satisfy HUP w.r.t. each individual component, which is a same thing in my opinion. Also, you should remember that the HUP gets modified in more dimensions essentially because of Pythagorean theorem: the square of total error is a sum of squares of individual errors. Therefore your equation for $\Sigma$ should have 3 instead of 1 on the r.h.s. $\endgroup$
    – Marek
    Mar 18, 2011 at 11:52
  • $\begingroup$ A small liberal arts school library should be able to get any nearly any book through interlibrary loan. You may have to wait a week or more for a physical book, but paywalled research papers or ebooks should come through in days or even hours. $\endgroup$
    – Paul T.
    Nov 30, 2023 at 22:32

1 Answer 1

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First some preliminaries. Suppose you have hermitian operators $A$ and $B$ and some state $\left| \psi \right>$. Denote by $\left<X\right>$ the expectation of $X$ in the state $\psi$, i.e. $\left<\psi\right| X \left| \psi\right>$. Denote by $\bar A := A - \left< A \right>$ and $\bar B := B - \left< B \right>$ the part of of $A$, resp. $B$ with vanishing expectation.

So, let's compute $\left< \bar A^2 \right> \left<\bar B^2 \right>$. According to Cauchy-Schwarz inequality that this is always greater or equal than $\left|\left< \bar A \bar B \right>\right|^2$ (just plug in $\psi$ and interpret these expressions as scalar products). Now, we can express the product as sum of hermitian and antihermitian component $$\left|\left< \bar A \bar B \right>\right|^2 = {1\over4}\left< [A,B]/i \right>^2 + {1\over 4}\left< \{\bar A, \bar B\} \right>^2$$

(here we used the fact that $[\bar A, \bar B] = [A,B]$). If the commutator is just a number times identity operator then we can discard the expectations and after removing the anticommutator term (because it doesn't have any important interpretation and it doesn't spoil the inequality) we are left with HUP. But we're not interested in this application right now. Instead, we want to minimize the error term and that means we want equalities everywhere (it's not clear that it's possible to attain them, but let's assume this for a while). First, Cauchy-Schwarz inequality becomes equality if the vectors in the scalar product are colinear $$\bar B \left| \psi \right> = c \bar A \left| \psi \right>$$ Second, we want the expectation of anticommutator to vanish $$\left<\psi\right| \{\bar A, \bar B\} \left| \psi \right> = 0$$ So this gives us two equations for $\psi$. Let's see what we can get from them for $A = x$ and $B = p$. For simplicity let's assume that $\left<x\right> = \left<p\right> = 0$ (the general solution doesn't change anything much).

From first condition we obtain $$(p - cx) \left | \psi \right> = 0$$ which is a differential equation $$ (i \partial_x + cx) \psi(x) = 0$$ with a solution $\psi(x) = K \exp(-\alpha x^2)$ with ${\rm Re} \alpha > 0$ (so that this is indeed a vector from our Hilbert space) and $K$ being just a normalization constant. Finally from the anticommutator relation we get $\alpha = {1 \over 4(\Delta x)^2}$ and we're done.

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    $\begingroup$ What if we restricted the domain of x, (say $x>0$). How does the solution of the differential equation change in that case ? $\endgroup$
    – user35952
    Aug 31, 2016 at 9:58

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