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Physics LibreTexts

2.1: Wave Functions

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  • Position Space

    A quantum state is a complete description of a physical system in the physical model we know as quantum theory. Contrary to popular belief, quantum mechanics does not just apply to the microscopic realm, though for most cases there are simpler, more effective models for describing macroscopic behavior. A quantum state is also not limited to a single particle – indeed we will look at some states that describe multiple particles. Naturally states of single particles are simpler to handle mathematically, and we have to start somewhere, so we will assume we are talking about single particle states until further notice. It is a postulate of quantum mechanics that the quantum state of a system contains every bit of physical information about that system. This is a lot of information!

    Mathematically, we describe every quantum state with a vector. Of course, with all the information stored in this quantity, it can't be any ordinary vector, and in fact we assume that quantum states are describable with uncountably-infinite-dimensional vectors in a Hilbert space (see Section 1.2 for more on these kinds of vectors). It is therefore typical to represent a quantum state with a ket: \(\left|\Psi\right>\). It is important to note that this state vector does not exist at any particular position in space. One of the pieces of information it contains relates to the position of the particle, so in some sense it is "bigger than" secondary concepts like position in space!

    The state vector \(\left|\Psi\right>\) can be described with any complete (infinite) set of unit vectors, which in Section 1.2 we wrote generically as \(\left|\alpha\right>\). The most common choice for specific unit vectors to replace the generic \(\alpha\) kets are those related to position, typically (again, in one dimension) written as \(\left|x\right>\). We can also interpret \(\left|x\right>\) as the quantum state of a particle located at position \(x\). In general a particle is not so precisely located, but just as a standard vector of unit length might happen to lie along the \(y\)-axis and therefore be equal to \(\widehat j\), so too might the state of a particle happen to include as one of its attributes the position \(x\). In general, however, a particle's quantum state vector includes many components, which represent a mix of possible positions. The components of this vector is called the wave function in position space, and we get it using Equation 1.2.16:

    \[ \psi\left(x\right) = \left<\;x\;|\;\Psi\;\right> \]

    Here the phrase "position space" merely refers to the basis of unit vectors we chose to expand our Hilbert space vector into. We assume that this set of unit vector states is complete, which means that if we know \(\psi\left(x\right)\) for every value of \(x\) (i.e. we know the full wave function), then we can completely construct the quantum state vector \(\left|\;\Psi\;\right>\).

    The place where this abstraction meets physical reality is the interpretation of the wave function as a probability amplitude. Starting with Equation 1.2.23 in position space (replace the \(\alpha\) with an \(x\)):

    \[ \left<\;\Psi\;|\;\Psi\;\right> = \int \limits_{-\infty}^{+\infty}\psi^*\left(x\right)\psi\left(x\right)d\alpha = \int \limits_{-\infty}^{+\infty}\left|\psi\left(x\right)\right|^2dx \]

    Comparing this with Equation 1.3.2 we see that we can associate the magnitude-squared of the wave function with the probability density, provided we insist that the quantum state vector is normalized. Note that the wave function itself is not the probability density – we use the word "amplitude" because this function will satisfy a wave equation, and like all waves, it is the square of the amplitude that is proportional to the "intensity," which in this case is a probability density. So in words, our description of the meaning of the position space wave function is this:

    \( \left|\psi\left(x\right)\right|^2dx \) is the probability of the particle being found at a position between \(x\) and \(x+dx\)

    Suppose a particle is found at a specific position, which we will call \(x'\). What is the wave function of this particle in position space? Well, the Hilbert space vector is aligned with a single unit vector: \(\left|\;x'\;\right>\). So constructing the wave function in the usual way with Equation 2.1.1, and applying Equation 1.2.29, we get:

    \[ \psi\left(x\right) = \left<\;x\;|\;x'\;\right> = \delta\left(x - x'\right) \]

    This joins our probabilistic interpretation of quantum mechanics with our notion of what it means for a particle to be located at a single point – the probability amplitude of finding it at a position other than \(x'\) is zero. While this amplitude is infinite at that point, the integral of its magnitude-squared is 1, which means it is certainly there (probability = 1).

