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0:06
In the previous tutorial, we introduced the wavefunction, which is represented by the Greek letter psi. We talked about how to apply operators to the wavefunction, but we didn’t talk much about the wavefunction itself. So in this tutorial, let’s try to understand the wavefunction a bit better. We should note that much of this discussion will assume a working comprehension of concepts in linear algebra, such as vector spaces.
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So if your goal is a deep comprehension of quantum mechanics, my linear algebra tutorials are available should you need some review. If you’re up to speed, then we can begin to unpack a few things. The best way to go about understanding the wavefunction is to describe what it does. If we consider some quantum system, it can be described by a set of wavefunctions, and the wavefunctions which describe that particular system can be said
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to form something called a Hilbert space. A Hilbert space is a particular kind of vector space. Hilbert spaces have some specific rules and properties that we won’t fully explain here. For now let’s just talk about how to imagine the wavefunctions that form them. We can picture wavefunctions as vectors living in a vector space of potentially infinite dimensions. So for comparison, in a three-dimensional vector space, we have the Cartesian components, x, y, and z, forming a basis for the vector
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space. They are orthogonal, so their linear projections, or dot products, are always zero. That is, the dot product of x and y is zero, and the same goes for x and z, as well as y and z. But the dot product of x and x is not zero, nor is it for y and y, or z and z. These basis vectors are linearly independent, so we can’t write any of them in terms of the others, since no combination of x and y can give us the z vector, and so forth.
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Now back to wavefunctions. Each wavefunction is a vector in a Hilbert space, just like x, y, or z are vectors in your everyday Cartesian space. Everything we just said about x, y, and z, can be applied to the set of wavefunctions that comprise that Hilbert space. The difference is that a quantum system may be described by a large number of wavefunctions, much more than three, all of which are orthogonal with each other. And they are also linearly independent,
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meaning that none of them can be written in terms of the others, just like in the three-dimensional case. Now that we are conceiving of wavefunctions as vectors, it’s time to introduce some new notation, bra-ket notation. This notation utilizes two constructs, bras and kets. These use angle brackets and vertical lines to represent certain vectors, and we will use this notation with the wavefunction. When we use a ket this
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way, we are referring to the wavefunction in vector form, where a, b, and c can be considered coefficients that accompany each component of that wavefunction, no different from how we’ve used vectors to represent functions in linear algebra. So psi equals A psi one, plus B psi two, plus C psi three, with each of these representing basis vectors. This represents the sets of linearly independent wavefunctions that are solutions of the system.
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So once again, considering the ket to be a shorthand for the vector representation of this wavefunction, the bra can be considered the complex conjugate of the ket, or psi star, which is the transposed conjugate of this vector, therefore involving a, b, and c arranged in a row instead of a column, and with a star, indicating that each term is the complex conjugate of its corresponding term in the ket, or a star, b star, c star.
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Now we are ready to define a new operation called the inner product in a Hilbert space, which uses the bra-ket notation we just learned. The inner product of psi one and psi two is equal to the integral over all space of psi one star times psi two with respect to x. A few things to know regarding the inner product are as follows. The inner product of a wavefunction with itself is always one. And the inner product of a wavefunction with a different wavefunction
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that is not a linear combination of the first is always zero. This makes sense if we think of the inner product as the projection of one vector onto another. Any orthogonal projection is zero since we are dealing with linearly independent vectors, like how this unit vector in x does not contain any amount of the y or z vectors. But any projection of a vector onto itself must be one, as a vector is equal to itself.
