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'''Claimed - Rehaan Naik 11/27/2022''' | '''Claimed - Rehaan Naik 11/27/2022''' <br> | ||
The Born rule provides a link between the mathematical formalism of quantum theory and experiment, and as such is almost single-handedly responsible for practically all predictions of quantum physics. <br> | |||
It is an important result of quantum mechanics that describes the probability density of a measured quantum system. In particular, it states that the square of the wavefunction is proportional to the probability density function. This result is a continuous form of the norm squared of the inner product being the probability that a known state takes the value of another. <br> | |||
Discrete Probability<math> P_{\psi}(A = a) = |\bra{\phi_{a}}\ket{\psi}|^2 </math> <br> | |||
Continuous Probability Density Function <math> P = \left | \Psi(x,t) \right |^2 </math> <br> | |||
This result is particularly important because it allows physicists to test their predictions about quantum systems, making quantum mechanics a testable theory. | |||
==Notation and Formalism for Discrete Observables== | |||
Quantum systems, and all the values and possible states that they can hold, are associated with ''Hilbert Spaces''. These contain the states that a quantum system can have, and are vector spaces. Then, each quantum state can be represented as a vector in this Hilbert Space. (Because these are vectors in a vector space, linear combinations of quantum states also produce quantum states - within some constraints - we require that the norm (a generalized idea of length) be 1, making the state vector essentially a unit vector). <br> | |||
Then a state <math> \psi </math> is represented by the state vector <math> \ket{\psi} </math> <br> | |||
Now let's consider observables, such as position or momentum or angular momentum. Let us represent such an observable as <math> A </math>. Then for a discrete observable, there are a discrete set of outcomes that we represent as <math> {a} </math>. For discrete observables, <math> A </math> is a matrix, and we have a set of vectors <math> {\ket{\phi_a}} </math> that are eigenvectors of <math> A </math> corresponding to the eigenvalues <math> {a} </math>. <br> | |||
<math> A{\ket{\phi_a}} = a\ket{\phi_a} </math> <br> | |||
Then we have a neat little realization: <br> | |||
<math> Prob_{\phi_a}(A = a) = 1 </math> <br> | |||
The expression above refers to the probability of the state <math> \ket{\phi_a} </math> having the outcome <math> a </math> for the observable <math> A </math> | |||
==Born Rule as the Inner Product== | |||
For an arbitrary state <math> \ket{\psi} </math>, we now have the Born rule: <br> | |||
<math> Prob_{\psi}(A = a) = |\bra{\phi_a}\ket{\psi}|^2 </math> <br> | |||
The expression on the right is the norm squared of the ''inner product''. | |||
===Inner Product=== | |||
Note that both <math> \ket{\psi} </math> and <math> \ket{\phi_a} </math> are state ''vectors''. The inner product is a generalization of dot products, and is the multiplication of the conjugate transpose of the vector on the left and the vector on the right. <br> | |||
Conjugate transpose of <math> \ket{\psi} = \bra{\psi} </math> <br> | |||
A note: <br> | |||
<math> |\bra{\psi}\ket{\psi}|^2 = \bra{\psi}\ket{\psi} = 1 </math> | |||
The | ==The Wavefunction as a Continuous Representation of State== | ||
The wavefunction is a complex function. However, there is no physical interpretation of complex numbers. Converting a complex number to a real number involves taking the norm of the complex number. Now, see that the wavefunction <math> \psi(x) </math> is the continuous analogue of <math> \ket{\psi} </math> in the x basis. <math> \ket{\psi} </math> is a state vector for which the norm is finite, and the inner product is defined within the Hilbert Space. We know that the norm squared of the inner product of a state vector with itself is 1 since we have defined state vectors to have norm 1. Since the wavefunction is the analogue of the state vector, it makes sense that an analogue of the inner product is defined for it - the inner product of a state vector with itself is the multiplication of the the conjugate transpose of the state vector with itself. So, the norm squared of the wavefunction comes from the multiplication of the wavefunction with its complex conjugate. <br> | |||
