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Lectures on Quantum Mechanics 5: The Heisenberg Uncertainty Principle

This note has been used to help create the Heisenberg Uncertainty Principle wiki

Conjugate Variables

Before I derive the uncertainty principle, I would like to introduce the notion of conjugate variables. The uncertainty principle relates the standard deviation two conjugate variables, which was briefly mentioned in Lecture 3. In Lecture 1 we described the wave-packet of a free particle being

\[\Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\phi(p)}{e}^{i(px-Et)/ \hbar}dp.\]

If we let \(\Phi(p,t)={\phi(p)}{e}^{iEt/ \hbar}\), the integral becomes

\[\Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\Phi(p,t)}{e}^{ipx/ \hbar}dp.\]

Here we find that the position-space wavefunction \(\Psi(x,t)\) is indeed the Fourier transform of another wavefunction that is dependent on momentum instead. We will call \(\Phi(p,t)\) the momentum-space wavefunction.

Following this logic, we can recover the momentum-space wavefunction given the position-space wavefunction by taking the inverse Fourier transform:

\[\Phi(p,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\Psi(x,t)}{e}^{-ipx/ \hbar}dx.\]

So what? What did we learn by doing this? Well, what this bit of Fourier analysis has achieved is very interesting, because we see a direct relationship between two variables \(x\) and \(p\), that describe the same wave-packet in different spaces. In this example, a free particle is measured in the position space by \(\Psi(x,t)\), but measured in the momentum space by \(\Phi(p,t)\). Yet regardless which space we choose to measure the particle, the physics doesn't change. Thus by discovering this relationship of conjugate variables, we can make measurements on particles in physical experiments in two ways. In fact, this is how crystallography and much of solid-state physics is done (see Reciprocal Lattice).

Heisenberg's Uncertainty Principle (HUP)

From Lecture 3 we discussed the definition of standard deviation for some variable, say \(x\), \({\sigma}_{x}^{2} = \left<{(x-\left<x \right>)}^{2}\right>\). Back then we were concerned with numbers, but to derive the uncertainty principle, we need to write the standard deviation in terms of operators acting on wavefunctions.

Thus we consider the inner product


Let \(f=(\hat{x}-\left<\hat{x}\right>)\Psi\), thus \({\sigma}_{x}^{2}=\left<f|f\right>\).

Similarly we have \[{\sigma}_{p}^{2}=\left<(\hat{p}-\left<\hat{p}\right>)\Psi|(\hat{p}-\left<\hat{p}\right>)\Psi\right>=\left<g|g\right>\] where \(g=(\hat{p}-\left<\hat{p}\right>)\Psi\).

By the Cauchy-Schwarz inequality, we get \[{\sigma}_{x}^{2}{\sigma}_{p}^{2} = \left<f|f\right>\left<g|g\right> \ge {|\left<f|g\right>|}^{2}.\]

Now let's simplify \(\left<f|g\right>\):

\[ \begin{align*} &\left<f|g\right>=\left<(\hat{x}-\left<\hat{x}\right>)\Psi|(\hat{p}-\left<\hat{p}\right>)\Psi\right> \\ &=\left<\Psi|(\hat{x}-\left<\hat{x}\right>)(\hat{p}-\left<\hat{p}\right>)\Psi\right> \\ &=\left<\Psi|(\hat{x}\hat{p}-\hat{x}\left<\hat{p}\right>)-\hat{p}\left<\hat{x}\right>+\left<\hat{x}\right>\left<\hat{p}\right>)\Psi\right> \\ &=\left<\Psi|\hat{x}\hat{p}\Psi\right>-\left<p\right>\left<\Psi|\hat{x}\Psi\right>-\left<x\right>\left<\Psi|\hat{p}\Psi\right>+\left<x\right>\left<p\right>\left<\Psi|\Psi\right>\\ &=\left<\hat{x}\hat{p}\right>-\left<\hat{p}\right>\left<\hat{x}\right>-\left<\hat{x}\right>\left<\hat{p}\right>+\left<\hat{x}\right>\left<\hat{p}\right>\\ &=\left<\hat{x}\hat{p}\right>-\left<\hat{p}\right>\left<\hat{x}\right> \end{align*}\]

Similarly, \[\left<g|f\right>=\left<\hat{p}\hat{x}\right>-\left<\hat{x}\right>\left<\hat{p}\right>.\]

I know the bra-ket notation and expectation of an operator both use angle brackets, which makes the above algebra pretty confusing. Just remember that \(\left<\hat{x}\right>\) is the expectation of \(\hat{x}\), which is a number.

