The Born Rule: Difference between revisions
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==Normalizing the Wavefunction== | ==Normalizing the Wavefunction== | ||
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> | 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> | |||
So, the norm squared of At the position <math>x</math> and time <math>t</math>, <math> \left | \Psi(x,t) \right |^2 </math> is the probability density (the physical importance is found in the square of <math>\Psi</math>). By definition, an entire probability density function must have an area equal to one. Hence it follows that <br> | So, the norm squared of At the position <math>x</math> and time <math>t</math>, <math> \left | \Psi(x,t) \right |^2 </math> is the probability density (the physical importance is found in the square of <math>\Psi</math>). By definition, an entire probability density function must have an area equal to one. Hence it follows that <br> | ||
<math>\int_{-\infty}^{\infty}{ \left | \Psi(x,t) \right |^2 dx}=1 </math>, | <math>\int_{-\infty}^{\infty}{ \left | \Psi(x,t) \right |^2 dx}=1 </math>, | ||
Revision as of 21:19, 27 November 2022
Claimed - Rehaan Naik 11/27/2022
The Born Rule 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]
Normalizing the Wavefunction
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]
So, the norm squared of At the position [math]\displaystyle{ x }[/math] and time [math]\displaystyle{ t }[/math], [math]\displaystyle{ \left | \Psi(x,t) \right |^2 }[/math] is the probability density (the physical importance is found in the square of [math]\displaystyle{ \Psi }[/math]). By definition, an entire probability density function must have an area equal to one. Hence it follows that
[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]