<math> \left | \Psi(x,t) \right |^2 = \Psi^*(x,t)\Psi(x,t) </math> <br> | |||
==Probability Density Function== | |||
Clearly, since we have mandated that the inner product of the two state vectors is finite, so must be the norm squared of the wavefunction. (The wavefunction belongs to the Hilbert space within which the inner product is defined). <br> | |||
In a discrete sense, the norm squared of the inner product gives the probability for an outcome upon measurement of a state for a given observable. However, for a wavefunction, we deal with an infinite Hilbert Space, and the spectrum of the observable is continuous - so its norm squared cannot give values of probability for a specific. Instead we interpret it as giving the probability density function of the state (as represented by the wavefunction) on the chosen observable. <br> | |||
Now, to get the probability over the entire observable, we integrate from <math> -\infty \rightarrow +\infty </math>. It makes sense that this integral would be 1, since the summation of probability over all the possible outcomes is necessarily 1. Since we are integrating the norm-squared of the wavefunction, and we have the following result: | |||
<math>\int_{-\infty}^{\infty}{ \left | \Psi(x,t) \right |^2 dx}=1 </math>, | |||
<br> ''as a given particle in question must be located somewhere between '' <math> -\infty < x <\infty </math> <br> | |||
<math>\int_{-\infty}^{\infty}{ \left | \Psi(x,t) | ''Or more formally:'' | ||
<br>as a given particle in question must be located somewhere between <math> -\infty < x <\infty </math> <br> | |||
Or more formally: | |||
<math>P_{x\in[-\infty,\infty]}=1 </math> | <math>P_{x\in[-\infty,\infty]}=1 </math> | ||
We can say that <math> \Psi(x,t) </math> must be '''square integrable'''. (All this means is that <math>\int_{-\infty}^{\infty}{ \left | \Psi(x,t) \right |^2 dx} < \infty </math>) <br> | |||
===Calculating Probability=== | |||
This is now straightforward since we have the probability density function. <br> | |||
The probability that the outcome of measurement of observable <math> A </math> lies between the values <math> a_1 </math> and <math> a_2 </math> for the state <math> \psi </math> is represented formally as <br> | |||
<math> Prob_\psi(a_1 \leqslant A \leqslant a_2) </math> <br> | |||
As expected from how probability distributions are defined: <br> | |||
<math> Prob_\psi(a_1 \leqslant A \leqslant a_2) = \int_{a_1}^{a_2}{\left | \Psi(a) \right |^2 da} </math> <br> | |||
See that the wavefunction is shown as <math> \Psi(a) </math> instead of <math> \Psi(x,t) </math>. We make the conscious choice to '''change the basis of <math> \Psi </math> from <math> x </math> to <math> a </math> since we are calculating the probability on the observable <math> A </math>'''. <math> a </math> represents the ''generalized eigenvector'' of <math> A </math>. These ''generalized eigenvectors'' of <math> A </math> form a complete basis for the Hilbert Space within which the wavefunction is defined, so we can fully define the <math> \Psi(x,t) </math> in the <math> a </math> basis instead of the <math> x </math> basis. All this does is make calculations easier. | |||
==Examples== | |||
===Simple=== <br> | |||
Given a state <math> \ket{\psi} = \frac{1}{2}\ket{+} + \frac{i\sqrt{3}}{2}\ket{-} </math>, calculate the probability that a measurement of the state comes in the state <math> \ket{+} </math>. | |||
Now, from above, we can assume that since we have expressed <math> \ket{\psi} </math> in the basis <math> {\ket{+}, \ket{-}} </math>, they must be a complete basis of the Hilbert Space and are the eigenvectors of the observable. Regardless, the probability, as per Born's rule, is as follows: | |||
<math> |\bra{+}\ket{\psi}|^2 = |\bra{+}[\frac{1}{2}\ket{+} + \frac{i\sqrt{3}}{2}\ket{-}]|^2 </math> <br> | |||
A feature of quantum observables is that they are represented by Hermitian matrices, whose eigenvalues have orthonormal eigenvectors. Since the outcomes (eigenvalues) have the corresponding state vector as the basis vectors, the basis vectors are orthonormal to each other. <br> | |||
Remember that the inner product is a generalization of the dot product - so the inner product of orthogonal state vectors: | |||
<math> < \bra{+}\ket{-}>= |< \bra{+}\ket{-}>|^2 = 0 </math> | |||