    Momentum Space

    Position is not the only continuous variable that describes the state of a particle. The state vector contains information about all such variables, and the next most common such variable to be considered is momentum. We follow the procedure exactly as we did for position space – we associate our abstract Hilbert space continuous variable \(\alpha\) with momentum. Again we are working in one dimension, so this would be momentum along that dimension, which we will continue to refer to as the \(x\)-direction. As we may recall from our previous study of quantum mechanics, dealing with the units of momentum and position can be a pain, so it is preferable to represent the momentum in terms of the wave number, \(k\), where from the deBroglie relation we have: \(p=\dfrac{h}{\lambda} = \left(\dfrac{h}{2\pi}\right)\left(\dfrac{2\pi}{\lambda}\right) = \hbar k\). A Hilbert space unit vector that represents a state of a specific momentum corresponding to wave number \(k\) is written as \(\left|\;k\;\right>\), and the wave function in momentum space comes about the same way as it did for position space. [Note: From this point on, we will refer to wave number and momentum interchangeably, to avoid having to keep saying "...the wave number associated with the momentum..."]

    \[ \phi\left(k\right) = \left<\;k\;|\;\Psi\;\right> \]


    It is standard practice to use different variables for the wave function in position and momentum space, which has result in the change above to \(\phi\left(k\right)\). The reason for this is that using the same variable name implies that the functional form is the same, with \(k\) replacing \(x\). For example, if we have the function:

    \[\psi\left(x\right) = mx+b, \nonumber\]

    then one might naturally think that the function \(\psi\left(p\right)\) must be:

    \[\psi\left(k\right) = mk+b, \nonumber\]

    But this definitely not the case! For a single particle, the component of the wave function that multiplies the \(x^{th}\) unit vector in the position basis looks nothing like the component of the wave function that multiplies the \(k^{th}\) unit vector in the momentum basis, so the functions look entirely different. We use different variables for the wave function in these spaces to avoid this misunderstanding.

    The main drawback to this practice is that one might lose sight of the fact that these wave functions are components of the same quantum state vector, but are different only because they are part of a different basis.

    The probabilistic interpretation for this wave function naturally does not address the chances of finding the particle at any particular position in space, but rather the probability of finding it with a specific momentum:

    \( \left|\phi\left(k\right)\right|^2dk \) is the probability of the particle being measured to have a momentum between \(k\) and \(k+dk\)

    Also, as with position space, the momentum-space wave function of a particle with a single specific momentum \(k'\) is a delta function:

    \[ \phi\left(k\right) = \left<\;k\;|\;k'\;\right> = \delta\left(k - k'\right) \]

    Linking Position and Momentum Space

    Position and momentum spaces are just different bases of unit vectors that we can use to describe the same quantum state vector, so there must be some way to translate from one to the other. If we were talking about ordinary vectors, then going from one basis of unit vectors to another could be something as simple as a rotation of the cartesian axes, or perhaps a bit trickier case of computing a jacobian transformation between cartesian and polar coordinate systems. Both rotations and more general jacobian transformations convert a finite number of unit vectors into each other, so they can be represented by matrices. For Hilbert space unit vectors, it isn't surprising that an integral is involved. We can see specifically how to make this translation by using our old friend the resolution of the identity:

    \[ \phi\left(k\right) = \left<\;k\;|\;\Psi\;\right> = \left<\;k\;\right| \left[ \int \limits_{-\infty}^{+\infty} \left|\;x\;\right> \left<\;x\;\right| dx \right] \left| \;\Psi\;\right> = \int \limits_{-\infty}^{+\infty} \left<\;k\;|\;x\;\right>\psi\left(x\right) dx\]

    Okay, fine, but how does this tell us how to translate between the functions \(\psi\left(x\right)\) and \(\phi\left(p\right)\)? Namely, what is \( \left<\;p\;|\;x\;\right>\)? We interpret it as an inner product between two different types of unit vectors, so it must be some kind of scalar function, but what does it look like? Here is where we need to draw upon some of those quantum mechanics concepts we previously studied. It will be stated without proof (because really it is a postulate of quantum mechanics) that in order for there to exist the uncertainty principle between position and momentum, and for particles to behave like waves in double-slit apparatuses, we need to postulate that this quantity is:

    \[ \left<\;k\;|\;x\;\right> = \frac{1}{\sqrt{2\pi}}e^{-ikx} \]

    Plugging this back in above gives us our translation between position and momentum space wave functions:

    \[ \phi\left(k\right) = \frac{1}{\sqrt{2\pi}}\int \limits_{-\infty}^{+\infty} e^{-ikx}\psi\left(x\right) dx\]

    This is known in mathematics as a fourier transform.

