Position Operator: From Ket vectors to wavefunctions

Suppose we are interested in measuring the position of a quantum particle. The observable that will do the job is the position operator. The eigenvalue equation for position operator is

\hat x \left|x\right\rangle=x \left|x\right\rangle

Here \left|x\right\rangle are position eigenkets and x are the corresponding eigenvalues. This operator is different from the one we have already encountered for the energy and will need slightly different treatment. The eigenvalues of this operator are continuous. The completeness relation thus changes into an integration instead of a summation.

Completeness relation

We write the completeness relation for the eigenkets of position operator as

\int_{-\infty}^{\infty}dx \left|x\right\rangle\left\langle x\right|=\hat I

We can use this completeness relation to write any arbitrary ket as

\left|\psi\right\rangle=\hat I \left|\psi\right\rangle=\int_{-\infty}^{\infty}dx \left|x\right\rangle\left\langle x|\psi\right\rangle

Wavefunctions and probability density

The scalar product/expansion coefficient/probablity amplitude  \left\langle x|\psi\right\rangle is called (position space) wavefunction.  We write it as

\psi(x)=\left\langle x|\psi\right\rangle

Wavefunctions are the scalar product of ket vectors with position eigenkets.

The modulus square of the wave function is again related to the probability of the measurement outcome of the position meaurement of the particle. However due to the continuous nature of the eigenvalues of this operator, we need to modify the corresponding postulate (Postulate 4 discussed in the class). The postulate that gives the probabilities of measurement outcome of position measurement of the quantum particle can be stated as follows:

The probability of obtaining the position of the particle in a narrow strip of width dx centered at x=x_0 is given by |\psi(x_0)|^2dx

This can be seen in the figure below

ProbabilityDensity

 

The quantity plotted as the function of x is  |\psi(x)|^2 and is called probability density. Obviously the probability of finding a particle in a location is higher if the probability density is high over there. We can now also answer the questions like “What is the probability of finding the particle between x=a and x=b. This probability will be \int_a^b|\psi(x)|^2dx.

Scalar product as an integral

Another use of the completeness relation is to write the scalar product in the form of an integral. Consider the norm of the vector \left|\psi\right\rangle

\left\langle\psi|\psi\right\rangle=\left\langle\psi|\hat I|\psi\right\rangle=\int_{-\infty}^{\infty}\left\langle\psi|x\right\rangle\left\langle x|\psi\right\rangle dx=\int_{-\infty}^{\infty} \psi^*(x)\psi(x) dx=1

More generally the scalar product of two vectors can be written as

\left\langle\phi|\psi\right\rangle=\left\langle\phi|\hat I|\psi\right\rangle=\int_{-\infty}^{\infty}\left\langle\phi|x\right\rangle\left\langle x|\psi\right\rangle dx=\int_{-\infty}^{\infty} \phi^*(x)\psi(x) dx

Orthogonality relation

The orthogonality condition is also modified in case of continuous spectrum. It is given as follows for the position eigenkets

\left\langle x^\prime|x\right\rangle=\delta(x^\prime-x)

The expression on right is called Dirac delta function. Its value is zero if the argument is non-zero. If the argument is zero then the value of the function is undefined. However the integration over this undefined value is unity. We write it

\int \delta(x)=1.

 

 

 

 

8 thoughts on “Position Operator: From Ket vectors to wavefunctions

  1. sir, its very informative and understandable but its little bit difficult to understand the mathematical expression
    kindly if you explain these relation by example or problems
    thanks

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    1. Orthogonality comes from orthogonal which can be understood as being perpendicular.
      Consider the unit vectors along x and y axis. They are perpendicular to each other. We say they are orthogonal to each other. Another way to say it is that orthogonal vectors are linearly independent from each other i.e. they can not be written in terms of each other. like you can not write the unit vector along x axis as the linear combination of unit vectors along y and z axis.

      The unit vectors we use in quantum mechanics are also orthogonal to each other. And we write their orthogonality condition in the way it is written in the post.

      Does it help?

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  2. Sir first of all i appreciate your efforts. Indeed, it will be really helpful for us we can easily pick and understand things but sir I’m little bit confused in orthoganality relation. Can you explain it?
    First thing i don’t understand why the derac delta is zero or non zero at specific condition? can you briefly explain this orthoganality relation in case of continuous spectrum.. I’ll be thankful to you.

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  3. Orthogonality means that two vectors share nothing in common. Example think of a vector along z axis, and another along x axis. These two will be orthogonal to each other. We check this orthogonality by taking the scalar product of the two vectors. Thus scalar product of any vector along z axis, and any other along x axis will be zero.

    The same concept is extended to ket vectors and functions. The position eigenkets belonging to different position eigenvalues are the vectors that do not share anything. Their scalar product is zero. The orthogonality is done over here.

    The next issue is the scalar product of a position ket with itself. This is a bit complicated. It is not defined. These ket vectors do not have a norm. However the integral over the scalar product of a ket vector with itself is unity.

    The above two properties of position kets can be expressed elegantly by writing their scalar product as Dirac delta.

    I hope this helps.

    (In any case you do not need to worry too much about the norm of a position eigenket in your first encounter with quantum mechanics.)

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