And the inner product of a vector with itself: | |||
<math> < \bra{+}\ket{+}>= |< \bra{+}\ket{+}>|^2 = 1 </math> | |||
Thus, the probability <math> = |\bra{+}\frac{1}{2}\ket{+} + \bra{+}\frac{i\sqrt{3}}{2}\ket{-}]|^2 = |\frac{1}{2}\bra{+}\ket{+}|^2 = |\frac{1}{2}|^2 = \frac{1}{4} </math> | |||
===Middling=== | |||
===Difficult=== | |||
==Connectedness== | |||
How is this topic connected to something that you are interested in? | |||
How is it connected to your major? | |||
Is there an interesting industrial application? | |||
==History== | |||
Max Born, in his 1926 paper argued that the amplitude of the wavefunction that Schrödinger had created was related to probability that you will find the particle at that position if you detect it experimentally. Born claimed that he had simply generalized from photons, the quantum “packets of light” that Einstein proposed in 1905. Einstein, Born said, had interpreted “the square of the optical wave amplitudes as probability density for the occurrence of photons. This concept could at once be carried over to the wavefunction.” The occurrence of photons became generalized as the position of a measured particle whose wavefunction is being described. | |||
Interestingly, and disappointingly for our understanding of Quantum Mechanics, Born introduced his eponymous rule in the context of scattering theory, but did not derive his results from fundamental principles. His connection of the amplitude to the probability density function was somewhat of a guess, and did not come from a mathematically rigorous base. It did however, work remarkably well. | |||
==See also== | |||
==References== | |||
Ball, Phillip. “The Born Rule Has Been Derived from Simple Physical Principles.” Quanta Magazine, 13 Feb. 2019, [https://www.quantamagazine.org/the-born-rule-has-been-derived-from-simple-physical-principles-20190213/] <br> | |||
Landsman, N.P. The Born Rule and Its Interpretation. [https://www.math.ru.nl/~landsman/Born.pdf] <br> | |||
Townsend, J S. A Modern Approach to Quantum Mechancs. Saisalito, Ca, University Science Books, 2000. |
Latest revision as of 04:13, 29 November 2022
Claimed - Rehaan Naik 11/27/2022
The Born rule provides a link between the mathematical formalism of quantum theory and experiment, and as such is almost single-handedly responsible for practically all predictions of quantum physics.
It is an important result of quantum mechanics that describes the probability density of a measured quantum system. In particular, it states that the square of the wavefunction is proportional to the probability density function. This result is a continuous form of the norm squared of the inner product being the probability that a known state takes the value of another.
Discrete Probability[math]\displaystyle{ P_{\psi}(A = a) = |\bra{\phi_{a}}\ket{\psi}|^2 }[/math]
Continuous Probability Density Function [math]\displaystyle{ P = \left | \Psi(x,t) \right |^2 }[/math]
This result is particularly important because it allows physicists to test their predictions about quantum systems, making quantum mechanics a testable theory.
Notation and Formalism for Discrete Observables
Quantum systems, and all the values and possible states that they can hold, are associated with Hilbert Spaces. These contain the states that a quantum system can have, and are vector spaces. Then, each quantum state can be represented as a vector in this Hilbert Space. (Because these are vectors in a vector space, linear combinations of quantum states also produce quantum states - within some constraints - we require that the norm (a generalized idea of length) be 1, making the state vector essentially a unit vector).
Then a state [math]\displaystyle{ \psi }[/math] is represented by the state vector [math]\displaystyle{ \ket{\psi} }[/math]
Now let's consider observables, such as position or momentum or angular momentum. Let us represent such an observable as [math]\displaystyle{ A }[/math]. Then for a discrete observable, there are a discrete set of outcomes that we represent as [math]\displaystyle{ {a} }[/math]. For discrete observables, [math]\displaystyle{ A }[/math] is a matrix, and we have a set of vectors [math]\displaystyle{ {\ket{\phi_a}} }[/math] that are eigenvectors of [math]\displaystyle{ A }[/math] corresponding to the eigenvalues [math]\displaystyle{ {a} }[/math].
