# Is there a heuristic explanation for the derivation of Heisenbergs Uncertainty Principle from String Theory?

Heisenberg famously derived his uncertainty principle by considering the disturbance that a measurement would have on a small enough system.

Of course in the mathematical formalism of Quantum Mechanics the relationship is derived from more basic principles.

How does String Theory account for it? Heuristically, and mathematically?

• – Dilaton Jun 21 '13 at 2:25
• String theory is just one quantum-mechanical theory. Why should the HUP need a different justification in ST than in any other quantum-mechanical theory? – Ben Crowell Jun 21 '13 at 2:33
• String theory assumes quantum mechanics. It doesn't explain, underlie, or supersede it. – Michael Brown Jun 21 '13 at 3:03
• @MoziburUllah No, quantum mechanics is completely general. It can talk about point particles, fields, strings, composite structures of arbitrary complexity, you name it. Sometimes people distinguish the different possibilities in the name, e.g. quantum field theory, string theory, quantum chemistry etc. But all of these theories share the exact same underlying quantum mechanics that you learn in an undergrad physics degree (sometimes with extra window dressing to make complicated structures look simpler, but without changing the quantum foundation). – Michael Brown Jun 21 '13 at 4:30
• @MoziburUllah If I may attempt to rephrase what the others are driving at... There are two different uses of the term "quantum mechanics" here. Their use is "in general a theory of non-commuting operators on a Hilbert space, subject to the following assumptions..." while yours is "the application of such non-commuting operators and linear algebra to point particles in $\mathbb{R}^3$, perhaps tensor-producted with finite-dimensional Hilbert spaces for spin, etc." Undergrad courses emphasize the latter, but if done right all the elements of the more general theory are in fact present. – user10851 Jun 21 '13 at 14:49

• In string theory, the HUP has an additional term such that it reads $\Delta x = \frac{\hbar}{\Delta p} + \alpha' \frac{\Delta p}{\hbar}$ due to the minimal lenth introduced. – Dilaton Jun 21 '13 at 3:07
• @Dilaton: Don't you mean $\Delta x\geq\frac{\hbar}{\Delta p}+\alpha'\frac{\Delta p}{\hbar}$? I don't think it can be "equal". – Abhimanyu Pallavi Sudhir Jun 21 '13 at 3:27