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The way to represent these relationships at the same time mathematically is to show the inner product of psi N and psi M as being equal to something called the Kronecker delta, which is presumed to be zero when N and M are unequal, and one when N and M are equal. Let us reiterate that for a Hilbert space, the inner product is equivalent to the dot product, and inner products give a scalar as the result of the combination of two vectors
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from a vector space. So we can pretty much think of the inner product as the dot product for wavefunctions. The way we are using the dot product in this context is indeed the same as what we are familiar with, even though we are using integrals and functions rather than sums and vectors, it is simply that wavefunctions within a Hilbert space don’t correlate with physical reality. Ket of psi is a vector in the Hilbert space, while psi of x is the projection
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of that vector onto the space we are familiar with, such that we can describe observables like position and momentum. With an enhanced view of the wavefunction, let’s now return to the concept of the probability density function that we introduced in the previous tutorial. First, let’s recall the mathematical meaning of the probability density function with the following equation, involving the product of the wavefunction and its complex
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conjugate, represented by psi star. So P of x and t equals psi star times psi, which equals the absolute value of psi of x squared. Now let’s expand a bit. The probability density function tells us that at some instant t, the probability P of finding a particle between x and x plus dx is equal to psi star times psi. We should note that the probability density is always a real number. To find the probability that a particle will be found
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in some region of space, we can calculate the following integral of the probability density function, from x zero to x one. Then let’s understand that if we integrate the probability density function from negative infinity to positive infinity, this will equal one, meaning that the probability of finding the particle somewhere in space is 100%. If we were applying this to a classical particle, we would get a very narrow peak centered in
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a specific location, like x = 5. Integrating only from 4 to 6 we would still get a result of one, since the particle is located within these boundaries. If we were to integrate over some other interval, we would get a result of zero, which makes sense, since the particle is not there, but is somewhere else. But with a quantum particle, we get a different result. Perhaps it is still located at x = 5, but its presence is much more diffuse. In this
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example, the probability density function extends like a bell curve, let’s say from 1 to 9. Integrating from 4 to 6 will no longer give us one, we will get some fraction of one, as will be the case with any other interval that contains some of the probability density function. Integrating over this interval will give us something close to one, let’s say 0.93, but there is still some probability when performing a measurement of finding the
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particle towards either edge of this bell curve. This is why normalization is so important, it ensures a result of one, which is the equivalent of knowing with certainty that the particle is located within a particular interval, because it’s the largest possible interval, all space. It tells us that this particle exists, even though we don’t know exactly where. Now we need to introduce some more terminology. We can say that wavefunctions which are eigenfunctions
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of observables with physical meaning are normalized. Again, by normalized, we mean that if we calculate the integral of the probability density function over all space, which as we know is the integral over all space of the modulus squared of the wavefunction, it will always be precisely one. Rather predictably, two examples of observables with physical meaning that will have eigenfunctions in the form of normalized wavefunctions are position, x, and momentum, p.
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In the next tutorial we will look at the inner product of psi and another wavefunction of the same vector space, like psi one, psi two, and so forth, which correspond to different eigenenergies of a quantum particle, and this will be equal to zero, due to the linear independence of all wavefunctions forming such a vector space. But first, let’s continue to discuss normalization. It is the case that the Schrodinger equation
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is linear, which means that in finding a solution for this equation, if we arrive at a wavefunction phi one that doesn’t satisfy the normalization condition, that is the inner product of the wavefunction with itself does not equal one, but rather equals some constant C that is not equal to one, then we need to normalize it. We need to find a way to make C equal to one. Now it is the case that if a wavefunction is a solution to the Schrodinger equation,
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then any linear combination of this wavefunction will also be a solution. So given that phi one is a solution, then phi two, which is defined as equal to A phi one, must also be a solution. So since the inner product of phi two with itself would look like this, we can also just replace phi two with A phi one, and we can then pull the constants A star times A out of the integral. This will equal one because we have decided that it
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should equal one, because we need a normalized wavefunction, and that is the condition of normalization. Now we can find what A must be for this to be true. So recall that the expression for phi one gave us a value of C. Let’s therefore replace this entire integral in the previous expression with C, since they are equivalent. That gives us A star times A times C equals one. Recalling that A star times A equals the absolute value of A quantity
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squared, since these terms differ only in the sign of their imaginary terms, we can then divide both sides by C, take the square root, and A will be equal to one over root C, where C represents the inner product of phi one with itself. The main thing we should take from this is that all wavefunctions with physical implications must be normalized. In other words, the definition of the probability density function implies
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that wavefunctions which are solutions of the Schrodinger equation must be normalized, and we will talk more about the Schrodinger equation in the next tutorial. Now that we have a better understanding of wavefunctions, let’s see what we can do with them. First let’s talk about the superposition principle. This states that quantum systems can be described by combinations or superpositions of wavefunctions, which still must satisfy
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the normalization condition. One example of this is the famous double slit experiment, which we introduced on a conceptual level in a previous tutorial. Now let’s add some math. As we recall, given this grating with two slits, a quantum particle will exhibit wavelike behavior, passing through both slits, and generating an interference pattern at the detector beyond. The interesting feature of this experiment is the act of the particle
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passing through both slits simultaneously, and at that precise moment, we can describe the quantum state of the particle as the superposition of the particle being in the left slit and the particle being in the right slit. Mathematically, we would represent this as psi equals phi L plus phi R, essentially representing the sum of a wavefunction localized at one slit and a wavefunction localized at the other. Now let’s introduce two constants, A L and
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A R, which represent the probabilities of the particle being at either slit. If we desired, we could make this more complicated by adding more slits, and in turn, we would simply add more wavefunctions to this equation, each with their own respective constant. If there were four slits, we would describe the superposition this way, with four wavefunctions and four constants, where phi one represents the wavefunction of a particle at slit 1, and A 1 represents
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the probability of the particle being found at slit 1, and the same for the other three. What we must wrap our heads around as best we can, is that the superposition tells us that a particle can be at several points at the same time, and this is precisely what the double slit experiment demonstrates, as the particles do physically pass through both slits. It is a probabilistic way of viewing reality that is difficult for our classically
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trained brains to comprehend, but nevertheless, this is how the universe operates on this scale.