[math]\displaystyle{ A{\ket{\phi_a}} = a\ket{\phi_a} }[/math]
Then we have a neat little realization:
[math]\displaystyle{ Prob_{\phi_a}(A = a) = 1 }[/math]
The expression above refers to the probability of the state [math]\displaystyle{ \ket{\phi_a} }[/math] having the outcome [math]\displaystyle{ a }[/math] for the observable [math]\displaystyle{ A }[/math]
Born Rule as the Inner Product
For an arbitrary state [math]\displaystyle{ \ket{\psi} }[/math], we now have the Born rule:
[math]\displaystyle{ Prob_{\psi}(A = a) = |\bra{\phi_a}\ket{\psi}|^2 }[/math]
The expression on the right is the norm squared of the inner product.
Inner Product
Note that both [math]\displaystyle{ \ket{\psi} }[/math] and [math]\displaystyle{ \ket{\phi_a} }[/math] are state vectors. The inner product is a generalization of dot products, and is the multiplication of the conjugate transpose of the vector on the left and the vector on the right.
Conjugate transpose of [math]\displaystyle{ \ket{\psi} = \bra{\psi} }[/math]
A note:
[math]\displaystyle{ |\bra{\psi}\ket{\psi}|^2 = \bra{\psi}\ket{\psi} = 1 }[/math]
The Wavefunction as a Continuous Representation of State
The wavefunction is a complex function. However, there is no physical interpretation of complex numbers. Converting a complex number to a real number involves taking the norm of the complex number. Now, see that the wavefunction [math]\displaystyle{ \psi(x) }[/math] is the continuous analogue of [math]\displaystyle{ \ket{\psi} }[/math] in the x basis. [math]\displaystyle{ \ket{\psi} }[/math] is a state vector for which the norm is finite, and the inner product is defined within the Hilbert Space. We know that the norm squared of the inner product of a state vector with itself is 1 since we have defined state vectors to have norm 1. Since the wavefunction is the analogue of the state vector, it makes sense that an analogue of the inner product is defined for it - the inner product of a state vector with itself is the multiplication of the the conjugate transpose of the state vector with itself. So, the norm squared of the wavefunction comes from the multiplication of the wavefunction with its complex conjugate.
[math]\displaystyle{ \left | \Psi(x,t) \right |^2 = \Psi^*(x,t)\Psi(x,t) }[/math]
Probability Density Function
Clearly, since we have mandated that the inner product of the two state vectors is finite, so must be the norm squared of the wavefunction. (The wavefunction belongs to the Hilbert space within which the inner product is defined).
In a discrete sense, the norm squared of the inner product gives the probability for an outcome upon measurement of a state for a given observable. However, for a wavefunction, we deal with an infinite Hilbert Space, and the spectrum of the observable is continuous - so its norm squared cannot give values of probability for a specific. Instead we interpret it as giving the probability density function of the state (as represented by the wavefunction) on the chosen observable.
Now, to get the probability over the entire observable, we integrate from [math]\displaystyle{ -\infty \rightarrow +\infty }[/math]. It makes sense that this integral would be 1, since the summation of probability over all the possible outcomes is necessarily 1. Since we are integrating the norm-squared of the wavefunction, and we have the following result:
[math]\displaystyle{ \int_{-\infty}^{\infty}{ \left | \Psi(x,t) \right |^2 dx}=1 }[/math],
as a given particle in question must be located somewhere between [math]\displaystyle{ -\infty \lt x \lt \infty }[/math]
Or more formally:
[math]\displaystyle{ P_{x\in[-\infty,\infty]}=1 }[/math]
We can say that [math]\displaystyle{ \Psi(x,t) }[/math] must be square integrable. (All this means is that [math]\displaystyle{ \int_{-\infty}^{\infty}{ \left | \Psi(x,t) \right |^2 dx} \lt \infty }[/math])
Calculating Probability
This is now straightforward since we have the probability density function.