    We can of course transform in the opposite direction as well – from momentum space to position space. All it takes is to repeat what we did in Equation 2.1.6, this time using the resolution of the identity in the momentum basis:

    \[ \psi\left(x\right) = \left<\;x\;|\;\Psi\;\right> = \left<\;x\;\right| \left[ \int \limits_{-\infty}^{+\infty} \left|\;k\;\right> \left<\;k\;\right| dk \right] \left| \;\Psi\;\right> = \int \limits_{-\infty}^{+\infty} \left<\;x\;|\;k\;\right>\phi\left(k\right) dk\]

    We see that the function appearing in the integral is the same as before, except that the bra and ket are reversed. This is just the complex conjugate of what we used before, so:

    \[ \psi\left(x\right) = \frac{1}{\sqrt{2\pi}}\int \limits_{-\infty}^{+\infty} e^{+ikx}\phi\left(k\right) dk\]

    While we have viewed this as a link between two different bases used to describe the same quantum state, there is another common and useful interpretation as well. In a previous treatment of quantum mechanics you were exposed to fourier series, and the idea that any periodic function can be expressed as a linear combination of (perhaps an infinite number of) harmonic functions (sines and cosines). The wave function in position space can take on pretty much any form, meaning it can be any function of \(x\), as long as it is normalized – it doesn't need to be periodic. This fourier transform is the extension of the fourier series that "decomposes" any function into a linear combination (this time on a continuum, thus the integral) of harmonic functions, which in this case are packages in the exponential.

    In this fourier analysis viewpoint, the wave function in momentum space is the set of coefficients in the linear combination. That is, \(\phi\left(k\right)\) provides the weight for the contribution of the harmonic function with wave number \(k\). We sometimes refer to the function \(\phi\left(k\right)\) as the spectral content of the wave function \(\psi\left(x\right)\). It is a sort of "recipe" that tells us how much of each exponential to mix in to get the position-space wave function. If we are given the wave function and want to know its spectral content, then we only need to plug it into Equation 2.1.8.

    Plane Waves, Delta Functions, and the Uncertainty Principle

    Recall from an earlier study of waves that a complex plane wave with a momentum \(k\) in the \(+x\)-direction is given by the equation:

    \[ f\left(x,t\right) = Ae^{i\left(kx-\omega t\right)}\]

    If this represents a wave function (so \(f\left(x,t\right)\) is a probability amplitude), then it should be clear that there is no way to distinguish points along the \(x\)-direction from each other for this particle. That is, all the points from \(x=-\infty\) to \(x=+\infty\) are equally probable places to find the particle, as shown by the fact that the probability density doesn't depend upon \(x\):

    \[\left|f\left(x,t\right)\right|^2 dx = f^*\left(x,t\right)f\left(x,t\right)dx = \left[Ae^{i\left(kx-\omega t\right)}\right]^* \left[Ae^{i\left(kx-\omega t\right)}\right]dx=\left|A\right|^2dx \]

    Ignoring the time part of this wave function, a conclusion we draw from this is that we have absolutely no idea about the position of a particle that behaves as a plane wave, while we know that particle's momentum precisely. We can confirm this using tools we have already developed. The wave function in momentum space for a particle with a definite momentum \(k'\) is given by Equation 2.1.5. Plugging this into Equation 2.1.10 and employing the property given in Equation 1.2.30, we get:

    \[ \psi\left(x\right) = \frac{1}{\sqrt{2\pi}}\int \limits_{-\infty}^{+\infty} e^{+ikx}\delta\left(k - k'\right) dk = \frac{1}{\sqrt{2\pi}}e^{+ik'x} \]

    Sure enough, a particle with a wave function in momentum space that represents a single momentum has a wave function in position space that is a plane wave with that same momentum. The uncertainty principle is satisfied – perfect knowledge of the momentum results in no knowledge whatsoever of the position. Naturally, this works both ways, and the same can be demonstrated with the fourier transform in the opposite direction. In fact, we can derive a simple and very useful formula for the delta function simply by using the resolution of the identity:

    \[ \delta\left(x-x'\right) = \left<\;x'\;|\;x\;\right> = \int \limits_{-\infty}^{+\infty}dk \left<\;x'\;|\;k\;\right>\left<\;k\;|\;x\;\right> =  \int \limits_{-\infty}^{+\infty}dk\left[\dfrac{e^{+ikx'}}{\sqrt{2\pi}}\right]\left[\dfrac{e^{-ikx}}{\sqrt{2\pi}}\right] = \frac{1}{2\pi}\int \limits_{-\infty}^{+\infty}dk \;e^{ik\left(x'-x\right)}\]

    One last useful property of the delta function is the "filtering" effect it has on a function multiplied by it:

    \[ f\left(x\right) \delta\left(x-x'\right) = f\left(x'\right) \delta\left(x-x'\right) \]

    This is clearly true because the parts of \(f\left(x\right)\) that are not at the position \(x'\) multiply the zero part of the delta function, which means you might as well only multiply the part of the function that coincides with the spike, and ignore the rest.