# The Uncertainty Principle

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Hi guys!

Back after a lot of days ! :)

So I was studying quantum mechanics and gor interested to find out the derivation of uncertainty principle. In the course of doing this, I found there were two kinds of derivations.

One is relating the uncertainty principle as a basic fundamental property of Fourier transform and deriving it from Fourier transform.

The other one uses operators and derives the relation by exploiting the non-commutavity of two operators.

My question is since the two derivations yield the same result, there must be some deep subtle connection between the two, or, between Fourier transformation and non-commutative operators.

Cheers!

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5 minutes ago, Sriman Dutta said:

Hi guys!

Back after a lot of days !

So I was studying quantum mechanics and gor interested to find out the derivation of uncertainty principle. In the course of doing this, I found there were two kinds of derivations.

One is relating the uncertainty principle as a basic fundamental property of Fourier transform and deriving it from Fourier transform.

The other one uses operators and derives the relation by exploiting the non-commutavity of two operators.

My question is since the two derivations yield the same result, there must be some deep subtle connection between the two, or, between Fourier transformation and non-commutative operators.

Cheers!

You are absolutely right. There is a deep connection. The Fourier transform of a function can be seen as the momentum representation of that function.

It's based on the concept of considering functions as vectors in an infinite-dimensional space. Analogously to how the i-th component of a vector can be obtained by using the scalar product by the $$\boldsymbol{e}_{i}$$ vector of the basis,

$v_{i}=\boldsymbol{v}\cdot\boldsymbol{e}_{i}$

you can obtain the p-component of a function by integrating (analogous to the scalar product):

$\hat{f}\left(p\right)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}dxf\left(x\right)e^{ipx/\hbar}=\left\langle f\left|p\right.\right\rangle$

where the matrix coefficients that give you the change of bases are the exponentials:

$\left\langle x\left|p\right.\right\rangle =\frac{e^{ipx/\hbar}}{2\pi\hbar}$

You must learn the concepts of operators, eigenvalues, and eigenvectors to complete the details. Do you know about those?

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18 hours ago, joigus said:

You are absolutely right. There is a deep connection. The Fourier transform of a function can be seen as the momentum representation of that function.

It's based on the concept of considering functions as vectors in an infinite-dimensional space. Analogously to how the i-th component of a vector can be obtained by using the scalar product by the $$\boldsymbol{e}_{i}$$ vector of the basis,

$v_{i}=\boldsymbol{v}\cdot\boldsymbol{e}_{i}$

you can obtain the p-component of a function by integrating (analogous to the scalar product):

$\hat{f}\left(p\right)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}dxf\left(x\right)e^{ipx/\hbar}=\left\langle f\left|p\right.\right\rangle$

where the matrix coefficients that give you the change of bases are the exponentials:

$\left\langle x\left|p\right.\right\rangle =\frac{e^{ipx/\hbar}}{2\pi\hbar}$

You must learn the concepts of operators, eigenvalues, and eigenvectors to complete the details. Do you know about those?

Yes I know that the position wavefunction and momentum wavefunction are a Fourier pair.

But their corresponding operators x^ and p^ (I'm representing the operators here) are non-commutative.

So what's the connection between the two facts?

PS: Forget just x and p. The uncertainty principle is valid for any pair of Fourier pair of variables a and b.

and same it is true for any pair of non-commutative operators A and B.

How's all this get in together?

Edited by Sriman Dutta
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1 hour ago, Sriman Dutta said:

How's all this get in together?

The connection comes from the Robertson version of the uncertainty relations. The one you can find on Wikipedia is the Robertson-Schrödinger version.

I will give you a proof of the Robertson version which does not involve the anti-commutator.

Say you have any two operators $$A$$ and $$B$$, which in general do not commute.

Say your system is in a pure quantum state $$\left|\psi\right\rangle$$.

The mean square deviation is defined as,

$\left(\triangle_{\left|\psi\right\rangle }A\right)^{2}=\left\langle A^{2}\right\rangle _{\left|\psi\right\rangle }-\left\langle A\right\rangle _{\left|\psi\right\rangle }^{2}=\left\langle \psi\left|A^{2}\right|\psi\right\rangle -\left\langle \psi\left|A\right|\psi\right\rangle ^{2}$

and similarly for $$B$$.

Now define operators $$A'$$ and $$B'$$ centred on their respective average values:

$A'=A-\left\langle A\right\rangle _{\left|\psi\right\rangle }$

$B'=B-\left\langle B\right\rangle _{\left|\psi\right\rangle }$

It's easy to see that,

$\left[A',B'\right]=\left[A,B\right]$

Now you formally build the 1-parameter family of operators:

$C=A'+i\lambda B'$

This operator is not Hermitian, but it is always true that,

$\left\Vert C\left|\psi\right\rangle \right\Vert ^{2}=\left\langle \psi\left|C^{\dagger}C\right|\psi\right\rangle \geq0$

This gives you a polynomial condition in $$\lambda$$:

$\left(\triangle_{\left|\psi\right\rangle }A\right)^{2}+\left(\triangle_{\left|\psi\right\rangle }B\right)^{2}\lambda^{2}+i\lambda\left\langle \left[A,B\right]\right\rangle _{\left|\psi\right\rangle }\geq0$

If this polynomial must always be above the real axis, the discriminant must be negative or zero:

$\left(\left\langle i\left[A,B\right]\right\rangle _{\left|\psi\right\rangle }\right)^{2}-4\left(\triangle_{\left|\psi\right\rangle }A\right)^{2}\left(\triangle_{\left|\psi\right\rangle }B\right)^{2}\leq0$

Keep in mind that if $$A$$ and $$B$$ are Hermitian, so is $$i\left[A,B\right]$$.

This immediately gives you the Robertson version of the uncertainty relations for arbitrary operators $$A$$ and $$B$$:

$\triangle_{\left|\psi\right\rangle }A\triangle_{\left|\psi\right\rangle }B\geq\left|\left\langle \frac{i}{2}\left[A,B\right]\right\rangle _{\left|\psi\right\rangle }\right|$

So when two operators do not commute, you cannot define "dispersions" or "precisions" (mean square deviations) better than those given by the above.

I hope that helped. It's the simplest demonstration I know of the more general Robertson version for arbitrary operators.

Edited by joigus
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Thanks a lot. It all makes it clear

Plus I myself also derived the results and found the connection.

Thanks once again!

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17 hours ago, Sriman Dutta said:

Thanks a lot. It all makes it clear

Plus I myself also derived the results and found the connection.

Thanks once again!

You're welcome.

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