The probability that the outcome of measurement of observable [math]\displaystyle{ A }[/math] lies between the values [math]\displaystyle{ a_1 }[/math] and [math]\displaystyle{ a_2 }[/math] for the state [math]\displaystyle{ \psi }[/math] is represented formally as
[math]\displaystyle{ Prob_\psi(a_1 \leqslant A \leqslant a_2) }[/math]
As expected from how probability distributions are defined:
[math]\displaystyle{ Prob_\psi(a_1 \leqslant A \leqslant a_2) = \int_{a_1}^{a_2}{\left | \Psi(a) \right |^2 da} }[/math]
See that the wavefunction is shown as [math]\displaystyle{ \Psi(a) }[/math] instead of [math]\displaystyle{ \Psi(x,t) }[/math]. We make the conscious choice to change the basis of [math]\displaystyle{ \Psi }[/math] from [math]\displaystyle{ x }[/math] to [math]\displaystyle{ a }[/math] since we are calculating the probability on the observable [math]\displaystyle{ A }[/math]. [math]\displaystyle{ a }[/math] represents the generalized eigenvector of [math]\displaystyle{ A }[/math]. These generalized eigenvectors of [math]\displaystyle{ A }[/math] form a complete basis for the Hilbert Space within which the wavefunction is defined, so we can fully define the [math]\displaystyle{ \Psi(x,t) }[/math] in the [math]\displaystyle{ a }[/math] basis instead of the [math]\displaystyle{ x }[/math] basis. All this does is make calculations easier.
Examples
===Simple===
Given a state [math]\displaystyle{ \ket{\psi} = \frac{1}{2}\ket{+} + \frac{i\sqrt{3}}{2}\ket{-} }[/math], calculate the probability that a measurement of the state comes in the state [math]\displaystyle{ \ket{+} }[/math].
Now, from above, we can assume that since we have expressed [math]\displaystyle{ \ket{\psi} }[/math] in the basis [math]\displaystyle{ {\ket{+}, \ket{-}} }[/math], they must be a complete basis of the Hilbert Space and are the eigenvectors of the observable. Regardless, the probability, as per Born's rule, is as follows:
[math]\displaystyle{ |\bra{+}\ket{\psi}|^2 = |\bra{+}[\frac{1}{2}\ket{+} + \frac{i\sqrt{3}}{2}\ket{-}]|^2 }[/math]
A feature of quantum observables is that they are represented by Hermitian matrices, whose eigenvalues have orthonormal eigenvectors. Since the outcomes (eigenvalues) have the corresponding state vector as the basis vectors, the basis vectors are orthonormal to each other.
Remember that the inner product is a generalization of the dot product - so the inner product of orthogonal state vectors:
[math]\displaystyle{ \lt \bra{+}\ket{-}\gt = |\lt \bra{+}\ket{-}\gt |^2 = 0 }[/math]
And the inner product of a vector with itself:
[math]\displaystyle{ \lt \bra{+}\ket{+}\gt = |\lt \bra{+}\ket{+}\gt |^2 = 1 }[/math]
Thus, the probability [math]\displaystyle{ = |\bra{+}\frac{1}{2}\ket{+} + \bra{+}\frac{i\sqrt{3}}{2}\ket{-}]|^2 = |\frac{1}{2}\bra{+}\ket{+}|^2 = |\frac{1}{2}|^2 = \frac{1}{4} }[/math]
Middling
Difficult
Connectedness
How is this topic connected to something that you are interested in? How is it connected to your major? Is there an interesting industrial application?
History
Max Born, in his 1926 paper argued that the amplitude of the wavefunction that Schrödinger had created was related to probability that you will find the particle at that position if you detect it experimentally. Born claimed that he had simply generalized from photons, the quantum “packets of light” that Einstein proposed in 1905. Einstein, Born said, had interpreted “the square of the optical wave amplitudes as probability density for the occurrence of photons. This concept could at once be carried over to the wavefunction.” The occurrence of photons became generalized as the position of a measured particle whose wavefunction is being described. Interestingly, and disappointingly for our understanding of Quantum Mechanics, Born introduced his eponymous rule in the context of scattering theory, but did not derive his results from fundamental principles. His connection of the amplitude to the probability density function was somewhat of a guess, and did not come from a mathematically rigorous base. It did however, work remarkably well.
See also
References
Ball, Phillip. “The Born Rule Has Been Derived from Simple Physical Principles.” Quanta Magazine, 13 Feb. 2019, [1]
Landsman, N.P. The Born Rule and Its Interpretation. [2]
Townsend, J S. A Modern Approach to Quantum Mechancs. Saisalito, Ca, University Science Books, 2000.