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Now that we understand normalization and the probabilistic interpretation of quantum mechanics, we are ready to develop one of the most important topics in quantum mechanics, and that is the calculation of expectation values. Earlier we saw that the integral over all space of the probability density function gives us a value of one for a normalized wavefunction. This one, which is the maximum value possible, tells us that the probability of finding the particle somewhere in space is 100%, which
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makes sense, since the particle has to exist somewhere. Now, if we want to determine where it is most likely to find the particle, we have to perform a similar calculation. Given that we are talking about the location of the particle, we are describing its position, so we will need to use the position operator in conjunction with the probability density function. The operator will be placed like so, in between psi star
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and psi, such that it acts on psi, and this expression will be equal to x zero. X zero is an expectation value. It is a number, which in this case tells us the most probable location of the particle. Now let’s try to understand this in greater detail. In statistics, an expectation value is a weighted average of a variable. Think of it this way. You want to measure the variable x, but it does not have a definite value, it exists
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as a distribution of values, something which will be true all the time in quantum mechanics, as we are beginning to understand. The expectation value is the weighted sum of all possible values that x can take times their respective probabilities, so each value contributes to the sum in a way that is proportional to its respective probability. Now let’s apply this to quantum mechanics. Say we are performing an experiment many, many times. Perhaps we
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are throwing a quantum particle out the window, such that it lands on the ground below, and we take a picture of where it lands each time. Then say we superimpose all of the pictures for all the trials, so that we can see all of the final positions at once, and whenever it lands in the same spot again, that spot becomes darker. This singular image tells us the most probable position for the particle, with the darkness of each spot relating to
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how that spot is weighted. And we can consider the darkest spot to be representative of the expectation value for position at the measuring device, which in this case we are analogizing to be the ground itself. Once again, mathematically, we will write this as the integral over all space of psi star times the position operator acting on psi. This is how we define the expectation value of the x coordinate at a given time
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t, which is written this way, and read as the expectation value of x with wavefunction psi. Notice how we use both left and right brackets, with the variable in the middle. We can write an equivalent equation for the momentum operator, whereby the expectation value of p equals the integral over all space of psi conjugate times the operator p acting on psi. We can then rewrite this, using the definition of the momentum operator, which
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as we recall from the previous tutorial is negative i times h bar times the partial derivative with respect to x. This applies in a broader context as well. We can write all dynamical variables as functions of position and momentum. This means that it is possible to find the expectation value of any dynamical variable, Q, by calculating it this way, with the expectation value of Q of x and p being equal to this expression.
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Let’s apply this to one specific example, kinetic energy. In the classical physics series, we learned that the kinetic energy of an object, which from now on we will represent as T, is equal to one half m v squared. And since momentum equals mass times velocity, if we just put an additional mass term in both the numerator and denominator to keep the expression equivalent, we can change m squared v squared into mv quantity squared, or p squared, leaving
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us with p squared over 2m. Now we can take this classical equation and translate it into a quantum equation by replacing momentum with the momentum operator. Let’s then plug in our expression for the momentum operator, putting it out front, but twice, since the operator is squared. Now to simplify, negative i times negative i gives i squared, or negative one, so let’s just get rid of those and make the whole expression negative. Then we
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can pull h bar squared up here, and the partial derivatives multiply to give the second derivative with respect to x. This is the quantum operator for kinetic energy, which we will be using frequently. Now let’s find the expectation value for kinetic energy for a given wavefunction, psi. As we learned, to do this, we place the operator T in between the bra of psi and ket of psi. We can then replace the operator with the
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expression it represents, we get the integral of psi star times the kinetic energy operator acting on psi, dx. Recall that even though we are not writing it explicitly, psi depends on x, which is why we are able to differentiate it with respect to x. Now recognize that we have some constants, with h bar and mass, so let’s pull all of this out of the integral. That leaves psi star times the second derivative of psi with respect to x, dx. So we can now
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calculate the kinetic energy of a quantum particle, given its wavefunction. In this tutorial, we did several things. First, we more clearly defined what a wavefunction is, from both a conceptual and mathematical standpoint. We learned that wavefunctions are a special type of vector living in something called a Hilbert space. Next, we outlined the most important properties of wavefunctions. Namely that a quantum system
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is described by a set of wavefunctions that are linearly independent, meaning that their inner products are zero, except for their self-inner product. Then we stated that a wavefunction consistent with physical reality must be normalizable, meaning that its inner product with itself must be equal to one. From here we must simply know that for any wavefunction that is a solution to the Schrodinger equation, it is possible to normalize it.
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We will expand on this in the next tutorial. And finally, we learned how to use wavefunctions to perform one of the most important calculations in quantum mechanics, and that is determining expectation values. We learned what these are, and then calculated expectation values for position, momentum, and kinetic energy. That’s a lot to take in, so watch once more from the top if you need to, otherwise let’s move on to a deeper examination of the Schrodinger
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equation, where many of these concepts will be further clarified through example.
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