There is a useful property about complex numbers: \[{z}^{*}z \ge {\left[\frac{1}{2i}(z-{z}^{*})\right]}^{2}.\]

Therefore if we replace \(z\) with \(\left<f|g\right>\) and \({z}^{*}\) with \(\left<g|f\right>\), we get

\[\left<f|f\right>\left<g|g\right>\ge {\left[\frac{1}{2i}(\left<f|g\right>-\left<g|f\right>)\right]}^{2}.\]

This translates to \[{\sigma}_{x}^{2}{\sigma}_{p}^{2} \ge {\left(\frac{1}{2i}[\hat{x},\hat{p}]\right)}^{2}.\]

Remember from Lecture 4 we found out that \([\hat{x},\hat{p}]=i\hbar\). Hence, by substituting the above commutator relation and taking the square root, we arrive at the HUP \[{\sigma}_{x}{\sigma}_{p} \ge \frac{\hbar}{2}.\]

Visit my set Lectures on Quantum Mechanics for more notes.


  1. Prove that \[\int_{-\infty}^{\infty} {\Psi}^{*}\Psi dx = \int_{-\infty}^{\infty} {\Phi}^{*}\Phi dp\] given the position-space and momentum-space wavefunctions.

  2. Operators in momentum-space The position operator in momentum space is \[\hat{x} = i\hbar\frac{\partial}{\partial p}\] and the momentum operator is \[\hat{p} = p.\] Verify that \(\left[\hat{x},\hat{p}\right]=i\hbar\) is valid in momentum space.

  3. The position-momentum uncertainty principle is often written loosely as \[(\Delta x)(\Delta p) \ge \frac{\hbar}{2}.\] Show that the energy-time uncertainty holds \[(\Delta E)(\Delta t) \ge \frac{\hbar}{2}\] hence, proving \[{\sigma}_{E} {\sigma}_{t} \ge \frac{\hbar}{2}.\]

Note by Steven Zheng
1 year, 9 months ago

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Oh I love the comic :) Calvin Lin Staff · 1 year, 9 months ago

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@Calvin Lin Me too ! :D Priyansh Sangule · 1 year, 9 months ago

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The fact that position and momentum are "Fourier transforms" of each other leading to the position-momentum uncertainty principle has an analogy in acoustics. If we hear a pure tone, say \(440hz\), that corresponds to a single spike in the frequency spectrum (namely, \(440hz\)!), but a single acoustic spike, say like what happens when a hammer hits a solid surface, will correspond to a broad range in the frequency spectrum, i.e., it's composed of many frequencies. If we go with Born's probability amplitude model for particles, we can see we can't get unlimited "precision" in both position and momentum. This is similar to the problem of aperture and image resolution in optics, i.e. for higher resolutions, larger apertures are needed. The same kind of math keeps re-appearing across different disciplines in physics. Michael Mendrin · 1 year, 9 months ago

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@Michael Mendrin That is what I find astounding about physics. As more math is discovered, it enables us to describe more physics. Steven Zheng · 1 year, 9 months ago

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So where are the earlier lectures? Snehal Shekatkar · 1 year, 9 months ago

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@Snehal Shekatkar Here

Please like and reshare my notes and set! Steven Zheng · 1 year, 9 months ago

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@Steven Zheng Thanks :) Snehal Shekatkar · 1 year, 9 months ago

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I have never studied abt these thingss...! Thnks fr sharing this...! GM Harimangal Pandey · 1 year, 9 months ago

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@Steven Zheng, The HUP can also be connected to the motion of a ceiling fan. The better you know a particle's speed on the blade of a fan, the lousier is your knowledge about its position. By the way, you did quite a bit of explaining! Ameya Salankar · 1 year, 9 months ago

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WOW !!! Thanks @Steven Zheng it's nicely explained . BTW , you would really make a great teacher , don't you think so ? Azhaghu Roopesh M · 1 year, 9 months ago

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@Azhaghu Roopesh M I actually get that a lot! I do tutor math and physics.

I am currently taking QM at UBC. These notes are organized in order of mathematical foundation. My sources are the lectures of my professor, Griffiths "Introduction to Quantum Mechanics" and McMahon's "Quantum Mechanics Demystified." Certain side notes may come from a course I took on the Physics of Materials (a bit of quantum chemistry and solid state physics). Steven Zheng · 1 year, 9 months ago

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@Steven Zheng A brilliant effort , I must say . If I may, I suggest that you write some wiki on QM so as to encourage other Brilliant minds to look it up, since QM isn't a favourite topic for many . Azhaghu Roopesh M · 1 year, 9 months ago

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Excellent Explanation!!! Steven Zheng .

I wanted some clarification regarding the problem you recently posted. Don't you think the wavefunction must become zero at the end points to satisfy the requirements for it to be a valid wavefunction ? But if it is equally likely to find a particle throughout, then the wavefunction would be a non-zero constant throughout the width \(a\) which is a clear contradiction.What are your thoughts regarding this? Sudeep Salgia · 1 year, 9 months ago

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@Sudeep Salgia The wavefunction is zero on the boundaries of the infinite square well. "In the region" means \(0< x<a\) so it is still equally likely to find a particle throughout. Steven Zheng · 1 year, 9 months ago

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